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Found 263 results

  1. rich_web_development

    Display Everything on the same page

    Hi, I'm trying to get a form where you submit information about a home to a database and below the form it displays the homes with the information and a photo. Below the home information and photo I have a button to DELETE RECORD and a button to upload a photo. All of the php and forms are in one file so submitting the home info and deleting record are just have action="samefile.php". This is so that it stays on the same page and doesn't have to jump to another page. I cannot get the upload photo to stay on the same page. The only way I can get the upload photo to work is to send it to another file. This of course takes me to another page which I don't want. I want everything to stay on the same page. The code I have (in a file called "PicTest03.php) is as follows: <!DOCTYPE html> <head> </head> <html> <body> <form action="PicTest03.php" method="POST"><pre> Name <input type="" name="name"> Bedrooms <input type="text" name="bedrooms"> Length <input type="text" name="length"> Width <input type="text" name="width"> Serial Number <input type="text" name="serialno"> <input type="submit" value="ADD RECORD"> </pre> </form> <br><br><br> <?php require_once 'login.php'; $conn = new mysqli($hn, $un, $pw, $db); if ($conn->connect_error) die($conn->connect_error); if (isset($_POST['delete'])&& isset($_POST['serialno'])) { $serialno = get_post($conn, 'serialno'); $query = "DELETE FROM holidayhomes WHERE serialno='$serialno'"; $result = $conn->query($query); if (!$result) echo "DELETE failed: $query<br>" . $conn->error . "<br><br>"; } if (isset ($_POST['name']) && isset ($_POST['bedrooms']) && isset ($_POST['length']) && isset ($_POST['width']) && isset ($_POST['serialno'])) { $name = get_post ($conn, 'name'); $bedrooms = get_post ($conn, 'bedrooms'); $length = get_post ($conn, 'length'); $width = get_post ($conn, 'width'); $serialno = get_post ($conn, 'serialno'); $query = "INSERT INTO holidayhomes (name, bedrooms, length, width, serialno) VALUES ('$name','$bedrooms', '$length', '$width', '$serialno')"; $result = $conn->query($query); if (!$result) echo "INSERT failed: $query<br>" . $conn->error . "<br><br>"; } $query = "SELECT * FROM holidayhomes"; $result = $conn->query($query); if (!$result) die ("Database access failed: " . $conn->error); $rows = $result->num_rows; for ($j = 0 ; $j < $rows ; ++$j) { $result->data_seek($j); $row = $result->fetch_array(MYSQLI_NUM); echo <<<_END <pre> name $row[0] Bedrooms $row[1] Length $row[2] Width $row[3] Serial No $row[10] MainPic $row[4] <img src="uploads/$row[4]" width=200 height=200> </pre> <pre> <form action="PicTest03.php" method="POST"> <input type="hidden" name="delete" value="yes"> <input type="hidden" name="serialno" value="$row[10]"> <input type="submit" value="DELETE RECORD"> </form> </pre> <pre> <form action="mainphoto01.php" method="post" enctype="multipart/form-data"> Select main photo: <input type="hidden" name="serialno" value="$row[10]"> <input type="file" name="fileToUpload" id="fileToUpload"> <input type="submit" value="Upload Image" name="submit"> </form> </pre> <br><br> _END; } $result->close(); $conn->close(); function get_post($conn, $var) { return $conn->real_escape_string($_POST[$var]); } ?> </body> </html> The code for "mainphoto01.php is: <?php require_once 'login.php'; $conn = new mysqli($hn, $un, $pw, $db); if ($conn->connect_error) die($conn->connect_error); $target_dir = "uploads/"; $target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]); $uploadOk = 1; $imageFileType = pathinfo($target_file, PATHINFO_EXTENSION); if(isset($_POST["submit"])) { $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]); if($check !== false){ echo "FIle is an image - " . $check["mime"] . "." ; $uploadOk = 1; } else { echo "File is not an image."; $uploadOk = 0; } } if ($uploadOk == 0){ echo "Sorry, your file was not uploaded."; } else { if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)){ echo "The file " . basename( $_FILES["fileToUpload"]["name"]) . " has been uploaded."; } else { echo "Hello!"; } } $mainpic = basename( $_FILES["fileToUpload"]["name"]); if(isset($_POST["serialno"])) $serialno = $_POST['serialno']; $query = "UPDATE holidayhomes SET mainpic='$mainpic' WHERE serialno='$serialno'"; $result = $conn->query($query); if(!$result) echo "INSERT failed: $query<br>" . $conn->error . "<br><br>"; ?> No matter what I try I cannot get the above code to work if I keep it in the "PicTest03.php" file. Can someone please show what the code should look like so I can get everything to show on the same page? I have had a look as the PHP and AJAX and DATABASE http://www.w3schools.com/php/php_ajax_database.asp which I got the example working but it uses a drop down list. Is there a way to convert the code used for the example on http://www.w3schools.com/php/php_ajax_database.asp and instead of drop down list just have the information from the database shown under the form for submitting the home information?
  2. rich_web_development

    JSON data from web server running PHP and MySQL Example

    Hi, I'm trying to follow the JSON example at http://www.w3schools.com/json/json_example.asp I have adjusted the code a little but the code is as follows: jsonTest.html <!DOCTYPE html> <html> <head> <style> h1 { border-bottom: 3px solid #cc9900; color: #996600; font-size: 30px; } table, th , td { border: 1px solid grey; border-collapse: collapse; padding: 5px; } table tr:nth-child(odd) { background-color: #f1f1f1; } table tr:nth-child(even) { background-color: #ffffff; } </style> </head> <body> <h1>Customers</h1> <div id="id01"></div> <script> var xmlhttp = new XMLHttpRequest(); var url = 'jsonTestPHP.php'; xmlhttp.onreadystatechange=function() { if (this.readyState == 4 && this.status == 200) { myFunction(this.responseText); } } xmlhttp.open("GET", url, true); xmlhttp.send(); function myFunction(response) { var arr = JSON.parse(response); var i; var out = "<table>"; for(i = 0; i < arr.length; i++) { out += "<tr><td>" + arr.Name + "</td><td>" + arr.Bedrooms + "</td><td>" + arr.Length + "</td></tr>"; } out += "</table>"; document.getElementById("id01").innerHTML = out; } </script> </body> </html> jsonTest.php <?php header("Access-Control-Allow-Origin: *"); header("Content-Type: application/json; charset=UTF-8"); require_once 'login.php'; $conn = new mysqli($hn, $un, $pw, $db); if ($conn->connect_error) die($conn->connect_error); $result = $conn->query("SELECT name, bedrooms, length FROM holidyhomes"); $outp = "["; while($rs = $result->fetch_array(MYSQLI_ASSOC)) { if ($outp != "[") {$outp .= ",";} $outp .= '{"Name":"' . $rs["name"] . '",'; $outp .= '"Bedrooms":"' . $rs["bedrooms"] . '",'; $outp .= '"Length":"'. $rs["length"] . '"}'; } $outp .="]"; $conn->close(); echo($outp); ?> So I have changed the var url = "http://www.w3schools.com/website/customers_mysql.php"; from the jsonTest.html to var url = 'jsonTestPHP.php'; When I run this I get "Invalid character: var arr = JSON.parse(response);" I just want to be able to grab the data from my database and display it on screen using JSON. Can anyone help me understand why I'm getting this Invalid character error or tell me if I'm going about this the totally wrong way? Thanks
  3. rich_web_development

    UPLOAD THEN DISPLAY PHOTOS TO/FROM MySQL WITH PHP

    Hi, I'm trying to do a simple website where I can upload some information about homes and add photos to a MySQL database and then display the information. So I have an html form that sends the details of the home to a php file called details.php. The html form is as follows. <!DOCTYPE html> <html> <body> <form action="details.php" method="POST"> <pre> Name <input type="" name="name"> Bedrooms <input type="text" name="bedrooms"> Length <input type="text" name="length"> Width <input type="text" name="width"> Serial Number <input type="text" name="serialno"> <input type="submit" value="ADD RECORD"> </pre> </body> </html> The details.php look like the following: <?php require_once 'login.php'; $conn = new mysqli($hn, $un, $pw, $db); if ($conn->connect_error) die($conn->connect_error); if (isset($_POST['delete'])&& isset($_POST['serialno'])) { $serialno = get_post($conn, 'serialno'); $query = "DELETE FROM holidayhomes WHERE serialno='$serialno'"; $result = $conn->query($query); if (!$result) echo "DELETE failed: $query<br>" . $conn->error . "<br><br>"; } if (isset ($_POST['name']) && isset ($_POST['bedrooms']) && isset ($_POST['length']) && isset ($_POST['width']) && isset ($_POST['serialno'])) { $name = get_post ($conn, 'name'); $bedrooms = get_post ($conn, 'bedrooms'); $length = get_post ($conn, 'length'); $width = get_post ($conn, 'width'); $serialno = get_post ($conn, 'serialno'); $query = "INSERT INTO holidayhomes (name, bedrooms, length, width, serialno) VALUES ('$name','$bedrooms', '$length', '$width', '$serialno')"; $result = $conn->query($query); if (!$result) echo "INSERT failed: $query<br>" . $conn->error . "<br><br>"; } $query = "SELECT * FROM holidayhomes"; $result = $conn->query($query); if (!$result) die ("Database access failed: " . $conn->error); $rows = $result->num_rows; for ($j = 0 ; $j < $rows ; ++$j) { $result->data_seek($j); $row = $result->fetch_array(MYSQLI_NUM); echo <<<_END <pre> Name $row[0] Bedrooms $row[1] Length $row[2] Width $row[3] Serial No $row[10] MainPic <img src="$row[4]" width=auto height=auto><br><br> </pre> <pre> <form action="details.php" method="POST"> <input type="hidden" name="delete" value="yes"> <input type="hidden" name="serialno" value="$row[10]"> <input type="submit" value="DELETE RECORD"> </form> </pre> <pre> <form action="mainphoto.php" method="post" enctype="multipart/form-data"> Select main photo: <input type="file" name="fileToUpload" id="fileToUpload"> <input type="submit" value="Upload Image" name="submit"> </form> </pre> _END; } $result->close(); $conn->close(); function get_post($conn, $var) { return $conn->real_escape_string($_POST[$var]); } ?> ​So the above inserts the information (about a home entered in the html form) into the MySQL database then displays the information and an option to delete the homes information then I have a form to upload a photo for the home. For uploading the photo I have a php file called mainphoto.php which is as follows: <?php require_once 'login.php'; $conn = new mysqli($hn, $un, $pw, $db); if ($conn->connect_error) die($conn->connect_error); $target_dir = "uploads/"; $target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]); $uploadOk = 1; $imageFileType = pathinfo($target_file, PATHINFO_EXTENSION); if(isset($_POST["submit"])) { $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]); if($check !== false){ echo "FIle is an image - " . $check["mime"] . "." ; $uploadOk = 1; } else { echo "File is not an image."; $uploadOk = 0; } } if ($uploadOk == 0){ echo "Sorry, your file was not uploaded."; } else { if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)){ echo "The file " . basename( $_FILES["fileToUpload"]["name"]) . " has been uploaded."; } else { echo " off!"; } } $mainpic = basename( $_FILES["fileToUpload"]["name"]); $query = "INSERT INTO holidayhomes (mainpic) VALUES ('$mainpic')"; $result = $conn->query($query); if(!$result) echo "INSERT failed: $query<br>" . $conn->error . "<br><br>"; ?> How do I get the form to submit and add the name of the photo to the same record(row) as the homes information? So then when it iterates through the record(rows) in the MySQL database I can display the photo to the relevant information about the home?
  4. I have the following peace of code. I cant solve how to fetch all the records in the database. I have to use prepare but if I use place holders for $sql then it doesnt work I tried foreach and while but I dont know how to use it. Can someone please tell what code to use to fetch multiple records from a table? <?php // make connection to database $sql = "SELECT id_nr, name FROM table;"; $id_nr =1; if ($stmt = $conn->prepare($sql)) { $stmt->bind_param('i', $id_nr); $stmt->execute(); $stmt->store_result(); $num_of_rows = $stmt->num_rows; $stmt->bind_result($id_nr, $name); while ($stmt->fetch()) { echo ' id number : ' . $id_nr . '<br>'; echo ' name : ' . $name . '<br>'; } $stmt->free_result(); } $stmt->close(); mysqli_close($conn); } ?>
  5. Hello, I am creating a web comic aggregation website. I am currently working on the back end in PHP, which uses an RSS reader to update a MySQL database every so often. Than I want the front end to access this database every time a user accesses the website. However, it seems like this will inevitably cause simultaneous reading from and writing to the database, from the front and back ends, respectively. Is this going to cause problems down the line? Or can MySQL handle this without glitching? Thanks in advance
  6. Balderick

    query to check existance of a record

    How to find a record in a table? I now use : $sql = "SELECT col1 FROM table WHERE col1 = '$var' " ; if ($stmt = $conn->prepare($sql)) { $stmt->execute(); $stmt->bind_result($var); while ($stmt->fetch()) { echo $var; echo ' : this variable exists <br>'; } $stmt->close(); } to fetch $var. But this actually results in an echo of the input. The goal is not necessarily echoing it, but determining whether it exists or not and report that its not existing So is it possible with other query to determine if $var exists and then use that as TRUE ?
  7. Kevin Castro

    Selecting No of records in php & mysql

    i want to show the record from mysql database when the users are not paid the initial amount? my database look like this: i've a table called roombooking in that field i've several columns like name,mob ,paid_amount and balance_amount.. so i want to retrieve a data who are all not paid the initial amount.... can u please tell me how to show it? Thanks in advance
  8. mysteriousmonkey29

    trying to design an efficient database for my website?

    Hello, I am currently working on a web comic aggregation website. My general Design is an RSS reader that grabs the publishing date and image source link From various web comic RSS feeds each time they are updated, and uses this to update a database. Then the front end accesses this database info to be displayed in a user-friendly format each time someone accesses the website. I might eventually add a user account system to store user preferences about comic arrangements, which would also need to be put in the database. I am currently writing the back end in PHP and MySQL, and am trying to decide on a general database design. I googled how to design an efficient relational database, which provided a lot of information, but also filled me with fear that my database will be massively inefficient (because I've never done this before), and that it will eventually make my site slow and unusable. So I thought I would check my design on some forums before I proceed. My original plan was to have two databases: one for comics, then one for users later (am not too concerned about this one, because I'm sure there are a ton of examples for how to make an efficient username/password/info database). The comics database would have one table for each different comic, and each table would have a column for publishing date and image source link. However, after reading about database normalization, I thought of an alternative design: one table for comics, including their name and a primary ID,and then one table for publishing dates and one table for image source links, both including their own primary ID, and also the foreign key of the comic ID for each link or date. However, then I realized that the publishing dates table isn't going to make any sense because the publishing date is kind of like a key for the image source link. So basically I'm pretty confused as to how to generally arrange my tables and fields. I was wondering if anyone has any suggestions? I also have another concern. Some RSS feeds do not include publication dates, so it's possible that I am going to need to have some null entries for publishing dates. The MySQL manual told me to use the "NOT NULL" whenever possible, so I was wondering if not using it for this category will be a problem? Finally, there are a few miscellaneous things that I was wondering about. I have read that I can change table types, storage engines, and row format (compact, dynamic, compresse, etc.). I have also read that sometimes websites sacrifice normalization (and thus total storage space) for speed somehow. I was wondering if these more nitpicky details are important for my website? Although I plan on accumulating very long lists of individual comics (at least thousands, maybe tens of thousands for each different web comic), and the website could potentially become accessed very frequently if it became popular, it doesn't seem like what I am trying to do with my database is nearly as complicated as the kinds of things that most people are trying to do. So I was wondering how much of this optimization stuff is really worth the time, and how much of it is just going to make insignificant differences in the performance of my website? Help would be much appreciated
  9. How To show category wise posts using custom fields i am use acf plugin and i want show custom fields page i have 3 custom fields 1 album 2 artist 3 lyrics and 2 category 1st remix 2nd top song i want creat deffrant page for all 3 fields like 1st page album where i can show all album with 1st category with paging See Below Attached file For Best Understanding
  10. sevillano665

    "Require_once" produces "undefined variable"

    Hi guys! I have a little problem and it's that I get an error in a determined function where I call my connection and I don't know why. The errors are: 1) Notice: Undefined variable: conexionidiomas in C:\wamp64\www\idiomas\preguntas-frecuentes.php on line 10 2) Warning: mysqli_real_escape_string() expects parameter 1 to be mysqli, null given in C:\wamp64\www\idiomas\preguntas-frecuentes.php on line 11 This is what I have from "/Connections/conexionidiomas.php": # FileName="Connection_php_mysql.htm" # Type="MYSQL" # HTTP="true" $hostname_conexionidiomas = "p:localhost:3307"; $database_conexionidiomas = "idiomasbd"; $username_conexionidiomas = "root"; $password_conexionidiomas = "asdasdf"; $conexionidiomas = mysqli_connect($hostname_conexionidiomas, $username_conexionidiomas, $password_conexionidiomas, $database_conexionidiomas); and this in "preguntas-frecuentes.php": <?php require_once('Connections/conexionidiomas.php'); ?> <?php if (!function_exists("GetSQLValueString")) { function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") { if (PHP_VERSION < 6) { $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue; } $theValue = function_exists("mysqli_real_escape_string") ? mysqli_real_escape_string($conexionidiomas, $theValue) : mysqli_escape_string($conexionidiomas,$theValue); switch ($theType) { case "text": $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL"; break; case "long": case "int": $theValue = ($theValue != "") ? intval($theValue) : "NULL"; break; case "double": $theValue = ($theValue != "") ? doubleval($theValue) : "NULL"; break; case "date": $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL"; break; case "defined": $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue; break; } return $theValue; } } $editFormAction = $_SERVER['PHP_SELF']; if (isset($_SERVER['QUERY_STRING'])) { $editFormAction .= "?" . htmlentities($_SERVER['QUERY_STRING']); } if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form1")) { $insertSQL = sprintf("INSERT INTO tblfrecuentes (strPregunta,fchFecha) VALUES (%s,NOW())",GetSQLValueString($_POST['strPregunta'], "text")); $Result1 = mysqli_query($conexionidiomas,$insertSQL) or die(mysqli_error($conexionidiomas)); } ?> So, if someone could help me I'd appreciate it so much. Regards!
  11. Hi I want to bring this information to insert multiple records from select <?PHP error_reporting(0); include("connect.inc.php"); ?> <?php include "connect.inc.php"; if (isset($_POST['submit'])) { $province_id = $_POST['province_id'][$i]; $province_id = $_POST['travel_id'][$i]; $i = 0; foreach ( $_POST as $val) { mysql_query("INSERT INTO travel_list (province_id, travel_id) VALUES ('$province_id', '$travel_id')"); $i++; } } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <link href="css/bootstrap.min.css" rel="stylesheet" media="screen"> <script src="js/bootstrap.min.js"></script> <title>Untitled Document</title> </head> <body> <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.2/jquery.min.js"></script> <script src="respond.js"></script> <form action="index.php" method="post"> <!--ส่งค่า post ไปหน้าเดิม --> <table width="896" border="1"> <thead> <tr> <th width="239">จังหวัด</th> <th width="552">สถานที่ท่องเที่ยว</th> <th width="83"><input type="hidden" id="txtNum" value="1" size="2" /></th> </tr> <tr> <th width="239"> <select id="selProvince" name="province_id[]"> <!--Default จังหวััด--> <option value="">กรุณาเลือกจังหวัด</option> <?PHP $SelectPr="SELECT * FROM province"; $QueryPro=mysql_query($SelectPr); while($Pro=mysql_fetch_array($QueryPro)){ ?> <option value="<?=$Pro['province_id']?>"><?=$Pro['province_name']?></option> <?PHP } ?> </select> </th> <th width="552"><select name ="travel_id[]" id="selTravel"><option value="">กรุณาเลือกจังหวัด</option></select></th> <th width="83"><button type="button" id="btnP">เพิ่มรายการ</button></th> </tr> <tr><td colspan="3"><center>รายการที่เพิ่ม</center></td></tr> </thead> <tbody> </tbody> </table> <input name="submit" type="submit" value="add"> </form> <br><br><br> <!--ทอสอบ ค่าแสดงผล Muti Atrray--> </body> </html> I'm insert 1 record but I choose just one , but more data is imported into fifth data . Informationprovince_id, travel_id not insert into HELP ME HOT FIX CODE I choose just one , but more data is imported into fifth data .
  12. I need to check if value inserting in database while **import excel file** , if it has already value in database then it should get update. Below is producttab table value in database prdid | prdname 00A | prd1 00B | prd2 00C | prd3 00D | prd4 Below is EXCEL FILE data prdid | prdname 00A | prdnew 00B | prd2new 00E | prd8 00H | prd9 So if i upload above excel file then , 00A , 00B should get UPDATE IN producttab table as they are already present there... but 00E,00H should get insert below is what i have tried, value is getting insert properly only UPDATE IS NOT HAPPENING if(isset($_POST["Upload"])) { $fileinfo = pathinfo($_FILES["uploadFile"]["name"]); $filetype = $_FILES["uploadFile"]["type"]; $remark = NULL; //Validate File Type if(strtolower(trim($fileinfo["extension"])) != "csv") { $_SESSION['msg_r'] = "Please select CSV file"; header("location:importfile.php"); exit; } else { $file_path = $_FILES["uploadFile"]["tmp_name"]; } $row = 0; $tempFileName = time().".csv"; if ( is_uploaded_file( $file_path ) ) { $fileCopied = copy( $file_path , $tempFileName); if (($handle = fopen($tempFileName, "r")) !== FALSE) { fgetcsv($handle); while (($data = fgetcsv($handle, 6000, ",")) !== FALSE) { $num = count($data); for ($c=0; $c < $num; $c++) { $col[$c] = $data[$c]; } $col1 = $col[0]; // prdid $col2 = $col[1]; // prdname $sql = "SELECT prdid FROM producttab WHERE prdid = '".$col1."' "; $query = db_query($sql); $pfetech = db_fetch($query); if($col1 == $pfetech['prdid']){ $sqlup = "UPDATE producttab SET prdid = ".$pfetech['prdid'].", prdname = ".$col2." "; $sqlup .= " WHERE prdid = ".$pfetech['prdid']." "; $resultsqlupdate = mysql_query($sqlup); }else{ $query = "INSERT INTO producttab(prdid,prdname) VALUES('".$col1."','".$col2.")"; $s = mysql_query($query); } } fclose($handle); } echo "<center>File data imported to database!!</center>"; } } }
  13. I have a table which is either empty or has 1 record If the table is empty i cant retrieve info from the database and get an error message. my code: $sql = "SELECT id FROM users WHERE id = (SELECT MAX(id)FROM users)"; $result = $conn->query($sql); foreach ( $conn->query($sql) as $row ) { $id = $row['id']; the for each line gives an error message: how to solve this?
  14. Its PHP MYSQL : I have a table prodt , in which i first INSERT a value and with its LAST INSERT ID i do update for MAX + 1 as below , BUT I AM GETTING ERROR You can't specify target table 'prodt' for update in FROM clause $a = db_insert_id(); $sqllast = "UPDATE prodt SET pdname= ((SELECT pdname FROM ( SELECT MAX( pdname ) AS pdname FROM prodt WHERE oid = ".db_escape($oid)." ) AS pdname ) + 1 ), pcyn = ".db_escape(0)." WHERE id = ".db_escape($a)." AND oid= ".db_escape($oid)." "; $resultsqllast = db_query($sqllast); if((!$resultsqllast) || (db_mysql_affected_rows($db) <= 0)) { throw new Exception('Wrong SQL UPDATE' . $sqllast . ' Error: '.db_error_msg($db) . db_error_no()); } After research i tried below : $sqllast = "UPDATE prodt SET pdname= ((SELECT pdname FROM ( SELECT MAX( pdname ) AS pdname FROM ( SELECT * FROM prodt WHERE oid = ".db_escape($oid)." )AS pdname ) AS pdname ) + 1 ), pcyn = ".db_escape(0)." WHERE id = ".db_escape($a)." AND oid= ".db_escape($oid)." "; $resultsqllast = db_query($sqllast); if((!$resultsqllast) || (db_mysql_affected_rows($db) <= 0)) { throw new Exception('Wrong SQL UPDATE' . $sqllast . ' Error: '.db_error_msg($db) . db_error_no()); } But still its not working, getting same error message ... Thanks
  15. zioarnold

    PHP Form with MySQL queries

    <?php include "connetti.php"; session_start(); $query = "SELECT id_studente, nome, cognome, anno_maturita, voto_maturita FROM utenti_studenti WHERE utenti_studenti.confermato = 0"; $risultato = @mysql_query($query); if (mysql_num_rows($risultato) == 0) { echo("Nessun elemento trovato"); header("refresh:3;url=AreaAmministratore.phtml"); exit(); } ?> <html> <head> <title>Annuario Studenti</title> <link rel="stylesheet" href="cssUtils/aggiungi_properties.css"/> <script type="text/javascript" src="jsUtils/jsUtils_annuario/annuario_properties.js"></script> </head> <body> <form method='post' action='confermaStudenti.php' name='aggiornaStatoStudente'> <div align='center'> <table class='tg'> <tr> <th class='tg' style='color: #000;'>ID</th> <th class='tg' style='color: black'>Nome</th> <th class='tg' style='color: black'>Cognome</th> <th class='tg' style='color: black'>Anno Maturita</th> <th class='tg' style='color: black'>Voto Maturita</th> <th class='tg' style='color: black'>Accetta</th> </tr><?php while ($row = @mysql_fetch_assoc($risultato)) { echo " <tr align='center'> <td class='tg' style='color: black'>" . $row['id_studente'] . "</td> <td class='tg' style='color: black'>" . $row['nome'] . "</td> <td class='tg' style='color: black'>" . $row['cognome'] . "</td> <td class='tg' style='color: black'>" . $row['anno_maturita'] . "</td> <td class='tg' style='color: black'>" . $row['voto_maturita'] . "</td> <td class='tg' style='color: black'>" . "<input type=\"hidden\" name=\"rifiuta\"/><input type=\"checkbox\" name=\"accetta\"/>" . "</td> </tr>"; } ?> </table> </div> <div align='center'> <a href='AreaAmministratore.phtml'>Torna indietro.</a> <input type='submit' value='Prosegui'/> </div> </form> </html> <?php session_start(); include("connetti.php"); if (isset($_POST["accetta"])) { $accetta = 1; } else { $accetta = 0; } if ($accetta) { $cognome = $_POST['cognome']; $nome = $_POST['nome']; $id_studente = $_POST['id_studente']; $query = "UPDATE utenti_studenti SET confermato='1' WHERE nome='$nome' AND cognome='$cognome' AND id_studente='$id_studente'"; $risultato = @mysql_query($query) or die('<p align="center">Errore!</p>' . mysql_error()); echo("<script>alert('La modifica eseguita')</script>"); header("refresh:0;url='AreaAmministratore.phtml'"); } else { echo "<script>alert('Errore');</script>"; header("refresh:0;url='AreaAmministratore.phtml'"); exit(); } ?> Greetings guys, could you help me with those codes above? Cuz I can't understand why it doesn't work... So I have an table with the users that not confirmed. So to confirm them I've made checkbox, so on the other side I have a control if the checkbox is checked so I need to update some values on my database right ? Well it doesn't work....
  16. Hello Every One I Want Show all USer last Useges Data From date of Last Recharge to To Date and condition With user owner by i Have 3 Table invoice - Need data from column date(need last invoice Last Recharge date For Below Table Condition Start ),username( Uniq In all table) acct - > Need data from column acctstarttime( For Start Date ),SUM( acct.acctinputoctets + acct.acctoutputoctets ) AS data(For Sum Of Useges Data ),SUM( acct.acctsessiontime ) AS acctsessiontime,username(Uniq In all table) users - > Need data from column username( Main Table Match username on this table base),owner,lastlogoff,expiration,uptimelimit,comobolimit, My Code is Here But Not working SELECT DATE( `acctstarttime` ) AS acctstarttime, SUM( acct.acctinputoctets + acct.acctoutputoctets ) AS data, acct.username, SUM( acct.acctsessiontime ) AS acctsessiontime, users.username, users.owner, users.lastlogoff, users.expiration, users.uptimelimit, .comblimit, users.enableuser,invoices.date FROM invoices,acct JOIN users ON acct.username = rm_users.username WHERE users.owner = 'admin' and acctstarttime between 'invoice.date' and 'users.expiration' GROUP BY MONTH( `acctstarttime` ) , acct.username ORDER BY invoices.date DESC LIMIT 0 , 30 I Think Problem in Where condition i want acctstarttime between 'invoice.date(Desc or User last recharge Date)' and 'users.expiration' I m very Confused How I dow Any One Can Help Thanks in Extra
  17. Tejpal

    financial year calculation in php

    Hello everyone I want Show Some Recorded according financial year For Example from 01-04-2016 to 31-3-2017 01-04-2017 to 31-03-2018 my Code $pst = date('Y'); $pt = date('Y', strtotime('+1 year')); $sql="SELECT *FROM mytable where date BETWEEN CAST('$pst-04-01' AS DATE) AND CAST('$pt-03-31' AS DATE)"; Anyone Can explain how i done this
  18. shashib

    Group by value ( remarks ) below tr

    Below is products table : id | mid | wgh | remark| remkok | 1 3 1.5 r3ok 1 2 2 1.5 0 3 2 0.6 nice 0 4 1 1.2 okh 0 5 4 1.5 bye 0 6 4 2.4 okby 0 7 3 3.0 oknice 1 I want to display remark below tr of group by mid ....like below mid wgh 3 1.5 3.0 remarks : r3ok, oknice 4 1.5 2.4 remarks : bye, okby 2 1.5 0.6 remarks : , nice 1 1.2 remarks : okh **What i have tried as below :** $pid= null; while($row = mysql_fetch_array($result)) { $rowpkts = $row['mid']; echo "<tr class=\"undercl\">"; if($rowpkts != $pid){ echo'<td align="center" valign="top">'.$row["mid"].'</td>'; }else{ echo'<td align="center" valign="top"></td>'; } echo'<td align="center" valign="top">'.$row["wgh"].'</td>'; echo "</tr>"; // what i tried to build for remarks as below $remsql = "SELECT mid as onu , GROUP_CONCAT(`remark` ORDER BY `id` ASC SEPARATOR ', ') AS plrmks FROM products WHERE remkok= 1 GROUP BY `mid`"; $fetchremk = mysql_query($remsql); $rowresults = mysql_fetch_array($fetchremk); if($rowresults['onu'] == $pid ){ echo"<tr style='border-style:underline;'>"; echo'<td align="center" align="top">'.$rowresults["plrmks"].'</td>'; echo"</tr>"; } } $pid = $rowpkts; } But remarks is not coming proper ...i means its not display below mid=3 or mid=1.....
  19. ggkolfin

    Showing data that no longer exist

    Hello everyone, I am not sure if this is the right place to post since I do not know for sure if it is PHP that causes my issue, so I apologize in advance if I shouldn't have posted here. I am using WampServer Version 3.0.0 64bit (Apache: 2.4.17 | PHP: 5.6.16 | MySQL: 5.7.9) on windows10. I am updating a website and I am keeping the database the same, except from some small optimizations here and there. I have a query that selects all active rows of a specific category, then I fetch the results and show the rows: if(mysqli_stmt_prepare($stmt, "SELECT * FROM myTableName WHERE _category=? AND _approved='1' ORDER BY _id DESC LIMIT ?, ?")) { mysqli_stmt_bind_param($stmt, "iii", $category, $offset, $rowsPerPage); mysqli_stmt_execute($stmt); mysqli_stmt_bind_result($stmt,$id,$title,...,$image,...); //all fields are binded, including image while(mysqli_stmt_fetch($stmt)) { //show recipes } } The query is correct and everything shows up fine, except from the image returned by the query. My old database used to store the image with the path, eg "/this/is/the/path/to/image.jpg" but I have updated the image table column to only store the file name so now $image variable should contain only the file name, eg "image.jpg". But it does not. The $image variable still contains the full path, although the table is updated! I though it was some kind of caching, but clearing the browser's cache, going incognito, hard refreshing, changing browser did not change anything. I also tried running: SET GLOBAL query_cache_size = 0; in mySql terminal but that did not change anything as well. I also tested php.ini for Op Cache but it was already disabled: opcache.enable=0 I do not know what to do about this and it is driving me crazy! Any help will be very much appreciated. Thank you all, georgia
  20. Hello Everyone I M in Problem i Am creat my 1st project for client so plesese help me that how i done i need how i calculate field in php like this Pic total of Credit total of debit and pending balance and grand total So tell Me how i done Below i attecthed
  21. WesleyA

    cant move column position PHPMYADMIN

    In PHPmyadmin I can't move the position of the columns. So when I go to change/modify (not sure how it's called in English) I get the possibility to move column and the place it after another column then where its placed now. Though, it does print the query it goes like ALTER TABLE `my_tab` CHANGE `col1` `col1` TINYINT(1) NULL DEFAULT NULL AFTER `col5`; I have not tested in the mysql console, but in PHP the columns don't move (anymore) like they did do beforehand.
  22. Hi everyone! Thank you in advance for reading this and any help you are able to provide. This has been a bit of a long road but I'm learning along the way. Before I begin I am well aware of the dangers of SQL injection and understand that using prepared statements would decrease injection attacks for the following code. This is a PHP/MySQL test code to see if it works before actual implementation on a live site. With that said here we go: I have a database and it contains four tables (for the sake of security I gave them disney character names) named huey, dewey, lewey and uncledonald. I would like to have the values from the columns deweysays in the table dewey, hueysays from the table huey and leweysays from the table lewey to show up in thier corresponding deweysays, hueysays and leweysays columns in the table uncledonald. See attached pic to see visually what I mean. I've tried the following code and get the result I want (values added to all tables) but only once. After that I get data in the dewey, huey and lewey tables but nothing else in the uncledonald table. Here is the PHP: <?php //Let's see whether the form is submitted if (isset ($_POST['submit'])) { $con=mysqli_connect("localhost","root","root","provingground"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $sql = "INSERT INTO dewey (lot_id, deweysays) VALUES (0, '{$_POST['deweyspeak']}');"; $sql .= "INSERT INTO huey (cust_id, hueysays) VALUES (0, '{$_POST['hueyspeak']}');"; $sql .= "INSERT INTO lewey (personal_id, leweysays) VALUES (0, '{$_POST['leweyspeak']}');"; $sql .= "INSERT INTO uncledonald (deweysays) SELECT deweysays FROM dewey "; $sql .= "INSERT INTO uncledonald (hueysays) SELECT hueysays FROM huey "; $sql .= "INSERT INTO uncledonald (leweysays) SELECT leweysays FROM lewey "; // Execute multi query if (mysqli_multi_query($con,$sql)){ print '<p> The Ducks Have Spoken.</p>'; } else { die ('<p>Could not add entry because:<b>' . mysqli_error() . '</b>.</p><p>The query being run was: ' . $sql . '</p>'); } } mysqli_close($con); ?> Is there something missing in my $sql query to uncledonald? Is the script completely off? Lot of questions…Help please!
  23. dauruk0512

    doing event doesn't working

    http://prntscr.com/a3dz84 please check why it doesn't working ? i have been tested change date in local computer but nothing happen, thanks UPDATE `cu_employee` SET `cu_employee_sisa_cuti` = '12' WHERE 1
  24. Hello everyone, I'm brand new to PHP and MySQL and I'm trying to build a login/register form for my company's website. I've literally scavenged the internet for the past 3 days and have watched multiple tutorials but still can't get my database to link to the php file(s). If anyone can guide me in the right direction or provide a dummy proof tutorial, it would be greatly appreciated!!! (I would attach my php code but I literally have nothing and have gotten no where) Any advice would help. Thanks!
  25. So recently I got my hands on one of youtube videos, it's quit old, and the guy explaining how to create a form for uploading image and all data that is connected with it, to MySQL and then showing on your web site. So everything is ok, but webpage does not display any images, it just show me white picture frame with picture icon on the top. This is my connection file with database: <?php $hostname_phpimage = "***"; $username_phpimage = "***"; $password_phpimage = "***"; $database_phpimage = "***"; // Create connection $inkedmen_marko = mysql_pconnect($hostname_phpimage, $username_phpimage, $password_phpimage); // Check connection if ($phpimage->connect_error) { die("Connection failed: " . $phpimage->connect_error); } ?> Then the file that uploads the images: <?php require_once('php/connect.php'); if($_POST['submit']) { $name=basename($_FILES['file_upload']['name']); $t_name=$_FILES['file_upload']['tmp_name']; $dir='/home4/inkedmen/public_html/images'; $cat=$_POST['cat']; if(move_uploaded_file($t_name,$dir."/".$name)) { mysql_select_db($database_phpimage, $inkedmen_marko); $qur="insert into anglija (mid, cid, name, path) values ('','$cat','$name','/home4/inkedmen/public_html/images/$name')"; $res=mysql_query($qur,$inkedmen_marko); echo'file upl success'; } else { echo 'not uploaded'; } } ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1"> </head> <body> <form action="upload.php" method="post" enctype="multipart/form-data"> <input type="file" name="file_upload"/> cat_id <input type="text" name="cat" /><br/> <input type="submit" name="submit" value="upload"/> </form> </body> </html> Then this one shows me what countries I have in one of the data base tables, so there is two england and slovenia, they both have a separate id that I have to type when uploading an image: <?php require_once('php/connect.php'); mysql_select_db($database_phpimage, $inkedmen_marko); $qur="select * from cat"; $res=mysql_query($qur,$inkedmen_marko); ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1"> </head> <body> <?php while($row= mysql_fetch_array($res)) { ?> <h1><a href="image.php?cid=<?php echo $row['id']?>"><?php echo $row['name'] ?> </a></h1><br/> <?php } ?> </body> </html> And finally, when I push on each country it directs me to the pictures that are connected with that countries according to the id given: <?php require_once('php/connect.php'); mysql_select_db($database_phpimage, $inkedmen_marko); $id=$_GET['cid']; $qur="select * from anglija where cid='$id'"; $res=mysql_query($qur,$inkedmen_marko); ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1"> </head> <body> <?php while ($row=mysql_fetch_array($res)){ ?> <img src="<?php echo $row['path'] ?>" width="300px" height="200px"> <br/> <?php } ?> </body> </html> So in the end it directs me to the pictures, but I can't see the pictures, it's just white windows with icon on the top. The pictures uploads to the image folder, the id, names of the file, path also uploads good, but somehow I can't show the images. I know it should be something small, but I can't figure it out. Maybe someone have any ideas????
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