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Found 278 results

  1. Dear all. For an image uploaded to a folder/directory and stored its name, type, and size in a DB table, what could be the merits of also storing its temporary name, noting that such temporary name is deleted by the server upon moving the image to the desired folder/directory?
  2. Hi all, In fetching a result from a table (below code), 1 is also printed after the fetched record (fetched record below). There is no 1 in the table, nor have I accidentally included it in my script. What is it and why does it appear? <?php require 'db/connect.php'; if($result = $db->query("SELECT * FROM user")) { if ($count = $result->num_rows) { $rows = $result->fetch_assoc(); echo '<pre>', print_r($rows), '</pre>'; } } ?> Array ( [id] => 1 [first_name] => Bill [last_name] => John [bio] => I'm a web developer [created] => 2016-11-21 12:50:12 ) 1
  3. Hi all, <?php require 'db/connect.php'; $result = $db->query("SELECT * FROM people"); ?> I obtained a blank page running the above script, which indicates that everything is fine (path and table name are correct). I then removed the * from the query to see if it would report an error, but it did not. I added print_r($result); and run it. Nothing has changed. I included the * and run, it reported the number of rows and columns in the table. Why there has a been a lack of error reporting in this instance, noting that the ini file is configured to report all sorts of errors, even the warnings and notices (and it is reporting all that in other scripts)?
  4. Hi all, In one website, I read the following note: "You should also think about storing files locations on disk. Using MySQL for storing images is thought to be Bad Idea™. Handling SQL table with big data like images can be problematic." Do you agree? If not, why? What is the best practice in storing and displaying on a website a big number of images?
  5. I'm trying to select data from a MySQL database that is hosted on a webserver. I want to be able to retrieve the data from a table within the database and then illustrate it within a HTML table. There's an example on W3Schools that I've been following, but I'm unable to retrieve the data successfully. http://www.w3schools.com/php/php_ajax_database.asp Below is the source code: (HTML) <html> <head> //Javascript code <script> function showUser(str) { if (str == "") { document.getElementById("txtHint").innerHTML = ""; return; } else { if (window.XMLHttpRequest) { // code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp = new XMLHttpRequest(); } else { // code for IE6, IE5 xmlhttp = new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange = function() { if (this.readyState == 4 && this.status == 200) { document.getElementById("txtHint").innerHTML = this.responseText; } }; xmlhttp.open("GET","getuser.php?q="+str,true); xmlhttp.send(); } } </script> </head> <body> <form> <select name="users" onchange="showUser(this.value)"> <option value="">Select a person:</option> <option value="1">Peter Griffin</option> <option value="2">Lois Griffin</option> <option value="3">Joseph Swanson</option> <option value="4">Glenn Quagmire</option> </select> PHP File: (getuser.phd) <!DOCTYPE html> <html> <head> <style> table { width: 100%; border-collapse: collapse; } table, td, th { border: 1px solid black; padding: 5px; } th {text-align: left;} </style> </head> <body> <?php $q = intval($_GET['q']); $con = mysqli_connect('www.example.com','user_Admin','12345-678','my_DB'); if (!$con) { die('Could not connect: ' . mysqli_error($con)); } mysqli_select_db($con,"ajax_demo"); $sql="SELECT * FROM user WHERE id = '".$q."'"; $result = mysqli_query($con,$sql); echo "<table> <tr> <th>Firstname</th> <th>Lastname</th> <th>Age</th> <th>Hometown</th> <th>Job</th> </tr>"; while($row = mysqli_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['FirstName'] . "</td>"; echo "<td>" . $row['LastName'] . "</td>"; echo "<td>" . $row['Age'] . "</td>"; echo "<td>" . $row['Hometown'] . "</td>"; echo "<td>" . $row['Job'] . "</td>"; echo "</tr>"; } echo "</table>"; mysqli_close($con); ?> </body> </html> *MySQL table is attached I think the issue might exist from mysqli_select_db($con,"ajax_demo"); onwards inside the PHP file. Should I be referring to the table that contains the data inside the database? I have the PHP File hosted on my webserver, so I'm not sure why it won't retrieve that data when a person is selected from the list of options on the HTML page. Any help would be much appreciated.
  6. j.silver

    utf8 VS utf8mb4

    Dear all, I came across the following statement on one website: "Note: Since MySQL 5.5.3 you should use utf8mb4 rather than utf8. They both refer to the UTF-8 encoding, but the older utf8 had a MySQL-specific limitation preventing use of characters numbered above 0xFFFD." 1) Is there any drawbacks in using utf8mb4, or it is recommended for use from now on?
  7. Hi I have been working on building a custom post a result set to a database which thankfully I have now managed with some help, now I am working on the other end of this project the displaying the saved data back to users, From the start of this I have an online tournament hosting group that go to my form and post a list of player names with points earned per tournament. Once the form is submitted it's sent to my form_processing.php where the results are exploded and split into 2 arrays "$player_name, $points" Then on the INSERT I have an INSERT $sql = "INSERT into `bg_points` (`player_name`, `points`) values (?, ?) on duplicate key update points = points + ?"; to of course update current player points if the exist or add new records if not. Then finally saved into my DB table, now I am working on building the page to display these points back to the players so they can track their earned points and standings each month. which is easy enough I just call an MySQL SELECT * FROM statement on the DB table that's done, Now I need to sort these results in lowest to highest order by points so if player1 has 120 points player2 has 50 points player3 has 75 points player4 has 5 points then I need to be displaying these back as player4 player2 player3 player1 I have had a look online for some possible examples to help guide me into this and I either get HTTP 500 or just no effect at all. here is the very basic code I have put together so far which I can and will update and add into my site framework once I have it in order and working... <HTML> <head></head> <body bgcolor="#0000FF"> <?PHP $servername = "localhost"; $username = "***********"; $password = "*******"; $dbname = "***********"; // Create connection $conn = new mysqli($servername, $username, $password, $dbname); // Check connection if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } //output the saved data $sql = "SELECT * FROM `bg_points` "; $result = $conn->query($sql); if ($result->num_rows > 0) { // output data of each row while($row = $result->fetch_assoc()) { echo "id: " . $row["id"]. " $nbsp $nbsp Playerv Name: " . $row["player_name"]. " $nbsp $nbsp Points: " . $row["points"]. "<br>"; } } else { echo "0 results"; } $conn->close(); ?> </body> </html> thanks for any help or ideas in advanced
  8. Hi I am having trouble inserting data into my MySQL database. This is my PHP code ini_set('display_errors', 1); ini_set('display_startup_errors', 1); error_reporting(E_ALL); static $connection; if(!isset($connection)) { $connection = new mysqli("localhost","username","password"); } if ($connection->connect_error) { die("Connection failed: " . $connection->connect_error); } $stmt = $connection->prepare("INSERT into Product(product_title,product_price,product_availability,productImage_1,productImage_2,productImage_3,productImage_4,product_description,product_shipping,product_pickup) VALUES(?,?,?,?,?,?,?,?,?,?)"); $product_title = $_POST['title']; $product_price = $_POST['price']; $product_availability = $_POST['stock']; $null = NULL; $product_description = $_POST['description']; $product_shipping = $_POST['postage']; $product_pickup = $_POST['pickup']; $stmt->bind_param('sisbbbbsis',$product_title,$product_price,$product_availability,$null,$null,$null,$null,$product_description,$product_shipping,$product_pickup); $stmt->send_long_data(3, file_get_contents($_FILES['img1']['tmp_name'])); $stmt->send_long_data(4, file_get_contents($_FILES['img2']['tmp_name'])); $stmt->send_long_data(5, file_get_contents($_FILES['img3']['tmp_name'])); $stmt->send_long_data(6, file_get_contents($_FILES['img4']['tmp_name'])); $stmt->execute(); $stmt->close(); $connection->close(); echo "Product inserted successfully"; I am also getting these errors. Notice: Undefined index: title in /Applications/MAMP/htdocs/Bourke/insertproduct.php on line 61 Notice: Undefined index: price in /Applications/MAMP/htdocs/Bourke/insertproduct.php on line 62 Notice: Undefined index: stock in /Applications/MAMP/htdocs/Bourke/insertproduct.php on line 63 Notice: Undefined index: description in /Applications/MAMP/htdocs/Bourke/insertproduct.php on line 67 Notice: Undefined index: postage in /Applications/MAMP/htdocs/Bourke/insertproduct.php on line 68 Notice: Undefined index: pickup in /Applications/MAMP/htdocs/Bourke/insertproduct.php on line 69 Fatal error: Call to a member function bind_param() on boolean in /Applications/MAMP/htdocs/Bourke/insertproduct.php on line 71 Thanks in advance.
  9. <?php $host = 'localhost'; $username = 'root'; $password = ''; $datadase = 'registerfinal'; $connect = mysqli_connect($host, $username, $password) or die ('error to connect to datadase'.mysqli_error()); if ($connect) { echo 'mysqli connect succsessfull'; } echo '<br /><br />'; $selectdb = mysqli_select_db($connect, $datadase) or die ('unable to select datadase'.mysqli_error()); if($selectdb) { echo 'database selected succsessfully'; } if(isset($_POST['savedetails'])) { $firstname = $_POST['firstname']; $lastname = $_POST['lastname']; $username = $_POST['username']; $password = $_POST['password']; $repeat_password = $_POST['repeat_password']; $gender = $_POST['gender']; $country = $_POST['country']; if(isset($_POST['food'])) { $food = $_POST['food']; $favfood = ""; foreach($food as $meal ) { $favfood = $meal.","; print_r($favfood); } } if(isset($_POST['imageUpload'])) { $imageUploadname = $_FILES['imageUpload']['name']; $imageUploadsize = $_FILES['imageUpload']['size']; $imageUploadtmp_name = $_FILES['imageUpload']['tmp_name']; $imageUploadtype = $_FILES['imageUpload']['type']; $uploadFolder = "uploadFolder/"; $destinationName = rand(1000, 10000).$imageUploadname; move_uploaded_file($imageUploadtmp_name, $uploadFolder.$destinationName); echo "$imageUploadname"; echo "$imageUploadsize"; echo "$imageUploadtmp_name"; echo "$imageUploadtype"; echo "$destinationName"; } $sqltwo = "INSERT INTO `registerfinaltable` (`id`, `firstname`, `lastname`, `username`, `password`, `repeat_password`, `gender`, `food`, `country`, `imageUploadname`, `imageUploadsize`, `imageUploadtype`) VALUES (NULL, '$firstname', '$lastname', '$username', '$password', '$repeat_password', '$gender', '$favfood', '$country', '$destinationName', '$imageUploadsize', '$imageUploadtype')"; $results = mysqli_query($connect, $sqltwo) ; if($results){ echo "inserted successfully"; } } ?> <html> <head> <title>register</title> </head> <body> <form action = "" method = "post" enctype = "multipart/form-data" > <label>first name : <input type = "text" name = "firstname" /> </label> <br /><br /> <label>last name : <input type = "text" name = "lastname" /> </label><br /><br /> <label>username : <input type = "text" name = "username" /> </label><br /><br /> <label>password : <input type = "password" name = "password" /> </label><br /><br /> <label>repeat password : <input type = "password" name = "repeat_password" /> </label><br /><br /> <label>Male : <input type = "radio" name = "gender" value = "Male" /> </label><br /><br /> <label>Female : <input type = "radio" name = "gender" value = "Female" /> </label><br /><br /> <label>pizza : <input type = "checkbox" name = "food[]" value = "pizza"/> </label><br /><br /> <label>burger : <input type = "checkbox" name = "food[]" value = "burger"/> </label><br /><br /> <label>chips : <input type = "checkbox" name = "food[]" value = "chips"/> </label><br /><br /> <label>sausage : <input type = "checkbox" name = "food[]" value = "sausage"/> </label><br /><br /> <label>sandwich : <input type = "checkbox" name = "food[]" value = "sandwich"/> </label><br /><br /> <label>Image : <input type = "file" name = "imageUpload" /> </label><br /><br /> <select name = "country"> <?php $sql = 'SELECT * FROM `countrie` '; $querry = mysqli_query($connect, $sql); while($country = mysqli_fetch_array($querry)):; ?> <option value = "<?php echo $country['country']; ?>"><?php echo $country['country']; ?></option> <?php endwhile;?> </select> <br /> <input type = "submit" name = "savedetails" /> </form> <table border = "1" bgcolor = "" width = "100%"> <tr><th>id</th><th>Firstname</th><th>Lastname</th><th>Username</th><th>Password</th><th>Password 2</th><th>Gender</th><th>Fav. Food</th> <th>Image</th> <th>Country</th><th>imageUploadname</th><th>imageUploadsize</th><th>imageUploadtype</th></tr> <?php $sqldata = "SELECT * FROM registerfinaltable"; $querysqldata = mysqli_query($connect, $sqldata); while($rows = mysqli_fetch_array($querysqldata) ):; ?> <tr> <td><?php echo $rows['id'];?></td> <td><?php echo $rows['firstname'];?></td> <td><?php echo $rows['lastname'];?></td> <td><?php echo $rows['username'];?></td> <td><?php echo $rows['password'];?></td> <td><?php echo $rows['repeat_password'];?></td> <td><?php echo $rows['lastname'];?></td> <td><?php echo $rows['gender'];?></td> <td><?php echo $rows['food'];?></td> <td><?php echo $rows['country'];?></td> <td><?php echo $rows['imageUploadname'];?></td> <td><?php echo $rows['imageUploadsize'];?></td> <td><?php echo $rows['imageUploadtype'];?></td> <?php endwhile;?> </tr> </table> </body> </html>
  10. My goal is to store javascript code into a database. My first idea was to use htmlspecialchars; store it in mysql in a table column and later retrieve it with htmlspecialchars_decode. All this to prevent injection / hacking. But online I read one or two warnings that it wouldnt work, which I assume is so (I didnt test it, but it seems quite obvious afterwards) . So my question is: is it possible to have a user store javascript in a database and use it in a php script for specific purposes in a secure way?
  11. Hi i am new for the PHP, i had the experienced before to do a login system page but what i want it is HTML and PHP file to separate i dont want PHP with HTML together to 1 file , cause professional people will more likely to use 2 file rather than 1 file easy to maintain and edit how to use the external PHP file coding without re-design whole page cause normally what i tested before was link or turn to other new page and result will come out but the design was totally gone so have any ways and suggestion to do that ... in the end, im sorry i not well in english and my knowledge of PHP still quite new , thanks you so much
  12. HI i trying to create a website of forum textarea editor are necessary too for recommendation which editor are best for textarea such as summernote, wysihtml5 .... and how is work to store the data/value/text with styling effect in mysql database ? Thanks all of you
  13. HI i trying to create a website of forum textarea editor are necessary too for recommendation which editor are best for textarea such as summernote, wysihtml5 .... and how is work to store the data/value/text in mysql database ? Thanks all of you
  14. HI i trying to create a website of forum textarea editor are necessary too for recommendation which editor are best for textarea such as summernote, wysihtml5 .... and how is work to store the data/value/text in mysql database ? Thanks all of you
  15. <?php $serv = "localhost:81"; $user = "root"; $password = ""; try { $con = new PDO("mysql:host=$serv;dbname=test", $user, $password); $con->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); echo "Successfully connected."; } catch (PDOException $e) { echo "Connection failed: " . $e->getMessage(); } ?> Here I'm trying to use PDO to connect to my SQL database "test". However, I receive this error each time I connect: SQLSTATE[HY000] [2002] No connection could be made because the target machine actively refused it. Am I connecting with the wrong server, user or password? Because I fail to see here the mistake.
  16. 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?
  17. 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
  18. 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?
  19. 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); } ?>
  20. 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
  21. 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 ?
  22. 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
  23. 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
  24. 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
  25. 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!
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