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Found 988 results

  1. 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
  2. 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.
  3. <?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>
  4. 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?
  5. Is it possible to add a 1 vote per IP restriction for this poll without using php? If not with IP restriction, how could a repeated voting of the same person be avoided? How can a Results link be added? Where should it refer? Thank you!
  6. 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
  7. 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
  8. 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
  9. Hi, I have a (simple) website and I have a 'contact me' page where its a basic form, name / email address / comments. and the necessary validation using JS. I have done all this on Notepad++. The site (I think) looks good, and is responsive. But now I want to be able to process the form, where I get the form content emailed to myself. Is PHP the way to go? Do I really need a server? I have tried google search, watched multiple videos but I dont know if I need to download eclipse with the PHP plug in / XAMPP? Is this overkill for a form with 3 fields? Totally confused! Any help appreciated.
  10. 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
  11. 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?
  12. 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
  13. I want to use preg_replace to input only a-zAZ with a limit of 30 characters I have a piece of code I used before but it works with preg_match. example function: <?php // input form to get the name // use the function here $name = valid_name($name); // function to validate name input function valid_name($data){ $data = ltrim($data); $data = rtrim($data); $data = preg_replace("/^(?:[A-Za-z][A-Za-z\-]{2,30}[A-Za-z]|[A-Za-z])$/", '' , $data); return $data; } ?> To give an impression of how I use it; a part of the confirmation form <html> <form action="mypage.php" method="post"> <p><input type="text" name="name" value="<?php echo $name; ?>" STYLE="background-color: #708DCC; color: white; height: 20px; width: 200px; border: none; font-size: 14px;" readonly ></p> <!-- rest of the form --> </html> The problem is in the preg_replace part. an input with chars like: ^&$ etc is not replaced, how to do this? if anyone can help I would be really happy.
  14. Hello, I am trying to create a proof of concept webpage that changes text in response to a button press using an MVC pattern (or at least as I understand it), and Ajax to avoid reloading the page. (I would like to implement Ajax in a larger MVC program I am working on but thought I would try to get it to work small-scale first). From playing around with examples here and here: https://www.sitepoint.com/the-mvc-pattern-and-php-1/ http://www.w3schools.com/php/php_ajax_php.asp I have the program working with each component individually (it works with the MVC pattern if I don't mind reloading the page to update the text, or it works without reloading the page if I don't mind essentially scrapping the MVC pattern). However, I'm trying to get both to work at once. I have combined the two examples so that the view uses Ajax to call the appropriate controller function, which successfully modifies the model (I'm sure this part works from debugging the program). However, when I try to refresh the content of the page using the output function of the view, nothing happens without reloading the page. Here is my code so far: <html> <head> <meta charset="UTF-8"> <!--ajax attempt--> <script> function callTextChange () { var xmlhttp = new XMLHttpRequest(); //if uncommented, this changes the text, but it doesn't fit with my MVC pattern /*xmlhttp.onreadystatechange = function() { if (this.readyState == 4 && this.status == 200) { document.getElementById("text").innerHTML = "changed with purely Ajax, without using MVC"; } };*/ xmlhttp.open("GET", "index.php?action=changeText", true); xmlhttp.send(); } </script> </head> <body> <?php class Model { public $text; public function __construct() { $this->text = 'default'; } function changeText () { $this->text = 'changed'; } } class View { private $model; public function __construct(Model $model) { $this->model = $model; } public function output() { //regular MVC method using button as a link //return $this->model->text.'<a href="?action=changeText"><button>change text</button></a>'; //attempted ajax method using button on click attribute to make an Ajax call return '<p id="text">'.$this->model->text.'</p>'.'<button onclick="callTextChange()">change text</button>'; } } class Controller { private $model; public function __construct(Model $model) { $this->model = $model; } function changeText() { $this->model->changeText(); } } $model = new Model(); $controller = new Controller($model); $view = new View($model); if (isset($_GET['action'])) { $controller->{$_GET['action']}(); } echo $view->output(); ?> </body> Any idea how to do what I'm trying to do? Is this even possible? Help would be much appreciated
  15. 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?
  16. 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); } ?>
  17. Hello, I am currently working on a PHP program that uses an RSS reader to update a MySQL database with comic information from various web comics. After some research, it seems like it would be a better idea to do this using a Cron job or something similar, as opposed to just a sleep loop within PHP. However, my script needs to pass information to itself from the last time it was run. I want it to compare information taken from each RSS page from the last time it ran to see if the page is updated, then only to add information to the database if it has indeed been updated. Unfortunately, I'm not sure how to do this/if this is even possible using a Cron job or Windows task scheduler (I am currently developing on Windows). Any idea how to solve this problem? I suppose I could just use a sleep loop as I was originally considering, but I have read in various locations that this is apparently a very glitchy and unreliable way of doing what I want to do, so I would like to avoid it if possible. Thanks in advance
  18. I am using Jquery-ui's Date picker to pick what date I want. But now I need to store that date I type/pick from DatePicker into php variable. I tried doing this myself but it didn't work, It wasn't being saved in variable without pressing a button. Do I need to use Ajax? If yes, can you help me with that? Thank you. <script> $( function() { $( "#datepicker" ).datepicker(); } ); </script> <input type="text" id="datepicker" name="datepicker" placeholder="Search by date" />
  19. Hello W3Schools! Its been a while I am trying to help a friend setting up he's webshop. It is already online but trying to make it look more nice according to webdesign template - this is hes website shop (www.thai-online.dk) i have already changed A LOT on this design, now i mostly need the menu... as you see the menu navigation has a lot of categories, and he told me that he wanted to add a lot more in the future if this shop could run online, so i am trying to figure out a solution for this. I already like the design myself, except the menu, so i have come to a solution that i am working on and hopefully you can help me out, work with me. As you see i currently have added 13 Categories, now my idea is to code some PHP into the menu on OpenCart so that if the menu design is ABOVE "1100px" in width AND has more than 7-8 categories on the menu, it will only show the 7-8 categories and at the end of it all show a little plus sign "+", when clicked on the plus sign a little side-menu will appear out from that little plus sign. The reason for using the 1100px width thing is to know if its on a pc orr on a mobile (if you know a better way please tell me, this is just to show you what i want) i know for sure the "plus sign" already has been made somewhere, but how to actually do it? We are using opencart version 2.0.1.2 or something like that... this is the code i found on the .tbl file: <?php foreach ($categories as $category) { ?> <?php if ($category['children']) { ?><li class="dropdown"><a href="<?php echo $category['href']; ?>" class="dropdown-toggle" data-toggle="dropdown"><span><?php echo $category['name']; ?></span> <img width="8" height="8" class="menu-close" src="catalog/view/theme/atr374opc2101/image/menu_close.gif" /><img width="8" height="8" class="menu-open" src="catalog/view/theme/atr374opc2101/image/menu_open.gif" /></a> <div class="dropdown-menu"> <div class="dropdown-inner"> <?php foreach (array_chunk($category['children'], ceil(count($category['children']) / $category['column'])) as $children) { ?> <ul class="list-unstyled"> <?php foreach ($children as $child) { ?><!-- --><li><a href="<?php echo $child['href']; ?>"><?php echo $child['name']; ?></a></li><!-- --><?php } ?> </ul> <?php } ?> </div> <a href="<?php echo $category['href']; ?>" class="see-all"><span><?php echo $text_all; ?> <?php echo $category['name']; ?></span></a></div> </li><?php } else { ?><!-- --><li><a href="<?php echo $category['href']; ?>"><span><?php echo $category['name']; ?></span></a></li><?php } ?> <?php } ?> On another note, we are also using BootStrap (latest version) to make the design responsive. And the reason why i use the 1100px width, like i said before is because of knowing if the screen is that wide AND to tell if i should be adding the category addon... if the screen however is less than 1100px (or so...), the menu should be normal. Reason why i do this is because i already like the responsive design, i only need to edit the PC-Screen design a little bit so the menu can fit without problems. Hope you all can help out one way or another, still struggling with this little thing Thanks in advance and again, its been a while, so..... HELLO // rootKID
  20. 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 ?
  21. My first question: I have a message like this: notice: undefined offset 0 I use an input form to check with dns_get_record and DNS_MX whether the domain of an email exists. I can check the emails if the domain does exist, but a wrong domain like: mail@verynonexistingdomain.com throws in the notice undefined offset 0 so does an empty input. <html> <center> <br> email input :<br> <form action="#" method = "post"> <input type ="text" name ="mail"> <br> <input type="submit" name = "click" value ="zend"> </form> <?php $domain =''; $data =''; var_dump($_POST['click']); var_dump($_POST['mail']); if (isset($_POST['mail'])) { echo $_POST['mail']; $input = $_POST['mail']; } // // function check_address($input){ list($user, $domain) = explode('@', $input); $data= dns_get_record($domain, DNS_MX); if($data[0]['host']==$domain&&!empty($data[0]['target'])) { return $data[0]['target']; } else { }} echo check_address($input); Echo '<br><br>'; echo $input . " : " ; if(check_address($input) ) { echo('<br>This MX records exists; I will accept this email as valid.<br> '); } else { echo('<br>No MX record exists; Invalid email.<br>'); } echo '</center>'; ?> </html> I had it working without the form part so with a constant variable, but when I added the form it threw the error. I tried several options, but cant find the logic behind executing a good input without throwing an error and a bad domain giving this notice. Anyone any idea?
  22. Hello everyone. I tried to install and configure PHP and MySQL for use on my own website using XAMPP. However, the Apache module gives me an error each time I try to start it. 10:40:53 [main] Starting Check-Timer 10:40:53 [main] Control Panel Ready 10:41:07 [Apache] Problem detected! 10:41:07 [Apache] Port 80 in use by "Unable to open process" with PID 4! 10:41:07 [Apache] Apache WILL NOT start without the configured ports free! 10:41:07 [Apache] You need to uninstall/disable/reconfigure the blocking application 10:41:07 [Apache] or reconfigure Apache and the Control Panel to listen on a different port 10:41:07 [Apache] Attempting to start Apache app... 10:41:07 [Apache] Status change detected: running 10:41:10 [Apache] Status change detected: stopped 10:41:10 [Apache] Error: Apache shutdown unexpectedly. 10:41:10 [Apache] This may be due to a blocked port, missing dependencies, 10:41:10 [Apache] improper privileges, a crash, or a shutdown by another method. 10:41:10 [Apache] Press the Logs button to view error logs and check 10:41:10 [Apache] the Windows Event Viewer for more clues 10:41:10 [Apache] If you need more help, copy and post this 10:41:10 [Apache] entire log window on the forums As the log said, I have to find and uninstall, disable or reconfigure the application which is blocking Port 80 or reconfigure Apache and the Control Panel to listen to a different port. The question is: What should I do? Should I change port or should I find the blocking application? And if the latter, which application could be blocking Port 80?
  23. I have a thing I dont get. I found this regex part to check for correct subdomains. function check_subdomain($data){ if (!empty($data)){ $data = preg_match("/^(?:[A-Za-z0-9][A-Za-z0-9\-]{0,61}[A-Za-z0-9]|[A-Za-z0-9])$/", $data); return($data); } else { } } echo check_subdomain($data); but the strange thing is that echo check_subdomain($data) outputs the input and adds a 0 when wrong and a 1 when correct. why is this done and how to avoid it?
  24. Hello. Ive recently joined this forum. a week or so ago I found the W3CSS pages & they are great. http://www.w3schools.com/w3css/default.asp Ive found the HTML / CSS code for pagination, & i thinki want to use it, however I am a bit puzzled. http://www.w3schools.com/w3css/w3css_pagination.asp example: <li><a href="#">2</a></li> <li><a href="#">3</a></li> can anyone advise what else is needed to use this system ? How does my website KNOW if i clicked page 1, page 2, or page 3 ? all the targets point to a hash-tag # Basically my main site runs on PHP, so i would appreciate any assistance in understanding a) how the script / website knows what value is selected how to pass this value to a php script as named value. - Eg $page Also - it would be good if the pagination page be updated to point to a few tutorials. Thank you.
  25. 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
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