lbaxterl Posted March 22, 2009 Share Posted March 22, 2009 Hi Ive just came across this syntax error and have no idea on how to go about fixing it hope someone can help? Im using Php myadmin # Server version: 5.0.45 # Protocol version: 10 Here is the error: "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ') VALUES('test my a', 'img/photo/tab_bg.gif', 'img/thumb/tab_bg.gif', '')' at line 1" <?php require 'config.php'; require 'functions.php'; require 'common.php'; if(isset($_FILES['fupload'])) { $filename = addslashes($_FILES['fupload']['name']); $source = $_FILES['fupload']['tmp_name']; $target = $path_to_image_directory . $filename; $description = $_POST['description']; $category = addslashes($_POST['category']); $source = $path_to_image_directory . $filename; $thumb = $path_to_thumbs_directory . $filename; // Validates the form input if(strlen($_POST['description']) < 4) $error['description'] = '<p class="alert">Please enter a description for your photo. </p>'; if($filename == '' || !preg_match('/[.](jpg)|(gif)|(png)|(jpeg)$/', $filename)) $error['no_file'] = '<p class="alert">Please select an image, dummy! </p>'; if(!$error) { move_uploaded_file($source, $target); $q = "INSERT INTO photo (description, source, thumb, category,) VALUES('$description', '$source', '$thumb', '$category')"; $result = $mysqli->query($q) or die(mysqli_error($mysqli)); if($result) { echo "Success! Your file has been uploaded"; } createThumbnail($filename); } // end preg_match } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html> <head> <meta http-equiv="content-type" content="text/html; charset=UTF-8" /> <link rel="stylesheet" href="css/default.css" /> <title>My Photos</title> </head> <body> <h1>My Photos</h1> <ul><?php getPhotos(); ?></ul> <h1>Upload a Photo</h1> <form enctype="multipart/form-data" method="post" action="admin.php"> <input type="hidden" name="MAX_FILE_SIZE" value="30000" /> <p><input type="file" name="fupload" /></p> <p><label for="description">Enter a Description: </label> <textarea rows="6" cols="50" id="description" name="description"></textarea></p> Please choose a category: <select name="select"> <option>Sports</option> <option>London</option> <option>Macro</option> <option>Landscapes</option> <option>Local</option> </select> <p><input type="submit" value="Upload Photo" name="submit" /></p> </form> Link to comment https://forums.phpfreaks.com/topic/150617-syntax-error-please-help/ Share on other sites More sharing options...
Maq Posted March 22, 2009 Share Posted March 22, 2009 Echo out your query to see what's in it: $q = "INSERT INTO photo (description, source, thumb, category,) VALUES('$description', '$source', '$thumb', '$category')"; echo $q; //echo it out to see what's actually in it Link to comment https://forums.phpfreaks.com/topic/150617-syntax-error-please-help/#findComment-791173 Share on other sites More sharing options...
trq Posted March 22, 2009 Share Posted March 22, 2009 This is an sql syntax error, not php, you have an extra comma after category. Should be... $q = "INSERT INTO photo (description, source, thumb, category) VALUES ('$description', '$source', '$thumb', '$category')"; Link to comment https://forums.phpfreaks.com/topic/150617-syntax-error-please-help/#findComment-791175 Share on other sites More sharing options...
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