englishcodemonkey Posted December 21, 2008 Share Posted December 21, 2008 Ok i have a database called "image" with which i am storing images in blob format.i am currently trying to display the images that are in the database in this table. this is my current code for index.php: <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <title>Browse Upload Files</title> </head> <body bgcolor="white"> <?php include 'db.inc'; $query = "SELECT id, shortName, mimeName FROM image"; /*if (!($connection = @ mysql_pconnect($hostName, $username, $password))) showerror(); if (!mysql_select_db("files", $connection)) showerror(); if (!($result = @ mysql_query ($query, $connection))) showerror();*/ ?> <h1>Image database</h1> <h3>Click <a href="insert.php">here</a> to upload an image.</h3> <?php //require 'disclaimer'; if ($row = @ mysqli_fetch_array($result)) { ?> <table> <col span="1" align="right"> <tr> <th>Short description</th> <th>File type</th> <th>Image</th> </tr> <?php do { ?> <tr> <td><?php echo "{$row["shortName"]}";?></td> <td><?php echo "{$row["mimeName"]}";?></td> <td><?php echo "<img src=\"view.php?file={$row["id"]}\">";?></td> </tr> <?php } while ($row = @ mysqli_fetch_array($result)); ?> </table> <?php } // if mysql_fetch_array() else echo "<h3>There are no images to display</h3>\n"; ?> </body> </html> db.inc: <?php // These are the DBMS credentials $hostName = "****"; $username = "****"; $password = "****"; // Show an error and stop the script function showerror() { if (mysqli_error()) die("Error " . mysqli_errno() . " : " . mysqli_error()); else die("Could not connect to the DBMS"); } // Secure the user data by escaping characters // and shortening the input string function clean($input, $maxlength) { $input = substr($input, 0, $maxlength); $input = EscapeShellCmd($input); return ($input); } ?> and view.php: <?php include 'db.inc'; $file = clean($file, 4); if (empty($file)) exit; if (!($connection = @ mysqli_pconnect($hostName, $username, $password))) showerror(); if (!mysqli_select_db("files", $connection)) showerror(); $query = "SELECT mimeType, fileContents FROM image WHERE id = $file"; if (!($result = @ mysqli_query ($query,$connection))) showerror(); $data = @ mysqli_fetch_array($result); if (!empty($data["fileContents"])) { // Output the MIME header header("Content-Type: {$data["mimeType"]}"); // Output the image echo $data["fileContents"]; } ?> when i upload this to my host and view the index.php file it says there are no images to display, but i have images there! Thanks and sorry if this is a stupid mistake, just need another pair of eyes to take a look! thanks! Link to comment https://forums.phpfreaks.com/topic/137903-calling-images-from-mysql-database-please-help/ Share on other sites More sharing options...
englishcodemonkey Posted December 21, 2008 Author Share Posted December 21, 2008 ok so i'm trying another way still with no luck, i have the "<img src="index.php" />" on one page and then the code for index.php is below: <?php $img = "P7180104.JPG"; $q = "SELECT fileContents from image where id='1'"; $r = @mysqli_query ($dbc, $q); $imagepath="$r"; $image=imagecreatefromjpeg($imagepath); header('Content-Type: image/gif'); imagejpeg($image); ?> if i just type the filepath after "$imagepath" then it will display the image, but it will not find the file path from the database. i'm thinking maybe i have performed the query incorrectly? is "where id='1'" correct? Thanks Link to comment https://forums.phpfreaks.com/topic/137903-calling-images-from-mysql-database-please-help/#findComment-720801 Share on other sites More sharing options...
Recommended Posts
Archived
This topic is now archived and is closed to further replies.