Php mysql image link

I have a mysql table that has id, name, type, size, content, and link. I have pages download.php and showimage.php. I want the link to affect the image so when the user clicks it sends them their. I only created showimage.php cause I had ran out of options on how to link and was hoping i could extract the link somehow from download.php. Heres the code for both.

Download.php

<?php
include 'config3.php';
include 'opendb.php';

$query = "SELECT name, type, size, content, link FROM `upload` ORDER BY RAND() LIMIT 1";

$result = mysql_query($query) or die('Error, query failed'); 
while($row= mysql_fetch_array($result)){
$content = $row['content']; 
$size= $row['size']; 
$type= $row['type'];
 $link= $row['link'];
header('Content-type: $type');
echo $content;
}
include 'closedb.php'; 
?>

Showimage.php

<html>
<head>
<title>Image Test</title>
</head>
 
<body>
<h1>Displaying image from database</h1>
 
<a href="download.php"><img src="download.php" height="250" width="250"/></a> </body>
 
</html>

Download.php code above only shows the image. I've tried playing with download.php to echo "<a href='".$link."'/><img src='".$content."' /></a>"; but when i do this the image turns into a huge jarbled mess. Any suggestions?

RazorRamon

Junior Poster in Training

74 posts since Sep 2010

Reputation Points: 10

Solved Threads: 0

4 Months Ago

Here this will help. Its a good place for tutorials.

http://cookbooks.adobe.com/post_Display_an_image_stored_in_a_database__PHP_-16637.html

fobos

Posting Whiz in Training

297 posts since Feb 2009

Reputation Points: 29

Solved Threads: 52

4 Months Ago

The tutorial you gave shows how to present the image.. which I'm already doing with the code above.. My problem is that the table also has a column for links. When the user clicks on the picture i want it to send the user to the appropriate address.

RazorRamon

Junior Poster in Training

74 posts since Sep 2010

Reputation Points: 10

Solved Threads: 0

4 Months Ago

Is the image stored as blob in your DB or is it a filepath?

Uploading image to blob: http://www.phpriot.com/articles/images-in-mysql/7
Extracting image and displaying it: http://www.phpriot.com/articles/images-in-mysql/8

(I don't like the use of mysql_result though)

diafol

Rhod Gilbert Fan (ardav)

Moderator

7,800 posts since Oct 2006

Reputation Points: 1,170

Solved Threads: 1,080

4 Months Ago

I figured it out.. after banging my head on the keyboard for 12 hrs... lol. I'll post the code for the sake of all the broken computer equipment out there. Here's it is.

Showimage.php

<?php
include 'config3.php';
include 'opendb.php';

$query = "SELECT id, name, type, size, content, link " .
         "FROM  `upload` ORDER BY RAND() LIMIT 1";
		 $result = mysql_query($query) or die('Error, query failed');
while($row= mysql_fetch_array($result)){
	$content = $row['content'];
		$id= $row['id'];
	$size= $row['size'];
	$type= $row['type'];
	$link= $row['link'];
	}
?>
<html>
<head>
<title>Image Test</title>
</head>
 
<body>
<h1>Displaying image from database</h1>
<a href="<?php echo $link?>" />
<img src="download.php?id=<?php echo $id?>" height="250" width="250"/>
</a>
</body>
 
</html>

Download.php

<?php
include 'config3.php';
include 'opendb.php';

if(isset($_GET['id'])){
 $id=$_GET['id']; 
}else{
echo 'did not get id'; 
}

$query = "SELECT id, name, type, size, content, link " .
         "FROM `upload` WHERE id='$id'";

$result = mysql_query($query) or die('Error, query failed');
while($row= mysql_fetch_array($result)){
	$content = $row['content'];
		$id= $row['id'];
	$size= $row['size'];
	$type= $row['type'];
	$link= $row['link'];


header('Content-type: $type'); 
echo $content;
exit();
}


include 'closedb.php'; 


?>

RazorRamon

Junior Poster in Training

74 posts since Sep 2010

Reputation Points: 10

Solved Threads: 0

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