problem in displaying multiple images

[swethak]

Hi,.

when i display the image by using below code i got the error as image displayed in the form of garbage value format.plz tell that what's the mistake in my code.

[PHP]
<?php
mysql_connect(" localhost", "root", "root") or die(mysql_error ());
mysql_select_db ("rainlist") ;
$result = mysql_query(spr intf("SELECT * from pix WHERE pid = 1", $_GET['pid']));
echo "SELECT * from images WHERE pid=".$_GET['id'];
$row = mysql_fetch_arr ay($result);
header("Content-type: image/jpeg");
$image=$row['imgdata'];

echo $image;
?>
[/PHP]

[swethak]

problem in displaying multiple images

hi,

i wrote a code to display multiple images.But it display only first image.plz tell that what's the problem in my code.

[code=php]
<?php
header("Content-type: image/jpeg");
if (! mysql_connect(" localhost","roo t","root")) {
$errmsg = "Cannot connect to database";
}
mysql_select_db ("rainlist") ;
$gotten = mysql_query("se lect * from pix");
while($row = mysql_fetch_arr ay($gotten)) {
echo $row['imgdata'];
}
?>
[/code]

[Markus]

I've never had this problem, as I've never been in this situation, but my guess is: because you're using Content-type: image/jpeg you'll only be able to display one image.

[Markus]

You've posted this again

Don't double post.

[Atli]

You've posted this again

Don't double post.

The two posts are indeed pretty much the same. As I see it, they describe two different attempts to fix the same problem.
I have merged them into this one.

Please try to keep all posts regarding the same problem in the same thread. Even if your attempts to fix the original problem are causing completely different errors, just post them in the same thread so it doesn't get scattered all over the place.

[Atli]

The reason why the code in the first post is generating garbage data is that you are printing data before you tell the browser what type of data to expect (that is, you are printing the query before you set the content-type and print the data). When you do this, the browser will automatically expect HTML data, which will display your image data as random hex characters.

For example, this will cause the output you described:
[code=php]
echo "Hello!";
header("Content-Type: image/jpeg");
readfile("somei mage.jpg");
[/code]
While this, would show an image:
[code=php]
//echo "Hello!";
header("Content-Type: image/jpeg");
readfile("somei mage.jpg");
[/code]
Note, that the only change is that I commented out the echo statement.