Unexpected T_VARIABLE in image database script

[Philth]

No doubt this has been covered lots in detail. I think I've almost got is sorted but for this one error.

"Parse error: syntax error, unexpected T_VARIABLE in /home/fhlinux162/g/greatestalbumsever.com/user/htdocs/php/image.php on line 11"

Usually something to do with a bracket missing or something, but I can't see what it is.

My code is as follows. $albumNumber is being passed from via a html link to this script. "content" is the row in my table storing the image data

<?
if(isset($_GET['albumNumber'])) 
{
mysql_connect("__MUNGED__","__MUNGED__","__MUNGED__"); 
mysql_select_db("freelance"); 
 
$albumNumber = $_GET['albumNumber'];
$query = "SELECT content FROM Albums WHERE albumNumber = '$albumNumber'";
 
$result = mysql_query($query)
$content=$r["content"];
 
header("Content-Type: image/jpeg"); 
echo $content;
}
?>
 
 

Any suggestions would be most welcome. Thanks!

[Markus]

Use code tags please. Makes things easier.

DONT PUT YOUR DETAILS UP EITHER! Anyone can get into your mysql now!

Change your password asap.

[pbmods]

Heya, Philth.

Changed thread title to better describe the problem (did you know that threads whose titles do not follow the Posting Guidelines actually get FEWER responses?).

Please use CODE tags when posting source code:

[CODE=php]
PHP code goes here.
[/CODE]

You're missing a semicolon.

[Markus]

<?
if(isset($_GET['albumNumber'])) 
{
mysql_connect("__MUNGED__","__MUNGED__","__MUNGED__"); 
mysql_select_db("freelance"); 
 
$albumNumber = $_GET['albumNumber'];
$query = "SELECT content FROM Albums WHERE albumNumber = '$albumNumber'";
 
$result = mysql_query($query) // <<<< missing ';' as pbmods said.
$content=$r["content"];
 
header("Content-Type: image/jpeg"); 
echo $content;
}
?>
 
 

[Philth]

Cheers guys!

Obvously wasn't paying too much attention when posting this - thanks for the help.

[pbmods]

Heya, Philth.

Glad to hear you got it working! Good luck with your project, and if you ever need anything, post back anytime :)

[Philth]

Well, couldn't check last night, but it seems to have dealt with the error issue, now it doesn't actually do anything - no error, no action - don't know which is worse.

I'm using the following as my html image script...

<img src='php/image.php?albumNumber=$albumNumber'/>

Which is successfully picking up the album number variable.

The image date is definately in the database, it just doesn't want to pick it up.