PHP Code:

$imagefilename = 'new';
$imagefilename = '../userimages/'.$imagefilename.'2.jpg'; 


im trying to join a variable to a string, is the code above enough to do this?

my current output just excludes the variable in between the two strings?

thanks

Is "new" part of the file name or a folder?

'new' is the old image name
currently the image is being save as 2.jpeg into /userimages
but it should be new2.jpeg

Ok
Your code should work in that case.

right ok, in that case ill look into the $_Get method too find any mistakes,

thanks for your help mate!

$_GET
PHP is case sensitive on variable names

yeah thats how ive got it,
ill just show you abit more of my code:

variables passed
code.php?thumb=thumb.jpg&filename=userimages/picnik_exported_file123618638723548.jpg

so in the php im trying to crop the filename to just picnik_exported_file123618638723548
then add 3crop.jpg to the end of it
eg. 3picnikexportedfile123618638723548crop.jpg
but the image is save as 3crop.jpg

PHP Code:

$imagefilename = $_GET['filename']; 
$imagefilename = substr($imagefilename, 11, 35); 
$imagefilename = rtrim($imagefilename); 
$imagefilename = ereg_replace("[^A-Za-z0-9]", "", $imagefilename);

 $imagefilename = '../userimages/3'.$imagefilename.'crop.jpg'; 


then further down im saving the image with:

PHP Code:

imagejpeg($dst_r, $imagefilename); 


i sorted it, should have used the POST method,
thanks

Haha, that mistake has cost me hours of frustration (over many projects).