Topic: [SOLVED] Interpreting ‘$user_’ as a variable rather than ‘$user’.

[cs.punk]

Ok guys, a naming problem.

Code: [Select]

$file_dir = "photos/originals/$user_$user_id_$file_count.jpg";
move_uploaded_file ($file_tmp, $file_dir);
 
I want a file like : ‘user_23_1.jpg’…? But it seems to be interpreting  ‘$user_’ as a variable rather than ‘$user’.

Thanks.

[papaface]

Code: [Select]

$file_dir = "photos/originals/".$user."_".$user_id."_".$file_count.".jpg";

[.josh]

That's because _ is a valid char for a variable name.  Wrap your variable in braces

{$user}_...

[cs.punk]

Thanks papaface and Crayon Violent!