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Stupid question: php, variables and misplaced quote marks..

S

scoob8000

2[H]4U
Joined
May 4, 2002
Messages
2,834
Very tired and I'm not thinkinging right..

I know I'll be embarassed when I see the answer..

Code: 
shell_exec('convert -background white -fill black -font arial -pointsize 32 label:$output sig.jpg');

What am I missing with my $output that the content of that variable is being passed but rather "$output" itself...
 
You're using single quotes. The variable won't be "interpreted" as a variable unless you use double quotes.

Best yet, make the variable obvious to other readers by usings a sprintf() or concatenating the string.
 
scoob8000 said:
Very tired and I'm not thinkinging right..

I know I'll be embarassed when I see the answer..

Code: 
shell_exec('convert -background white -fill black -font arial -pointsize 32 label:$output sig.jpg');

What am I missing with my $output that the content of that variable is being passed but rather "$output" itself...
Click to expand...

use double quotes or if you must use single,

Code: 
shell_exec('yourtexthere' . $output . 'yourtexthere');
 
I finally got it..

Too stupid of a mistake to admit. It's alwyas those little things that trip you up.

Thanks for the replies!
 
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