> Can someone tell me how to turn this line (which works):
>
> open(LABEL,">/path/to/the/file.gif");
>
> into something like this:
>
> open(LABEL,">/path/to/the/$file.gif");
>
> The gif file should be named according to the value of the $file
> variable. I never know what to do with quotes in these situations.
>
Hope this is what you are searching for.
$file="/path/to/the/file.gif";
open(LABEL,">$file");
I don't think so, unless I'm misreading your code. I want the .gif
file to be named according to the current value of the $file variable.
- Grant
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