Shell_exec with path, command option and variable

Question

I'm new to php. In my php file, I have this command :

shell_exec("\"C:\path_with_spaces_to_.exe\" -c -n 1 C:\path_to_image\$image_name 2>&1");

Weird thing is my $image_name is never getting read. I believe maybe I miss some fundamental things. I tried many ways, but It's been 3 hours here and still stuck.

Any suggestion about what mistake do I have here ?

B. Desai · Accepted Answer · 2017-07-19 04:42:15Z

1

Its because you have to escape \ other wise it will escape $. Try this solution:

echo("\"C:\path_with_spaces_to_.exe\" -c -n 1 C:\path_to_image\\$image_name 2>&1");

answered Jul 19, 2017 at 4:42

B. Desai

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Lyn Over a year ago

For other people, who had the same problem, there is another solution I found. But, @B. Desai answer still neater. Here is another solution for it using dot (.): shell_exec('"C::\path_with_spaces_to_.exe\" -c -n 1 C:/path_to_image/' . $image_name . ' 2>&1')

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