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I'm running the following exec method in php. 

exec("exec /bin/bash ./test.sh $file $fileY 2>&1", $output, $return);

However when I print out the $return variable I get 1 as a result which is an error code.

How do I diagnose this to find out why this error code is appearing, I've tried a number of different methods to figure it out like setting permissions for everything to 077 but none of it is working.

Can anyone advise me on what to do?

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  • What is the result of /bin/bash ./test.sh $file $fileY 2>&1 && echo $?? Commented Jun 29, 2016 at 13:58
  • Do you mean adding && echo $? to the end of the first parameter of the exec method? Commented Jun 29, 2016 at 14:01
  • No ! executing it in the terminal is what I meant Commented Jun 29, 2016 at 14:03
  • This is an [ intersting read ] for this one as you're passing arguents. Commented Jun 29, 2016 at 14:19
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