PHP: cannot run nested shell scripts

Question

I hope you can help me.

I am dealing with a problem that I cannot solve. This is my issue. I am trying to exec a bash script through PHP. I tried with the method

exec()

with 3 arguments, arg1, arg2, and arg3.

php code

<?php exec("./randomScript.sh arg1 arg2 arg3"); ?>

randomScript.sh

.... # random code which exploits the three arguments -> executed normally..
.... # random code which exploits the three arguments -> executed normally..

./secondScript.sh $arg1 $arg2 $arg3 #<- this is the script that is not running (is not even started).

I have tried to change the permission (I've got full permission), change the way I call the randomScript.sh (through absolute path), but nothing occurred. Besides, I tried with:

shell_exec()

method, but nothing changed. Of course, if I run the secondScript.sh via terminal everything works fine.

I cannot understand which is the problem. Could you please help me?

Thanks in advance.

See also https://stackoverflow.com/questions/12199353/how-can-i-debug-exec-problems and https://tldp.org/LDP/Bash-Beginners-Guide/html/sect_02_03.html — mario, Oct 02 '20 at 08:20
I do not see anywhere in your examples for you to even try to capture the output? How are you debugging this? How do you know you are at the stage to call the `exec` command? you can use `var_dump`, `print_r`, save even to a variable and try to output it, none of that is evident.. You can as well use `Output Buffering` to capture what is happening... no wonder you do not know why it is not working... Learn to Debug.. lowers the amount of headaches ;) — Ron, Oct 02 '20 at 10:44

Pascal Tovohery · Answer 1 · 2020-10-02T10:37:50.610

0

You should escape your string command before passing it in exec function, i think it may help you Escapeshellcmd()

...
$escaped_command = escapeshellcmd("./randomScript.sh arg1 arg2 arg3");
exec($escaped_command);
...

Escapeshellarg()

For system command, i recommand using system

...
$escaped_command = escapeshellarg("./randomScript.sh arg1 arg2 arg3");
system($escaped_command);
...

edited Oct 02 '20 at 10:37

answered Oct 02 '20 at 10:31

Pascal Tovohery

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score 0 · Answer 2 · answered Oct 02 '20 at 10:33

In the inner shell, the arguments are not called like that (because it is invoked by the shell, not by PHP).

./secondScript.sh "$1" "$2" "$3"

or just

./secondScript.sh $*

At this stage, just remember that spaces, quotes, and dollar signs in the arguments you pass must be avoided at all costs (the solution is escaping them, but how to do it exactly is tricky).

You also might want to do instead:

$ret = shell_exec("./randomScript.sh 'arg1' 'arg2' 'arg3' 2>&1");

so that in $ret you might find error messages and output from the inner shell.

score 0 · Answer 3 · answered Oct 02 '20 at 12:09

You also need to make sure that your PHP code does not change working dir, and both shell scripts have execute permissions, and filesystems allows to exec files on it.

I would avoid exec("./script $arg ...");, I would rather specify the interpreter to use and full path to the script like exec("sh /home/user/project/script.sh $arg ...");

PHP: cannot run nested shell scripts

3 Answers3