php - get value of variables defined in shell script

Question

How can I get value of a variable defined in a shell script? (the script is a configuration file, only contains variable-definitions) I cannot use "source" command in exec:

echo exec('var=2; echo $var'); //writes "2"
echo exec('source config.sh; echo $var'); //writes ""

How can I get value of variables defined in a shell script?

Have you checked it is the correct path? Try with absolute path to see if config.sh is requested properly. — fedorqui
– fedorqui, Commented Apr 18, 2013 at 9:15
yes, I use absolute path in my code (the problem exists that way too) - and the path is correct, the command runs well in terminal — nagy.zsolt.hun
– nagy.zsolt.hun, Commented Apr 18, 2013 at 9:18
I found a related question -> stackoverflow.com/questions/10302958/… — fedorqui
– fedorqui, Commented Apr 18, 2013 at 9:22

ndasusers · Accepted Answer · 2013-04-18 13:31:47Z

2

userThis worked for me just now on the terminal:

echo var=2 > shellscript
echo "<?php echo exec('source /home/user/shellscript; echo $var'); ?>" > phpscript.php
ls | grep script
phpscript.php
shellscript
php phpscript.php 
2

So I wrote the shellscript, then a php to read the shell script and ran the php script. It did just what you wanted.

Maybe you can add the other exec components to get a clue?

<?php 
echo exec('source /home/user/shellscript; echo $var',$out,$ret); 
echo "\nout0: ". $out[0]. "\nret: $ret\n";
?>

This shows me:

~]$ php phpscript.php 
2
out0: 2
ret: 0

answered Apr 18, 2013 at 13:31

ndasusers

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