okay so heres my bash command i want to send a variable to
perl -pi -e 's/ : /:/g' /opt/lampp/htdocs/"variable needs to go here"
i then have my variable in php wich is
$filename
how do i edit my bash script to accept the variable and how do i pass the variable and execute the bash from php any help would be appreciated. i just cannot figure out how i would make the bash ready to accept a variable and how to make php send it and execute th script
In php use the exec() function to invoke the bash script like this:
$filename = escapeshellarg($filename);
exec("perl -pi -e 's/ : /:/g' /opt/lampp/htdocs/".$filename);
Try using shell_exec in your php script to execute your shell script and pass your variable, like so:
$cmd="perl -pi -e 's/ : /:/g' /opt/lampp/htdocs/" . escapeshellarg($variable);
$r=shell_exec($cmd);
escapeshellarg is used to escape any potentially dangerous characters in $variable, to prevent a command line injection attack.
There is another solution, not so elegant, but it works.
You can echo the value from the php script, and catch it in a variable in the bash script (using backticks).
Don't forget to first disable error output in the php script, to eliminate the danger of having an output with notices, warnings...
Eg.
<?php
ini_set("display_errors", "Off");
ini_set("log_errors", "On");
$filename = 'someValue';
echo $filename;
?>
#!/bin/bash
#execute the php file
FILENAME=`php file.php`
perl -pi -e 's/ : /:/g' /opt/lampp/htdocs/$FILENAME
