I'm trying to execute a linux command through a PHP command-line script, which is no problem using the exec command.
The problem is, the command I am executing (mysqldump) outputs an error message if something is wrong (for example user/password is incorrect). I can't seem to be able to capture this error in order to log it. It just prints this error to the screen.
How do I cause this error not to be printed to the screen, but instead to put it in a variable for use in my script?
Thanks!
Use popen to run the process. The example #2 on this page shows exactly what you're looking for:
<?php
error_reporting(E_ALL);
/* Add redirection so we can get stderr. */
$handle = popen('/path/to/spooge 2>&1', 'r');
echo "'$handle'; " . gettype($handle) . "\n";
$read = fread($handle, 2096);
echo $read;
pclose($handle);
?>
exec("mysqldump -u user -p passwod database > outputfile.sql 2> error.log");
You need to redirect stderr to stdout, so you can capture it. This example routes stdout to devnull (thus ignoreing it) and routes stderr to you:
exec('ls * 2>&1 1>/dev/null');
I'm not too hot in Unix (New Years Resolution...) but these functions look helpful:
shell_exec - returns result as a string.passthru - It looks like you can execute this like: passthru('command', $result); and then use $result.tried using backticks?
$var = `command`;
The following will route stderr messages to the same place as the normal output.
exec("mysql_dump blah 2>&1",$output,$return_val)
if($return_val !== 0)
echo "there was an error"
2>&1 means re-route stderr messages to the same place as stdout, and thus will be loaded into the output array.
Have you looked at the system() command? It's been a while since I did any PHP, but that rings a bell.
