我可以传回执行脚本的输出,但是如果脚本出错,则不会输出错误。
// This is a file that doesn't exists, for testing
$command = './path/to/non/existing/script.sh';
$commandOutput = exec($command, $commandOutput); // works but no error output
//passthru($command, $commandOutput); // works but error output was 127 not file not found
//$commandOutput = escapeshellcmd($command);
echo "The Output:\n|".$commandOutput."|\n";
var_dump($commandOutput);
The Output:
||
我想要错误消息的输出:
The Output:
|file not found|
如何或什么功能/参数将执行此操作?
最佳答案
您可以将stderr重定向到stdout,以便exec()等通过将2>&1
附加到命令来获取错误消息。
见http://tldp.org/HOWTO/Bash-Prog-Intro-HOWTO-3.html