Question

Hi I am attempting to execute a jar from PHP 5

Here is the code I am executing:

$command = '"java -jar WEB-INF\\lib\\FileReceiver.jar '.$address.' '.$service_port.' \"'.$command.'\" \"'.$filePath.'\""';
exec($command, $out);

But the jar is not being executed.
I had logged the $command variable in firebug and took the output and inserted into the code such as:

exec("java -jar WEB-INF\lib\FileReceiver.jar 127.0.1.1 2018 \"docs/document.txt \" \"C:\\apache-tomcat-7.0.26\\webapps\\test\\downloads\\doument.txt\"", $out);

gives the correct output. I dont understand why hard coding it works but the variable which contains the same information does not.

Could someone help me out?

Thanks

Answer 1

try like this

$command = escapeshellarg($address).' '.escapeshellarg($service_port).' '.escapeshellarg($command).' '.escapeshellarg($filePath).' ';

exec('java -jar WEB-INF\\lib\\FileReceiver.jar '.$command, $out);