PHP: 'java' is not recognized as an internal or external command

Question

I'm trying to run a .jar application from PHP using exec:

exec('java -jar parser.jar $inputstring 2>&1', $output);

I can get it to run on cmd, and java -version returns so I have java installed. It seems like this is a path issue, but I am at my wits end.

I've tried specifying the full path, adding C:\Program Files (x86)\Java\jdk1.8.0_91\bin; to System Variables, using shell_exec, no avail. I get the same error.

Any help would be greatly appreciated.

Show how you added the full path. there's spaces in there. if you didn't quote the path, e.g. shell_exec('"C:\Program files\...."), then you'll be attempting to run a program named c:\program.exe, with argument files\.... — Marc B
– Marc B, Commented Jul 8, 2016 at 15:06
I have: shell_exec('"C:\Program Files (x86)\Java\jdk1.8.0_91\bin\java.exe" -jar parser.jar $inputstring 2>&1'); If I don't have 2>&1 it's blank. With it I get the same 'java' not recognized error. — Anda
– Anda, Commented Jul 8, 2016 at 15:16
2>&1 redirects stderr to stdout, which is why you get the error message. shell_exec only returns stdout, so you'd never see the error message. — Marc B
– Marc B, Commented Jul 8, 2016 at 15:19
You need to escape your backslashes. shell_exec ('"C:\\Program Files\\... — noggin182
– noggin182, Commented Jul 12, 2016 at 20:52
Try to echo the $PATH environement variable and see whether java is really in your path. Does the PHP script run under the same user as your cmd line test? — Rolf
– Rolf, Commented Jul 19, 2016 at 7:40

Devraj Gupta · Accepted Answer · 2016-07-27 10:37:03Z

1

Use this....

exec('java -jar parser.jar '.$inputstring.'2>&1', $output);
or
shell_exec("java -jar parser.jar $inputstring 2>&1 $output");
or
string exec ( 'java -jar parser.jar'  [, array &$output [, int &$return_var ]] )

answered Jul 27, 2016 at 10:37

Devraj Gupta

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