php exec() & java send parameters

Question

I want to connect to java via php and print the result out in php. Now I want to send a parameter to java and return it to php to see how it works. How can I do this? I have this code so far:

PHP:

exec("java jar/name.jar", $output);
print_r($output);

Java:

public class Main {
    public static void main(String[] args) {
        System.out.print(args[0]);
    }
}

The result is

Array ( )

eis · Accepted Answer · 2012-12-16 16:31:34Z

10

Do

exec("/full/path/to/java -jar jar/name.jar 2>&1", $output);
print_r($output);

because

your PHP env probably doesn't have java executable defined in the path (and it shouldn't, either)
jar files are executed with java -jar
you'll want 2>&1 to be able to see error messages in the output (redirect from stderr to stdout)

answered Dec 16, 2012 at 16:31

eis

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2 Comments

duckduckgo Over a year ago

third point was very useful without it some program do not $output anything, dont know why !

eis Over a year ago

@AbhishekK because it prints the errors on stderr, and normally only stdout is captured, where normal output goes

Richard Deurwaarder · Accepted Answer · 2012-12-16 15:02:42Z

1

Your java program expects atleast one argument, but the php code doesn't give it, so change your

exec('java jar/name.jar', $output);

to

exec('java jar/name.jar some_value', $output);

Edit:

The reason why you got Array ( ) when you did not specify an argument is probably because the error's send by java are send to stderr, and $output only gets data from stdout

answered Dec 16, 2012 at 15:02

Richard Deurwaarder

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4 Comments

bw2801 Over a year ago

I did what you said but the array is also empty.

Richard Deurwaarder Over a year ago

@bw2801 are you sure you got the java jar/name.jar part correct? it would mean that relative to your php file the file name.jar is in the map jar. The way I tested it was: create test.php and Main.java in the same directory with your code. run javac Main.java and run php test.php (test.php had exec('java Main some_value', $output);

bw2801 Over a year ago

the folder "jar" is in the same directory as the php file.

Richard Deurwaarder Over a year ago

You can try to use php -a, this will open php in interactive mode, then type exec('java jar/name.jar some_value', $output); perhaps it'll give you some more information

Ray Paseur · Accepted Answer · 2012-12-16 15:01:19Z

0

Consider using a RESTful design. Put the Java program at a URL and read from the URL with PHP *file_get_contents()*. Anything the Java program writes to the browser output stream will be visible in the PHP variable that is loaded by that function.

answered Dec 16, 2012 at 15:01

Ray Paseur

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Comments

MatsLindh · Accepted Answer · 2012-12-16 15:05:18Z

0

If you really want to integrate native Java code into your PHP application, the PHP/Java Bridge may be able to solve that in a better way than having to go through exec()-ing java code.

Creating a small service for your Java functionality is however the preferred way to go, then exposing those features through a small web based API (with JSON as the exchange format preferably).

answered Dec 16, 2012 at 15:05

MatsLindh

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