how to install .jar file onto zend server CE and call it from a php script

Question

I have the zend javabridge working. Now I want to call a .jar file myWebTest.jar with an class of the same name. Then i want to call a function testWeb which returns a hello world string.

This in on 32bit Win 7 professional system

I added myWebTest.jar to the classpath located in /zend/zendserver/etc/java_bridge_server.ini:

[JAVA_BRIDGE_SERVER]
  CLASSPATH="C:\Program Files\Zend\ZendServer\bin\javamw.jar;C:\Program Files\Zend\ZendServer\bin\myWebTest.jar;."

This path is correct.

I have restarted zend server and the code I am trying is:

$jObj = new Java("myWebTest");

// Print date through the object
print $jObj->testWeb("jim");

The log:

1 {main}
  thrown in C:\Program Files\Zend\Apache2\htdocs\javaObject.php on line 4
[08-Dec-2011 11:44:48] PHP Fatal error:  Call to a member function testWeb() on a non-object in C:\Program Files\Zend\Apache2\htdocs\javaObject.php on line 6
[08-Dec-2011 11:45:51] PHP Fatal error:  Uncaught exception 'JavaException' with message 'Java Exception java.lang.ClassNotFoundException: myWebTest
java.lang.ClassNotFoundException: myWebTest
    at java.net.URLClassLoader$1.run(Unknown Source)
    at java.security.AccessController.doPrivileged(Native Method)
    at java.net.URLClassLoader.findClass(Unknown Source)
    at java.lang.ClassLoader.loadClass(Unknown Source)
    at sun.misc.Launcher$AppClassLoader.loadClass(Unknown Source)
    at java.lang.ClassLoader.loadClass(Unknown Source)
    at java.lang.Class.forName0(Native Method)
    at java.lang.Class.forName(Unknown Source)

in C:\Program Files\Zend\Apache2\htdocs\javaObject.php:4
Stack trace:
  #0 C:\Program Files\Zend\Apache2\htdocs\javaObject.php(4): *No Class!*->jbridge('myWebTest')
  #1 {main}
thrown in C:\Program Files\Zend\Apache2\htdocs\javaObject.php on line 4

I can't find a how to on this anywhere. TIA Jim

Answer 1

From the documentation of Java

object java(string $class_name [ , ... ])

Parameters
class_name: Class name to create
       ...: Additional arguments are treated as constructor parameters

Return Value: The Java object that was created, NULL otherwise 

Not to be confused, actually class name needs to be the full qualified Name of the class. As an example, if you have a class as below:

package experiment;

public class Test {
    ....
}

Here class name is: Test Fully qualified name: experiment.Test

So to instantiate this class in PHP, you need to write:

$test = new Java("experiment.Test");

---

Additionally according to your exception log

Java Exception java.lang.ClassNotFoundException: myWebTest

It shows that, it can not find any class named as myWebTest in your class path. Which means either there is no class named myWebTest in the jar file, or jar is not properly loaded.

Answer 2

The argument to Java() is the fully qualified class name of the class you want to access, not the name of the jar file. For example the php java bridge docs say,

java("java.lang.Long")->MAX_VALUE

This access the class Long, in the package java.lang

There is some info on java packages and class names here.