php-java-bridge-users Mailing List for PHP/Java Bridge

If I have one JAR-file with some packages and classes I can access all classes without any problems:
$myObj = new java("newweb.farben");//newweb=package;farben=class

but if I puck a JAR in another JAR-file the classes are not found anymore. ....





      

As long as you notify the Java application server about your changes, for
example by re-deploying your changes, this will work, yes.

Apr 27, 2009 10:41 vorm. schrieb am <
php...@li...>:

Hi Jost,
thank you for your answer.
1) Is it absolutely impossible to do it on the way I did it? I added the
JAR-file (with many classes) to WEB-INF/lib and tryed to access the class?
If not, sorry for the stupid question,but
2)"You need to create a library containing your classes and add it to Java's

 repository, along with all dependencies."Do you mean I should create the
folder "classes" with all jars (I already have lib for it and the java
code/classes/ I have in eclipse). Could you explain it please step by step?
Many thanks in advance!!!





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Hi Jost,
thank you for your answer. 
1) Is it absolutely impossible to do it on the way I did it? I added the JAR-file (with many classes) to WEB-INF/lib and tryed to access the class? If not, sorry for the stupid question,but
2)"You need to create a library containing your classes and add it to Java's

 repository, along with all dependencies."Do you mean I should create the folder "classes" with all jars (I already have lib for it and the java code/classes/ I have in eclipse). Could you explain it please step by step?
Many thanks in advance!!!

 


      

Hi,

You need to create a library containing your classes and add it to Java's
repository, along with all dependencies.

Or you can add your classes and all their dependencies to your
JavaBridge.war zip file and re-deploy that. You can find a video at the
bottom of our install section on our web site, which shows that.

Regards,
Jost Boekemeier

Apr 27, 2009 8:24 vorm. schrieb am <
php...@li...>:

Hello Jost,
the underlined part of the code does not work. Is it the right way to find a
class in the jar?

<?php

require_once("http://localhost:8081/Ordner/java/Java.inc");

$System = java("java.lang.System");



$myObj = new java("WebProjekt2.web.FacettenBrowser");

echo (String) $myObj->HalloWelt();

?>
I get such an errormessage:
root cause java.lang.RuntimeException: PHP Fatal error:  Uncaught
[[o:Exception]:"java.lang.Exception: CreateInstance failed: new
WebProjekt2.web.FacettenBrowser. Cause: java.lang.ClassNotFoundException:
WebProjekt2.web.FacettenBrowser VM: 1.6.0_10@http://java.sun.com/" at:
#-11
org.apache.catalina.loader.WebappClassLoader.loadClass(WebappClassLoader.java:1387)
#-10
org.apache.catalina.loader.WebappClassLoader.loadClass(WebappClassLoader.java:1233)
#-9 java.lang.ClassLoader.loadClassInternal(ClassLoader.java:320)
#-8 java.lang.Class.forName0(Native Method)
#-7 java.lang.Class.forName(Class.java:247)
#-6
php.java.bridge.SimpleJavaBridgeClassLoader.forName(SimpleJavaBridgeClassLoader.java:157)
#-5 php.java.bridge.JavaBridge.CreateObject(JavaBridge.java:455)
#-4 php.java.bridge.Request.handleRequest(Request.java:452)
#-3 php.java.bridge.Request.handleRequests(Request.java:490)
#-2 php.java.bridge.http.ContextRunner.run(ContextRunner.java:146)
#-1 php.java.bridge.ThreadPool$Delegate.run(ThreadPool.java:60)
#0 http://localhost:8081/Ordner/java/Java.inc(248): java_ThrowE in
http://localhost:8081/Ordner/java/Java.inc on line 215


 php.java.servlet.fastcgi.FastCGIServlet$CGIRunner.parseBody(FastCGIServlet.java:484)

 php.java.servlet.fastcgi.FastCGIServlet$CGIRunner.doExecute(FastCGIServlet.java:379)

 php.java.servlet.fastcgi.FastCGIServlet$CGIRunner.execute(FastCGIServlet.java:387)
       php.java.servlet.CGIServlet.handle(CGIServlet.java:399)
       php.java.servlet.PhpCGIServlet.handle(PhpCGIServlet.java:382)
       php.java.servlet.CGIServlet.doGet(CGIServlet.java:469)
       javax.servlet.http.HttpServlet.service(HttpServlet.java:617)
       javax.servlet.http.HttpServlet.service(HttpServlet.java:717)
note The full stack trace of the root cause is available in the Apache
Tomcat/6.0.18 logs.
you have to put the class "HelloWorld" into the "HalloWelt.web" package.

so it looks like
[code]
package HalloWelt.web;

import ....;

public class HelloWorld{

 public HelloWorld() { /* do something */ }
}
[/code]
2009/4/24 <php...@li...>

> Hello,
> in my jar file there are many classes, and the class which I need to
access
> can't be found.
> I tryed to solve it so:
> $myObj = new java("HalloWelt.web.HelloWorld");
> but it does not work.Please tell me  how can I access this class?
>
>
>
>  Thank you in advance!
>
>



------------------------------------------------------------------------------
Crystal Reports &#45; New Free Runtime and 30 Day Trial
Check out the new simplified licensign option that enables unlimited
royalty&#45;free distribution of the report engine for externally facing
server and web deployment.
http://p.sf.net/sfu/businessobjects
_______________________________________________
php-java-bridge-users mailing list
php...@li...
https://lists.sourceforge.net/lists/listinfo/php-java-bridge-users

Hello Jost,
the underlined part of the code does not work. Is it the right way to find a class in the jar?

<?php 

require_once("http://localhost:8081/Ordner/java/Java.inc");

$System = java("java.lang.System");

 

$myObj = new java("WebProjekt2.web.FacettenBrowser");

echo (String) $myObj->HalloWelt();

?>
I get such an errormessage:
root cause java.lang.RuntimeException: PHP Fatal error:  Uncaught [[o:Exception]:"java.lang.Exception: CreateInstance failed: new WebProjekt2.web.FacettenBrowser. Cause: java.lang.ClassNotFoundException: WebProjekt2.web.FacettenBrowser VM: 1.6.0_10@http://java.sun.com/" at:
#-11 org.apache.catalina.loader.WebappClassLoader.loadClass(WebappClassLoader.java:1387)
#-10 org.apache.catalina.loader.WebappClassLoader.loadClass(WebappClassLoader.java:1233)
#-9 java.lang.ClassLoader.loadClassInternal(ClassLoader.java:320)
#-8 java.lang.Class.forName0(Native Method)
#-7 java.lang.Class.forName(Class.java:247)
#-6 php.java.bridge.SimpleJavaBridgeClassLoader.forName(SimpleJavaBridgeClassLoader.java:157)
#-5 php.java.bridge.JavaBridge.CreateObject(JavaBridge.java:455)
#-4 php.java.bridge.Request.handleRequest(Request.java:452)
#-3 php.java.bridge.Request.handleRequests(Request.java:490)
#-2 php.java.bridge.http.ContextRunner.run(ContextRunner.java:146)
#-1 php.java.bridge.ThreadPool$Delegate.run(ThreadPool.java:60)
#0 http://localhost:8081/Ordner/java/Java.inc(248): java_ThrowE in http://localhost:8081/Ordner/java/Java.inc on line 215

	php.java.servlet.fastcgi.FastCGIServlet$CGIRunner.parseBody(FastCGIServlet.java:484)
	php.java.servlet.fastcgi.FastCGIServlet$CGIRunner.doExecute(FastCGIServlet.java:379)
	php.java.servlet.fastcgi.FastCGIServlet$CGIRunner.execute(FastCGIServlet.java:387)
	php.java.servlet.CGIServlet.handle(CGIServlet.java:399)
	php.java.servlet.PhpCGIServlet.handle(PhpCGIServlet.java:382)
	php.java.servlet.CGIServlet.doGet(CGIServlet.java:469)
	javax.servlet.http.HttpServlet.service(HttpServlet.java:617)
	javax.servlet.http.HttpServlet.service(HttpServlet.java:717)
note The full stack trace of the root cause is available in the Apache Tomcat/6.0.18 logs.
you have to put the class "HelloWorld" into the "HalloWelt.web" package.

so it looks like
[code]
package HalloWelt.web;

import ....;

public class HelloWorld{

 public HelloWorld() { /* do something */ }
}
[/code]
2009/4/24 <php...@li...>

> Hello,
> in my jar file there are many classes, and the class which I need to access
> can't be found.
> I tryed to solve it so:
> $myObj = new java("HalloWelt.web.HelloWorld");
> but it does not work.Please tell me  how can I access this class?
>
>
>
>  Thank you in advance!
>
>