Develop Reference

How do I use a classloader to get inputstream on a model.zip file where the zip file is wrapped in a .jar file in classpath

The issue:
We have a jetty web-app, and in the application-code, I am trying to access a zip within a jar in classpath. Here's the jar in the libs folder:
/path/to/app/x.x.0-SNAPSHOT/apps/libs/my-model.jar
where my-model.jar is really just an empty folder with model.zip file inside it. If I extract this jar, I get johnsnow/mymodel.zip
My application code tries to access this zip as:
getClass().getResourceAsStream("johnsnow/mymodel.zip")
but of course, I don't get a proper handle to this resource and wind up getting a nullpointer exception. What am I doing wrong? Shouldn't I be able to access a file within a jar file in classpath using the getClass().getResourceAsStream() method?
Footnote:
Because model.zip was too large, we decided against shipping it with the code base. Thus we pushed it into a nexus repository, and reference the jar via a gradle compile dependency as follows:
compile "com.company.group.nlp:my-model:1.0#jar"
The fact that building the distribution pulls this jar, and puts it in apps/libs tells me that gradle does its part (of downloading the dependency to a classpath). The issue remains that I can't seem to find a way to access mymodel.zip inside my-model.jar

Try adding a slash to the file path:
getClass().getResourceAsStream("/johnsnow/mymodel.zip");
It will tell java to start looking for the class from the root folder, not from the current class package.

Factual API and coldfusion

I took the java implementation of the Factual API (reference http://developer.factual.com/) and made a JAR file for factual. I did this by opening a new project in eclipse with the factual java files and then exporting to a new jar file.
I put that jar file in my coldfusion installation's /WEB-INF/lib/ folder.
After restarting Coldfusion, I tried to create a new cfobject like so
<cfscript>
// Initialize the Java class.
factualClass=CreateObject("java", "src.main.java.com.factual.driver.Factual");
</cfscript>
I get an error indicating that it cannot find the Factual class.
Can anybody give me some guidance?

(Summary from comments)
It sounds like you may be exporting the source files ie *.java rather than the compiled class files, ie *.class. In the Jar Export wizard, be sure to select the "Export generated class files and resources" option. (To automatically compile the project sources before expi, enable the setting: JAR packaging > Build projects if not build automatically option). If you prefer you can also find pre-compiled jars in the MVN repository.
put that jar file in my coldfusion installation's /WEB-INF/lib/
folder.
CF10+ also supports dynamic class loading via a new application level setting THIS.javaSettings.
// Initialize the Java class.
factualClass=CreateObject("java", "src.main.java.com.factual.driver.Factual");
Just as a point of interest, src/main/java/ is not actually part of the libary class name. It is a standard directory structure used in Maven projects. It is probably included when exporting the sources, but not the compiled classes.
You can always verify the correct path and class name either by examining the API ie javadocs or by viewing one the source files. Package declarations are always at the top of the source file, such as on line 1 of src/main/java/com/factual/driver/Factual.java:
package com.factual.driver; // ie "com.factual.driver"
.. and the class declaration on line 39.
public class Factual { // ie "Factual"
Combined that gives you the exact (case-sensitive) path to use with createObject:
factualClass=CreateObject("java", "com.factual.driver.Factual");

How to read an internal file in a web project

All, Forgive me I am still learning the Java development right row.
Say we have the structure of web project like below.
The src and config is under the Java Resources folder of web project.
src
...
|- a.b.c.package
|-test.java
...
config
|-1.xml
|-2.xml
...
configfolder
|-1.properties
|-2.properties
testfolder
|-test.properties
I want to know if I use the dom4j. How to read the xml file(1.xml) within the test.java. Thanks.
My Experiment
In the test.java. I found I can succeed to use the class.getClassLoader().getResourceAsStream("test.properties") to load test.properties in any folders or sub folder of src.
Does it mean getResourceAsStream can search the specified file in any of the folders of project recursively? I just can't understand it how it works. Thanks.

getResourceAsStream() method searches for resources with classpath as root. I suppose in your case 'testfolder' is source folder. I you would have your 1.properties in 'testfolder/mypath/1.properties' then you need to specify getResourceAsStream("mypath/1.properties")

Where to place a file if it is getting accessed using ClassLoader.getSystemResource in WebApplication

I am using one third party jar in my code. In the jar file , in one of the classes, when I opened the class using de-compiler, the code below is written:
java.net.URL fileURL = ClassLoader.getSystemResource("SOAPConfig.xml");
Now I am using this in my webapplication, where should I place this SOAPConfig.xml so that it will find the fileURL.
Note: I have tried putting this XML in WEB-INF/classes folder. But it is not working. Your help will be appreciated.
In Addition: In the explaination you have given, It is telling me not to use this code snippet inside the third party jar in this way...What is the exact usage of this statement

ClassLoader.getSystemResource will load the resource from the system classloader, which uses the classpath of the application as started from the command line. Any classloaders created by the application at runtime (i.e. the one that looks in WEB-INF/classes) are not on the system classpath.
You need to
Look through the script that starts your server, find out which directories are on the classpath there, and put your SOAPConfig.xml in one of those. If necessary, change the classpath in the script to look in a separate directory that's just used for your config file.
Track down the person who used ClassLoader.getSystemResource in the library, kick them squarely in the nuts, and tell them never to do that again.

Getting a resource in a different project using classloader

Using the ClassLoader#getResource(), I need to access a file that is present in a project other than the one where my current code resides. How can this be done?
I'm using eclipse.
Directory Structure:
Root
|-project1
| |-package
| |-myResourceFile
|-project2
|-package
|-myCodeFile
I'm trying to get myResourceFile from myCodeFile, using myCodeFile.class.getClassLoader().getResource("../../project1/package/myResourceFile") but its always returning null. I do not want to add project1 to the classpath of project2. Though adding that also did not work.
With regards,

It's a bad idea to attempt to read files from another project like that because it ties you to exactly that directory structure. You already did the first step in decoupling the projects by using getResource() instead of using the java.util.File API so you can go the full way as well.
In Eclipse you can add other projects to a projects' build path (Project Properties -> Java Build Path -> Projects). You should be able to read the other projects' files now.

If you are using maven, then you can specify project1/package as a resource folder in the pom.xml of project2. You can theen use Classloader getResource method to get the resouce
http://maven.apache.org/plugins/maven-resources-plugin/examples/resource-directory.html