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Tags/Keywords : Using-Java-Technology - Search-Class-in-JAR - Coding-Tips-on-Java Example
Author : Amit
Date (Year/Month/Date): 2009-01-26
Class file search in Jar file using Java Technology

 



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How to search for a class
file from a bunch of Jar
files in a directory file
system?

I have created some of
simple to use tools using
Java Technology for making
developer work simpler in
some form or the other.
While working on some
assignment of adding a
required JAR file from a
list of many Jar files from
a directory, so as to be able
to overcome a class not found
exception, by placing the
required Jar file to classpath
environment variable. So next
step is to find the needed Jar
file from so many Jar files is
a tough job, and neither did
I want to put all the Jar
files in classpath nor to
open each and every Jar file
to see if the required class
file is present or not. Then
I realize the need of a piece
of code that does this job
for me by going through each
and every Jar files in the
directory and looking inside
it for the Class file I wanted
it to search. Once found just
shows me the Jar file and stop
searching further.




Java Technology has a package of classes from
java.util.zip and java.util.jar to deal with
Zip and Jar files entries. This gave me
an idea of writing this class to look for this
objective.
Step includes:
1. Take a directory path and class file name as input.
2. List all files available in that directory.
3. Iterate through the list of files and check
   if the file is a Jar file or not.
4. If a Jar file, then create an
   java.util.jar.JarFile object and
   enumerate through the entries (here entries are
   the class files).
5. For each entry (which is nothing but
   java.util.zip.ZipEntry),
 check if the toString returns a string equivalent to the class
 file name.
6. If yes, just return from the iteration and
   print the jar file name on console.


While writing code, step 1 has following code:

 public TestJar(String path, String classToSearch)

directory path and class to search as arguments
to constructor.

step 2:

 java.io.File f = new java.io.File(path);
 java.io.File[] files=f.listFiles();

create a File object for the path supplied, and list
all files in the directory as listFiles.
 step 3:
 for(int i=0;i<files.length;i++)
 {
 if(isJarFile(files[i])){

isJarFile is a method to check whether the file name
it takes as argument is a Jar file or not.
 public boolean isJarFile(File file)
 {
   StringTokenizer st=
       new StringTokenizer(file.getName(),".");
   String ext=null;
   while(st.hasMoreTokens())
   {
      if( ((String)st.nextToken()).equals("jar"))
      {
        return true;
      }
   }
   return false;
 }

step 4:
 fileName=files[i].getName();
 JarFile zf=new JarFile(path+"/"+ fileName);

create a JarFile object with an argument as the directory
path with the file name.

step 5:
 Enumeration e=zf.entries();
 while (e.hasMoreElements()) {
 ZipEntry ze=(ZipEntry)e.nextElement();

Extract each ZipEntry inside the Jar file.

step 6:
if((ze.toString()).equals(classToSearch+".class")){
System.out.println("\nThe jar file name is
:"+path+"/"+fileName+" : "+ze);
System.exit(0);;
If class to search is same as the ZipEntry toString, then
show this Jar file and exit.

This is purely my imagination and user or reader should
take appropriate precaution in handling these kind of
code writing in a project or at work place (if interested).
Writer of this article never claims that this will work
or is risk free.



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Author of this article/writeup has expressed his/her willingness
to help or guide users with any technical difficulties he/she faces
while working with the example code environment setting up, running
and resolving any such exception raised during compile or at runtime.
You may ask for any technical doubt or seek technical help related to
this article by using following form to reach for technical help
from the Author for FREE.

This article's Author shall be reading your request and responding
within reasonable time (no resolution timeframe defined as such).

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