Document class file and information in Jar files.

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This is pretty cool.  The purpose of this VB Script is to help you document where JAR (Java ARchive) files and specifically java class files are located so that you can address issues seen with a client or that you can speak intelligently with a developer when researching a customer issue.  Most times when documenting issues, it is helpful to know where the problem exists and if you are able to replicate the issue: sounds like QA Testing for bug reports, doesn't it?  

It can help with research and documentation for issues seen when working cases for Java and Tomcat type of files. You will find when supporting Java products is that you can take a jar file and simply treat it as a zip file.  What is good about this is that you can put the information from your findings and the documented results in to  your Support Desk or SharePoint site so that you can search the information for what class under which jar file is giving the user a challenge..  I cannot take full credit for the ideas behind the process as I have borrowed pieces from http://www.robvanderwoude.com and I am using a COM Object (.dll) from http://www.xstandard.com to look inside these jar files.

Here is an example that you may see in real life - In troubleshooting, you may see a message like:
Caused by: java.sql.SQLException: [Microsoft][SQLServer 2000 Driver for JDBC]Connection reset by peer: socket write error
      at com.microsoft.jdbc.base.BaseExceptions.createException(Unknown Source)

So..  How do I know what Jar file that is in?  What class file do I need to look at?

Sure, you can go into Apache Tomcat (<Apache_Directory>\work\Catalina\localhost\PCC\org\apache\jsp\jsp ) and look at the problem .java file there or you can pull this information from an archive that you have generated from the csv file to help you in your troubleshooting with an issue.

What you will need to do is put in the location that you want to have the script go through and pull up the contents of the jar files..
Set oFSO = CreateObject("Scripting.FileSystemObject")

Set fileOutput = oFSO.CreateTextFile(Replace(WScript.ScriptFullName, WScript.ScriptName, "") & "Jar_Files_Log.csv", True)

Folders "C:\Tomcat 5.0"


Function Folders(strFolder)
	Set oFSO = CreateObject("Scripting.FileSystemObject")
	Set objFolder = oFSO.GetFolder(strFolder)
	For Each Folder In Split(strFolder, ",")
		'WScript.Echo Folder
		Set objFolder = oFSO.GetFolder(Folder)
		''WScript.Echo objFolder.Path
		Set colFiles = objFolder.Files
		' -- Please note: This is necessary to show files
		' -- In a base folder (no subfolders)
		For Each objFile In colFiles
			If Right(objFile, 4) = ".jar" Then
				'WScript.Echo objFile.Name
				'WScript.Echo objFile.Path
				ZipContents objFile.Path
			End If
		' -- Recurse Subfolders
End Function

Function ZipContents(ZipFile)
	' This function uses X-standards.com's X-zip component to extract files from a ZIP file.
	' Arguments:
	' myZip         [string] the fully qualified path to the ZIP file
	' myTargetDir   [string] the directory where the extracted files will be located
	' myFileSpec    [string] the file(s) to be extracted, wildcards allowed
	' Written by Rob van der Woude
	' http://www.robvanderwoude.com
	' The X-zip component is available at:
	' http://www.xstandard.com/en/documentation/xzip/
	' For more information on available functionality read:
	' http://www.xstandard.com/printer-friendly.asp?id=C9891D8A-5390-44ED-BC60-2267ED6763A7
	Dim objZip, objItem
	Set objZip = CreateObject("XStandard.Zip")
	For Each objItem In objZip.Contents(ZipFile)
		'WScript.Echo ZipFile & "," & objItem.Path & objItem.Name
		If InStr(objItem.Name, ".") Then
			fileOutput.WriteLine Char(34) & ZipFile & Chr(34) & "," & objItem.Path & objItem.Name & "," & objItem.Size & "," & objItem.Date
		End If
	Set objZip = Nothing
	Set objItem = Nothing
End Function 
Sub ShowSubFolders(objFolder)
	Dim colFolders, objSubFolder
	Set colFolders = objFolder.SubFolders
	For Each objSubFolder In colFolders
		'WScript.Echo objSubFolder.Path
		'WScript.Echo objSubFolder
		Set colFiles = objSubFolder.Files
		For Each objFile In colFiles
			If Right(objFile, 4) = ".jar" Then
				'WScript.Echo objFile.Name
				'WScript.Echo objFile.Path
				ZipContents objFile.Path
			End If
End Sub

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The results look similar to the following which will easily sorted in Excel (Note: This is just a sampling of the results generated):
C:\Tomcat 5.0\bin\bootstrap.jar,META-INF/MANIFEST.MF,277,8/28/2004 19:59
C:\Tomcat 5.0\bin\bootstrap.jar,org/apache/catalina/launcher/CatalinaLaunchFilter.class,1648,8/28/2004 19:59
C:\Tomcat 5.0\bin\bootstrap.jar,org/apache/catalina/loader/Reloader.class,248,8/28/2004 19:59

Have fun..

Author:Kent Dyer

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