Develop Reference

How can I retrieve CSV file inside of the Unzip folder?

I am trying to read and parsed all the CSV files, these files are inside of extracted zip file, but unfortunately, I'm only getting this (No such file or directory) - How can I get inside that folder and read those files? Take note that this folder is already unzipped.
I'm also thinking, that if that is not possible - is there any way that I can extract the zip file to files only, for example below:
Zip File: ZipFolder.zip
Expected: (No folder at all, automatically the files is on the target location already).
CSVFile1.csv
CSVFile2.csv
Can someone help me to get through this? your help is appreaciated.. Thank you!

Replace a single file from jar file

I need to replace a single class file from a jar file. The jar file is quite big and every time I don't want to extract it and replace manually. I want to automate this procedure. Can someone please help me on this.
When jar file is extracted below folders are created :
render
classes
com
I need to go inside "com/cgp/f1/cmmi/" folder and replace a class file inside it.
Things I tried :
zip file.jar com/cgp/f1/cmmi/services.class Services.class
jar uf file.jar com/cgp/f1/cmmi/ services.class
jar -uf file.jar com\cgp\f1\cmmi\ services.class
jar uf file.jar com/cgp/f1/cmmi/services.class services.class
The error I am getting is :
when using jar command
com\cgp\f1\cmmi\ : no such file or directory
when using zip command :
zip warning: name not matched: com\cgp\f1\cmmi\Services.class
can some one please guide me where I am going wrong.

Maybe the jar -uf found here could help you: How to update one file in a zip archive
If graphical apps are an option, you could use winrar or 7-zip to replace the class. You don't need to extract the jar file to make this work. Just open the jar with one of those apps, go to de directory where is the class file to be replaced, drag-and-drop the new file to replace the old one and save.

Java - Saving file from JAR to disk

I'm trying to save a YAML formatted file packaged in the JAR to the user's disk.
The file is named config.yml and sits directly under one of the project's source folders. I can confirm that the file is packaged in the JAR with WinRAR.
I am using the following code to get the file from the JAR:
File file = new File(this.getClassLoader().getResource("config.yml").getPath());
And I am using this code to copy the file to a directory:
FileUtils.copyFileToDirectory(file, this.getDataFolder());
The getDataFolder() method is implemented by a reliable 3rd party API.
However, when I use a file instance defined with the same getDataFolder() File instance and the path "config.yml":
new File(this.getDataFolder(), "config.yml")
The console logs a FileNotFoundException.
The file path given in the stack trace is this: "file:\C:\Users\Evan\Desk
top\Test%20Server\plugins\lottocrates-1.0.jar!\config.yml" which seems correct. I tried opening the file with run prompt, which I was able to open the JAR with, but not the config.yml file.

JAR - Listing files into a folder

I would like to get a list of file contained in a directory which is in a jar package.
I have an "images" folder, within it I have an Images class that should load all images from that directory.
In the past i used the MyClass.class.getResourceAsStream("filename"); to read files, but how do I read a directory?
This is what I tried:
System.out.println(Images.class.getResource("").getPath());
System.out.println(new File(Images.class.getResource("").getPath()).listFiles());
I tried with Images.class.getResource because I have to work with File and there isn't a constructor that accepts an InputStream.
The code produces
file:/home/k55/Java/MyApp/dist/Package.jar!/MyApp/images/
null
So it is finding the folder which I want to list files from, but it is not able to list files.
I've read on other forums that in fact you can't use this method for folders in a jar archive, so how can I accomplish this?
Update: if possible, i would like to read files without having to use the ZipInputStream

You can't do that easily.
What you need to do:
Get the path of the jar file.
Images.class.getResource("/something/that/exists").getPath()
Strip "!/something/that/exists".
Use Zip File System to browse the Jar file.
It's a little bit of hacking.

Reading a file into compiled jar by relative path

How would you read a file into a program that's compiled into a jar next to it through its local directory? The type read would be a simple .txt file.

It depends on what the usage of the program is. Do you know how the jar is supposed to be executed? When you try to run it, does it spit out a "usage: somejar firstarg secondarg" type message?
Also, if its a jar that you've compiled and you know how it should be executed, then you may have forgot to set its main class or manifest.
Check this: http://www.mkyong.com/java/how-to-make-an-executable-jar-file/

If you want to read a file that exists within an external .jar file, you will need to unzip the .jar file first in your code and then retrieve the file. You can do this using Java's zip APIs. See this answer if this is the case: Easiest way to unpack a jar in java
If you want to read a file that is in the same .jar file that your code is executing, you can get the file as a resource. See this answer: Get a resource using getResource()
If the file is simply in the exact same directory as the executable .jar, create a new file like so:
File input = new File("myfile.txt");