Develop Reference

Reading Excel from src folder using JXL

I am a beginner and having a hard time to get my program run correctly after exporting it as runnable JAR.
I need to read an Excel called "bstn.xls", which is in the dDefault package such as all other classes.
Here is how I did it.
get the file path with: URL excelURL = this.getClass().getClassLoader().getResource("bstn.xls");
Create a File with the URL File file = new File(new ResourceLoader().loader());
Create a workbook with the file wbook = Workbook.getWorkbook(file);
I got the error in german here, but it says: the System cannot find the given path.
Is JXL not able to read with the full qualified path? I can't find anything on the JXL site.

Since the file is in the classpath, you may use the getWorkbook method taking an InputStream parameter :
InputStream stream = this.getClass().getClassLoader().getResourceAsStream("bstn.xls");
wbook = Workbook.getWorkbook(stream);

Can't create .txt files

I have it like this:
file=new File(pathString+"/TileRecordings/",jTextField1.getText()+".txt");
and then after null checking etc...
file.createNewFile();
Thing is, if the texfield value was 'test' it would create 'text.txt' AS A FOLDER. Not an actual text file.
Is this a Ubuntu only thing? How do I force it to literally create a text file, not a folder named 'test.txt'.

Maybe try features from java.nio.file
for example:
Path temp = Paths.get(pathString+"/TileRecordings/"+jTextField1.getText()+".txt");
Files.write(temp, contentToSave.toString().getBytes());

Change file name and its extension before output it Java

Lets say I've got a file in D: which has a filepath as:
D:\work\2012\018\08\2b558ad8-4ea4-4cb9-b645-6c9a9919ba01
at the moment the name of this file is not meaningful. It doesn't have a format. I would like to have the ability to create file object from the above filepath in Java and change its name and format before writing it into bytes. The new path could be as following:
D:\work\2012\018\08\mywork.pdf
After the changes have been made, I should still be able to gain access to the original file per normal.
I already know the type of the file and its name so there won't be a problem getting those details.

Just rename the file:
File file = new File(oldFilepath);
boolean success = file.renameTo(new File(newFilepath));
You could also just give a filename and use the same parent as your current file:
File file = new File(oldFilepath);
file.renameTo(new File(file.getParentFile(), newFilename));

It's not really clear what exactly you want; I hope this answer is useful to you.
Suppose you have a java.io.File object that represents D:\work\2012\018\08\2b558ad8-4ea4-4cb9-b645-6c9a9919ba01, and that you want to have a File object that represents D:\work\2012\018\08\mywork.pdf. You already know the new filename mywork.pdf but you want to get the directory name from the original File object. You can do that like this:
File original = new File("D:\\work\\2012\\018\\08\\2b558ad8-4ea4-4cb9-b645-6c9a9919ba01");
// Gets the File object for the directory that contains the file
File dir = original.getParentFile();
// Creates a File object for a file in the same directory with the name "mywork.pdf"
File result = new File(dir, "mywork.pdf");

Reading .txt file from another directory

The code I am running is in /Test1/Example. If I need to read a .txt file in /Test1 how do I get Java to go back 1 level in the directory tree, and then read my .txt file
I have searched/googled and have not been able to find a way to read files in a different location.
I am running a java script in an .htm file located at /Test1/Test2/testing.htm. Where it says script src=" ". What would I put in the quotations to have it read from my file located at /Test1/example.txt.

In Java you can use getParentFile() to traverse up the tree. So you started your program in /Test1/Example directory. And you want to write your new file as /Test1/Example.txt
File currentDir = new File(".");
File parentDir = currentDir.getParentFile();
File newFile = new File(parentDir,"Example.txt");;
Obviously there are multiple ways to do this.

You should be able to use the parent directory reference of "../"
You may need to do checks on the OS to determine which directory separation you should be using ['\' compared to '/']

When you create a File object in Java, you can give it a pathname. You can either use an absolute pathname or a relative one. Using absolutes to do what you want would require:
File file = new File("/Test1/myFile.txt");
if(file.canRead())
{
// read file here
}
Using relatives paths if you want to run from the location /Test1/Example:
File file = new File("../myFile.txt");
if(file.canRead())
{
// read file here
}

I had a similar experience.
My requirement is: I have a file named "sample.json" under a directory "input", I have my java file named "JsonRead.java" under a directory "testcase". So, the entire folder structure will be like untitled/splunkAutomation/src and under this I have folders input, testcase.
once after you compile your program, you can see a input file copy named "sample.json" under a folder named "out/production/yourparentfolderabovesrc/input" and class file named "JsonRead.class" under a folder named "out/production/yourparentfolderabovesrc/testcase". So, during run time, Java will actually refer these files and NOT our actual .java file under "src".
So, my JsonRead.java looked like this,
package testcase;
import java.io.*;
import org.json.simple.JSONObject;
public class JsonRead{
public static void main(String[] args){
java.net.URL fileURL=JsonRead.class.getClass().getResource("/input/sample.json");
System.out.println("fileURL: "+fileURL);
File f = new File(fileURL.toURI());
System.out.println("fileIs: "+f);
}
}
This will give you the output like,
fileURL: file:/C:/Users/asanthal/untitled/out/production/splunkAutomation/input/sample.json
fileIs: C:\Users\asanthal\untitled\out\production\splunkAutomation\input\sample.json

It worked for me. I was saving all my classes on a folder but I needed to read an input file from the parent directory of my classes folder. This did the job.
String FileName = "Example.txt";
File parentDir = new File(".."); // New file (parent file ..)
File newFile = new File(parentDir,fileName); //open the file

Java File manipulation

So I have an application with a JFileChooser from which I select a file to read. Then I change some words and write a new file. The problem that I am having is that when I write the new file it's saved in the project directory. How do I save it in the same directory as the file that I chose using the JFileChooser. Note: I don't want to use the JFileChooser to choose the location. I just need to save the file in the same directory as the original file that I read.

You choose a file like this:
File fileToRead = JFileChooser.getSelectedFile();
Then you read and change the content and write it back to the same location with a different name:
File fileToWrite = new File( fileToRead.getParent(), "newName.txt" );