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Creating a XML file in java

Posted on 2007-11-26
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Last Modified: 2010-03-30
I have a program that allows the user to drag and drop files in to a window.  I then click a button and the program zips the files that have been dragged.  Is it possible that when the zip file button is clicked the program can create a xml file.  The file would look something like this

Jaziar    <------Username
Nov, 25, 2007    <-----Date
file 1
file 2                 ,<----- then the names of all the files in the window

I will attach the code that zips
import java.awt.*;
import java.awt.event.*;
import javax.swing.*;
import java.util.Vector;
import java.util.zip.*;
import java.io.*; 
 
class UploadDriver implements ActionListener {
    final Vector<String> remember = new Vector<String>(); 
    
 
    public UploadDriver () {
        javax.swing.JFrame frame = new javax.swing.JFrame( "FileDrop" );
	javax.swing.JButton doit = new javax.swing.JButton("Upload Files!"); 
	doit.addActionListener(this); 
 
	//javax.swing.border.TitledBorder dragBorder = new javax.swing.border.TitledBorder( "Drop 'em" );
        final javax.swing.JTextArea text = new javax.swing.JTextArea();
	frame.getContentPane().add( 
				   new javax.swing.JScrollPane( text ), 
				   java.awt.BorderLayout.CENTER );
        frame.getContentPane().add(doit, java.awt.BorderLayout.SOUTH); 
	
        new FileDrop( System.out, text, /*dragBorder,*/ new FileDrop.Listener()
        {   public void filesDropped( java.io.File[] files )
            {   for( int i = 0; i < files.length; i++ )
                {   try
                    {   
			text.append( files[i].getCanonicalPath() + "\n" );
			remember.add(files[i].getCanonicalPath()); 
		    }   // end try
                    catch( java.io.IOException e ) {}
                }   // end for: through each dropped file
            }   // end filesDropped
        }); // end FileDrop.Listener
 
        frame.setBounds( 100, 100, 300, 400 );
        frame.setDefaultCloseOperation( frame.EXIT_ON_CLOSE );
        frame.show();
    }
 
    
    public void actionPerformed(ActionEvent ae) {
	System.out.println("Time to zip it, zip it good"); 
	System.out.println("Looking to zip " + remember.size() + " files"); 
 
	byte[] buf = new byte[1024];
	
	try {
	    // Create the ZIP file
	    String target = "target.zip";
	    ZipOutputStream out = new ZipOutputStream(new FileOutputStream(target));
	    
	    // Compress the files
	    for (int i=0; i<remember.size(); i++) {
		FileInputStream in = new FileInputStream(remember.elementAt(i));
		
		// Add ZIP entry to output stream.
		out.putNextEntry(new ZipEntry(remember.elementAt(i)));
		
		// Transfer bytes from the file to the ZIP file
		int len;
		while ((len = in.read(buf)) > 0) {
		    out.write(buf, 0, len);
		}
		
		// Complete the entry
		out.closeEntry();
		in.close();
	    }
	    
	    // Complete the ZIP file
	    out.close();
	} catch (IOException e) {
	}
	System.out.println("You have been ZIPPED"); 
    }
}

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Question by:Jaziar
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3 Comments
 
LVL 27

Expert Comment

by:mrcoffee365
ID: 20353255
Sure.  Make a new .xml file outside the loop where you manipulate the list of files.  Write the 2 header lines to the top, then for each file, write the file name to the .xml file.

If you plan to use XML syntax, then surround the data with XML tags, e.g.
<filename>file1</filename>
0
 

Author Comment

by:Jaziar
ID: 20353331
I know this is some what of a second question, but I will create a new question if you experts think I should.

But if you look at the above code, where it is zipping the files it is putting in complete paths to the files

Example:

C://Admin/Desktop/File.txt
C://tmp/file2.txt

So when it unzips the full path is still part of the file.  Is it possible to only zip the file? and not the complete path

Example:
File.txt
file2.txt

That way when I move the zip, it is not dependent on the dir structure
0
 
LVL 27

Accepted Solution

by:
mrcoffee365 earned 500 total points
ID: 20353903
Use the file's name -- the last element in the absolute path -- as the new ZipEntry.

And yes, it's a very different question.  But if we don't have to spend much time on it, then we'll just include it here.
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