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Unable to create ,zip file

 
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Hi

I have written one function for crating zip file



This function creats a zip file of the bin folder in jboss

how do i create zip file for all the folders in my C:\\test folder which contains few folders and few files
 
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You can use java.io.File for either files or directories.
So you can create the source directory like so:

File sourceDir = new File("C:\\test");

and then get all the subdirectories or files as an array of files:

File[] subdirects = sourceDir.listFiles();

Then you can simply test each subdirectory to see if it is a directory or file.

The best way to manipulate a batch of directories and files is to use a recursive process, as in the method zipFiles shown below.


So you can call zipFiles, passing in sourceDir = new File("C:\\test");

BufferedInputStream origin = null;
FileOutputStream dest = new FileOutputStream("c:\\MyZip1.zip");
CheckedOutputStream checksum = new CheckedOutputStream(dest, new Adler32());
ZipOutputStream out = new ZipOutputStream(new BufferedOutputStream(checksum));
//out.setMethod(ZipOutputStream.DEFLATED);
byte data[] = new byte[BUFFER];

public static void main(String[] args){
try{
zipFiles(new File("C:\\test");
out.close();
System.out.println("checksum : " + checksum.getChecksum().getValue());
} catch(Exception e){
e.printStackTrace();
}
}


private static void zipFiles(File dir){
if(dir.isDirectory()){
File[] files = dir.listFiles();
for (int i = 0; i < files.length; i++) {
zipFiles(files[i]);
}//end for

}else{
System.out.println("Adding: "+files[i]);
FileInputStream fi = new
FileInputStream(files[i]);
origin = new
BufferedInputStream(fi, BUFFER);
ZipEntry entry = new ZipEntry(files[i]);
out.putNextEntry(entry);
int count;
while((count = origin.read(data, 0,BUFFER)) != -1) {
out.write(data, 0, count);
}
origin.close();
}
}

Julia
 
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