How can i zip files in Java and not include files paths

Question

For example, I want to zip a file stored in /Users/me/Desktop/image.jpg

I made this method:

public static Boolean generateZipFile(ArrayList<String> sourcesFilenames, String destinationDir, String zipFilename){
  // Create a buffer for reading the files 
  byte[] buf = new byte[1024]; 

  try {
   // VER SI HAY QUE CREAR EL ROOT PATH
         boolean result = (new File(destinationDir)).mkdirs();

         String zipFullFilename = destinationDir + "/" + zipFilename ;

         System.out.println(result);

   // Create the ZIP file  
   ZipOutputStream out = new ZipOutputStream(new FileOutputStream(zipFullFilename)); 
   // Compress the files 
   for (String filename: sourcesFilenames) { 
    FileInputStream in = new FileInputStream(filename); 
    // Add ZIP entry to output stream. 
    out.putNextEntry(new ZipEntry(filename)); 
    // 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();

   return true;
  } catch (IOException e) { 
   return false;
  }  
 }

But when I extract the file, the unzipped files have the full path.

I don't want the full path of each file in the zip i only want the filename.

How can I made this?

Answer 1

You're finding your source data using the relative path to the file, then setting the Entry to the same thing. Instead you should turn the source into a File object, and then use

putNextEntry(new ZipEntry(sourceFile.getName()))

that'll give you just the final part of the path (ie, the actual file name)

Answer 2

Do as Jason said, or if you want to keep your method signature, do it like this:

out.putNextEntry(new ZipEntry(new File(filename).getName())); 

or, using FileNameUtils.getName from apache commons/io:

out.putNextEntry(new ZipEntry(FileNameUtils.getName(filename))); 

Answer 3

You could probably get away with accessing source files via new FileInputStream(new File(sourceFilePath, sourceFileName)).

Answer 4

Here:

// Add ZIP entry to output stream. 
out.putNextEntry(new ZipEntry(filename)); 

You're creating the entry for that file using the whole path. If you just use the name ( without the path ) you'll have what you need:

// Add ZIP entry to output stream. 
File file = new File(filename); //"Users/you/image.jpg"
out.putNextEntry(new ZipEntry(file.getName())); //"image.jpg"