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

问题

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?

回答1:

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"

回答2:

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)

回答3:

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))); 

回答4:

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

回答5:

// easy way of zip a file 

import java.io.*;

import java.util.zip.*;

 public class ZipCreateExample{

    public static void main(String[] args)  throws Exception  {
            // input file 
        FileInputStream in = new FileInputStream("F:/ZipCreateExample.txt");;
        // out put file 
        ZipOutputStream out =new ZipOutputStream(new FileOutputStrea("F:/tmp.zip"));
         // name of file in zip folder 
        out.putNextEntry(new ZipEntry("zippedfile.txt")); 

        byte[] b = new byte[1024];

        int count;

        // writing files to new zippedtxt file
        while ((count = in.read(b)) > 0) {
            System.out.println();

         out.write(b, 0, count);
        }
        out.close();
        in.close();
    }
}

回答6:

try {
    String zipFile = "/locations/data.zip";
    String srcFolder = "/locations";

    File folder = new File(srcFolder);
    String[] sourceFiles = folder.list();

    //create byte buffer
    byte[] buffer = new byte[1024];

    /*
     * To create a zip file, use
     *
     * ZipOutputStream(OutputStream out) constructor of ZipOutputStream
     * class.
     */
    //create object of FileOutputStream
    FileOutputStream fout = new FileOutputStream(zipFile);

    //create object of ZipOutputStream from FileOutputStream
    ZipOutputStream zout = new ZipOutputStream(fout);

    for (int i = 0; i < sourceFiles.length; i++) {
        if (sourceFiles[i].equalsIgnoreCase("file.jpg") || sourceFiles[i].equalsIgnoreCase("file1.jpg")) {
            sourceFiles[i] = srcFolder + fs + sourceFiles[i];
            System.out.println("Adding " + sourceFiles[i]);
            //create object of FileInputStream for source file
            FileInputStream fin = new FileInputStream(sourceFiles[i]);

            /*
             * To begin writing ZipEntry in the zip file, use
             *
             * void putNextEntry(ZipEntry entry) method of
             * ZipOutputStream class.
             *
             * This method begins writing a new Zip entry to the zip
             * file and positions the stream to the start of the entry
             * data.
             */

            zout.putNextEntry(new ZipEntry(sourceFiles[i].substring(sourceFiles[i].lastIndexOf("/") + 1)));

            /*
             * After creating entry in the zip file, actually write the
             * file.
             */
            int length;

            while ((length = fin.read(buffer)) > 0) {
                zout.write(buffer, 0, length);
            }

            /*
             * After writing the file to ZipOutputStream, use
             *
             * void closeEntry() method of ZipOutputStream class to
             * close the current entry and position the stream to write
             * the next entry.
             */

            zout.closeEntry();

            //close the InputStream
            fin.close();

        }
    }

    //close the ZipOutputStream
    zout.close();

    System.out.println("Zip file has been created!");

} catch (IOException ioe) {
    System.out.println("IOException :" + ioe);
}

回答7:

If you zip two files of the same name but with different paths you will run into duplicate file entry errors, though.

来源：https://stackoverflow.com/questions/3018579/how-can-i-zip-files-in-java-and-not-include-files-paths