尝试使用java.util.zip.ZipFile解压缩存档时出现FileNotFoundException

Question

I have a silly problem i haven't been able to figure out. Can anyone help me? My Code is as:

String zipname = "C:/1100.zip";
    String output = "C:/1100";
    BufferedInputStream bis = null;
    BufferedOutputStream bos = null;
    try {
        ZipFile zipFile = new ZipFile(zipname);
        Enumeration<?> enumeration = zipFile.entries();
        while (enumeration.hasMoreElements()) {
            ZipEntry zipEntry = (ZipEntry) enumeration.nextElement();
            System.out.println("Unzipping: " + zipEntry.getName());
            bis = new BufferedInputStream(zipFile.getInputStream(zipEntry));
            int size;
            byte[] buffer = new byte[2048];

It doesn't create a folder but debugging shows all the contents being generated. In Order to create a folder i used the code

if(!output.exists()){ output.mkdir();} // here i get an error saying filenotfoundexception

            bos = new BufferedOutputStream(new FileOutputStream(new File(outPut)));
            while ((size = bis.read(buffer)) != -1) {
                bos.write(buffer, 0, size);
            }
        }
    } catch (Exception ex) {
        ex.printStackTrace();
    } finally {
        bos.flush();
        bos.close();
        bis.close();
    }

My zip file contains images: a.jpg b.jpg... and in the same hierarchy, I have abc.xml. I need to extract the content as is in the zip file. Any helps here.

Answer 1

There are a few problems with your code: Where is outPut declared? output is not a file but a string, so exists() and mkdir() do not exist. Start by declaring output like:

File output = new File("C:/1100");

Furthermore, outPut (with big P) is not declared. It be something like output + File.seprator + zipEntry.getName() .

 bos = new BufferedOutputStream(new FileOutputStream(output + File.seprator + zipEntry.getName()));

Note that you don't need to pass a File to FileOutputStream, as constructors show in the documentation .

At this point, your code should work if your Zip file does not contain directory. However, when opening the output stream, if zipEntry.getName() has a directory component (for instance somedir/filename.txt ), opening the stream will result in a FileNotFoundException , as the parent directory of the file you try to create does not exist. If you want to be able to handle such zip files, you will find your answer in: How to unzip files recursively in Java?