Spring: how to parse uploaded zip file?

Question

I uploaded my zip archive to the server and want to open .txt and .jpg files in it. I successfully get my archive in my Controller and get the name of each file via ZipEntry. Now I want to open it but for this I should get a full path to my file.

I haven't found how I can do that. Could you suggest some approach how to do that ?

Update

I try to use example have been suggested below but I am not be able open the file

ZipFile zFile = new ZipFile("trainingDefaultApp.zip");

I have got the FileNotFoundException

So I return to my start point. I have upload form in Java Spring application. In controller I had got a zip archive as byte[]

@RequestMapping(method = RequestMethod.POST)
public String create(UploadItem uploadItem, BindingResult bindingResult){
    try {
        byte[] zip = uploadItem.getFileData().getBytes();
        saveFile(zip);

Then I had got each ZipEntry

    InputStream is = new ByteArrayInputStream(zip);
    ZipInputStream zis = new ZipInputStream(is);

    ZipEntry entry = null;
    while ((entry = zis.getNextEntry()) != null) {
        String entryName = entry.getName();
        if (entryName.equals("readme.txt")) {
            ZipFile zip = new ZipFile(entry.getName()); // here I had got an exception

According to docs I did all right but as for me it is strange to pass the file name only and suspect that you successfully will open the file

What do you mean? You said you open it successfully; what do you not have a "full path" to? — Dave Newton, Jun 13 '12 at 13:45
I have opened CommonsMultipartFile that contains my zip archive as byte[]. I convert it to ZipInputStream and in loop have got a ZipEntry for each file in my archive. Now I want to read this .txt file and .jpg file but ZipEntry contains only names of this files. So I haven't a full path to this files and can't open it as new File("d:/some.txt") — Winte Winte, Jun 13 '12 at 13:50

score 1 · Answer 1 · edited May 19 '16 at 11:18

1

zipFile.getInputStream(ZipEntry entry) will return you the inputstream for the specific entry.

Check out the javadocs for ZipFile.getInputStream() - http://docs.oracle.com/javase/6/docs/api/java/util/zip/ZipFile.html#getInputStream(java.util.zip.ZipEntry).

Update:

I misread your question. For using the ZipInputStream, there is sample code on Oracle's website (http://java.sun.com/developer/technicalArticles/Programming/compression/) that shows you how to read from the stream. See the first code sample: Code

Sample 1: UnZip.java.

Copying here, it is reading from the entry and writing it directly to a file, but you could replace that with whatever logic you need:

ZipInputStream zis = new ZipInputStream(new BufferedInputStream(fis));
ZipEntry entry;
while((entry = zis.getNextEntry()) != null) {
   System.out.println("Extracting: " +entry);
   int count;
   byte data[] = new byte[BUFFER];
   // write the files to the disk
   FileOutputStream fos = new FileOutputStream(entry.getName());
   dest = new 
   BufferedOutputStream(fos, BUFFER);

   while ((count = zis.read(data, 0, BUFFER)) != -1) {
        dest.write(data, 0, count);
   }
}

edited May 19 '16 at 11:18

tharindu_DG

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answered Jun 13 '12 at 14:04

Paul Cichonski

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My bad, just saw that you have a ZipInputStream, not a ZipFile. There is sample code on Oracle's website (http://java.sun.com/developer/technicalArticles/Programming/compression/) that shows you how to read from the stream. See the first code sample: Code Sample 1: UnZip.java. – Paul Cichonski Jun 13 '12 at 14:12
Mmm..how can I use ZipFile? In consructor I should pass the File that I can't create without a full path to this file. In case when I pass the name of file as String I got FileNotFoundException – Winte Winte Jun 13 '12 at 14:27
I try this example. I have same excaeption. It says that file not found. For test I hardcoded the name of my archive "tetx.zip" but I have failed when try to create new FileInputStream("tetx.zip"). Is it wrong or the issue in another place? – Winte Winte Jun 13 '12 at 14:59
If you are able, you may want to post your code. I'm not sure what the issue is. – Paul Cichonski Jun 13 '12 at 15:17
Thanks for your help, Paul. I expend my question with code and my description. Probably it helps to unsderstand the cause of issue. – Winte Winte Jun 14 '12 at 07:14
link to sample code is broken. Possibly this is the one that was referenced, nice read! http://www.oracle.com/technetwork/articles/java/compress-1565076.html – Chacko Apr 20 '17 at 07:23

score 1 · Accepted Answer · answered Jun 14 '12 at 11:27

I resolve my uissue. The solution is work directly with ZipInputStream. Here the code:

    private void saveFile(byte[] zip, String name, String description) throws IOException {
    InputStream is = new ByteArrayInputStream(zip);
    ZipInputStream zis = new ZipInputStream(is);

    Application app = new Application();
    ZipEntry entry = null;
    while ((entry = zis.getNextEntry()) != null) {
        String entryName = entry.getName();
        if (entryName.equals("readme.txt")) { 
           new Scanner(zis); //!!!
           //... 
           zis.closeEntry();

Spring: how to parse uploaded zip file?

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