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    <p><code>zipFile.getInputStream(ZipEntry entry)</code> will return you the inputstream for the specific entry.</p> <p>Check out the <a href="http://docs.oracle.com/javase/6/docs/api/java/util/zip/ZipFile.html#getInputStream(java.util.zip.ZipEntry)" rel="nofollow">javadocs</a> for <code>ZipFile.getInputStream()</code> - <a href="http://docs.oracle.com/javase/6/docs/api/java/util/zip/ZipFile.html#getInputStream(java.util.zip.ZipEntry)" rel="nofollow">http://docs.oracle.com/javase/6/docs/api/java/util/zip/ZipFile.html#getInputStream(java.util.zip.ZipEntry)</a>.</p> <p><strong>Update:</strong></p> <p>I misread your question. For using the <code>ZipInputStream</code>, there is sample code on Oracle's website (<a href="http://java.sun.com/developer/technicalArticles/Programming/compression/" rel="nofollow">http://java.sun.com/developer/technicalArticles/Programming/compression/</a>) that shows you how to read from the stream. See the first code sample: Code </p> <ul> <li>Sample 1: UnZip.java. </li> </ul> <p>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:</p> <pre><code>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); } } </code></pre>
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    1. POSpring: how to parse uploaded zip file?
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    1. USPaul Cichonski
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    1. COMy 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.
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      1. USPaul Cichonski
    2. COMmm..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
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      1. USWinte Winte
    3. COI 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?
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