Unzipping a file from InputStream and returning another InputStream

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伪装坚强ぢ
伪装坚强ぢ 2020-12-14 06:33

I am trying to write a function which will accept an InputStream with zipped file data and would return another InputStream with unzipped data.

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  • 2020-12-14 06:58

    It is on scala syntax:

    def unzipByteArray(input: Array[Byte]): String = {
        val zipInputStream = new ZipInputStream(new ByteArrayInputStream(input))
        val entry = zipInputStream.getNextEntry
        IOUtils.toString(zipInputStream, StandardCharsets.UTF_8)
    }
    
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  • 2020-12-14 07:08

    If you can change the input data I would suggested you to use GZIPInputStream.

    GZipInputStream is different from ZipInputStream since you only have one data inside it. So the whole input stream represents the whole file. In ZipInputStream the whole stream contains also the structure of the file(s) inside it, which can be many.

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  • 2020-12-14 07:12

    Concepts

    GZIPInputStream is for streams (or files) zipped as gzip (".gz" extension). It doesn't have any header information.

    This class implements a stream filter for reading compressed data in the GZIP file format

    If you have a real zip file, you have to use ZipFile to open the file, ask for the list of files (one in your example) and ask for the decompressed input stream.

    Your method, if you have the file, would be something like:

    // ITS PSEUDOCODE!!
    
    private InputStream extractOnlyFile(String path) {
       ZipFile zf = new ZipFile(path);
       Enumeration e = zf.entries();
       ZipEntry entry = (ZipEntry) e.nextElement(); // your only file
       return zf.getInputStream(entry);
    }
    

    Reading an InputStream with the content of a .zip file

    Ok, if you have an InputStream you can use (as @cletus says) ZipInputStream. It reads a stream including header data.

    ZipInputStream is for a stream with [header information + zippeddata]

    Important: if you have the file in your PC you can use ZipFile class to access it randomly

    This is a sample of reading a zip-file through an InputStream:

    import java.io.FileInputStream;
    import java.util.zip.ZipEntry;
    import java.util.zip.ZipInputStream;
    
    
    public class Main {
        public static void main(String[] args) throws Exception
        {
            FileInputStream fis = new FileInputStream("c:/inas400.zip");
    
            // this is where you start, with an InputStream containing the bytes from the zip file
            ZipInputStream zis = new ZipInputStream(fis);
            ZipEntry entry;
                // while there are entries I process them
            while ((entry = zis.getNextEntry()) != null)
            {
                System.out.println("entry: " + entry.getName() + ", " + entry.getSize());
                        // consume all the data from this entry
                while (zis.available() > 0)
                    zis.read();
                        // I could close the entry, but getNextEntry does it automatically
                        // zis.closeEntry()
            }
        }
    }
    
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  • 2020-12-14 07:14

    Unless I'm missing something, you should absolutely try and get ZipInputStream to work and there's no reason it shouldn't (I've certainly used it on several occasions).

    What you should do is try and get ZipInputStream to work and if you can't, post the code and we'll help you with whatever problems you're having.

    Whatever you do though, don't try and reinvent its functionality.

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