如何將 ZipInputStream 轉換為 InputStream？

Question

我有代碼，其中 ZipInputSream 轉換為字節 []，但我不知道如何將其轉換為輸入流。

private void convertStream(String encoding, ZipInputStream in) throws IOException,
        UnsupportedEncodingException
{
    final int BUFFER = 1;
    @SuppressWarnings("unused")
    int count = 0;
    byte data[] = new byte[BUFFER];
    while ((count = in.read(data, 0, BUFFER)) != -1) 
    {
       // How can I convert data to InputStream  here ?                    
    }
}

Answer 1

ZipInputStream 是 InputStream 的子類。

http://download.oracle.com/javase/6/docs/api/java/util/zip/ZipInputStream.html

Answer 2

這是我解決這個問題的方法。 現在我可以將單個文件從 ZipInputStream 獲取到內存作為 InputStream。

private InputStream convertZipInputStreamToInputStream(ZipInputStream in, ZipEntry entry, String encoding) throws IOException
{
    final int BUFFER = 2048;
    int count = 0;
    byte data[] = new byte[BUFFER];
    ByteArrayOutputStream out = new ByteArrayOutputStream();
    while ((count = in.read(data, 0, BUFFER)) != -1) {
        out.write(data);
    }       
    InputStream is = new ByteArrayInputStream(out.toByteArray());
    return is;
}

Answer 3

請在下面找到將從 ZIP 存檔中提取所有文件的函數示例。 此功能不適用於子文件夾中的文件：

private static void testZip() {
    ZipInputStream zipStream = null;
    byte buff[] = new byte[16384];
    int readBytes;
    try {
        FileInputStream fis = new FileInputStream("./test.zip");
        zipStream = new ZipInputStream(fis);
        ZipEntry ze;
        while((ze = zipStream.getNextEntry()) != null) {
            if(ze.isDirectory()) {
                System.out.println("Folder : "+ze.getName());
                continue;//no need to extract
            }
            System.out.println("Extracting file "+ze.getName());
            //at this moment zipStream pointing to the beginning of current ZipEntry, e.g. archived file
            //saving file
            FileOutputStream outFile = new FileOutputStream(ze.getName());
            while((readBytes = zipStream.read(buff)) != -1) {
                outFile.write(buff, 0, readBytes);
            }
            outFile.close();                
        }

    } catch (Exception e) {
        System.err.println("Error processing zip file : "+e.getMessage());
    } finally {
        if(zipStream != null)
            try {
                zipStream.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
    }
}

Answer 4

以下適用於FileOutputStream ：

try (ZipInputStream zin = new ZipInputStream(response.getInputStream())) {
        for (ZipEntry zipEntry = zin.getNextEntry(); zipEntry != null; zipEntry = zin.getNextEntry()) {
            File f = new File(dir, zipEntry.getName());
            try (FileOutputStream fout = new FileOutputStream(f)) {
                 int len;
                 byte[] buffer = new byte[1024];
                 while ((len = zin.read(buffer)) > 0) {
                     fout.write(buffer, 0, len);
                 }
                zin.closeEntry();
            }
        }
    }

Answer 5

ZipInputStream允許直接讀取 ZIP 內容：使用getNextEntry()進行迭代，直到找到要讀取的條目，然后從ZipInputStream中讀取。

如果您不想只讀取 ZIP 內容，但需要在傳遞到下一步之前對流應用額外的轉換，則可以使用PipedInputStream和PipedOutputStream 。 這個想法將與此類似（從內存中編寫，甚至可能無法編譯）：

import java.io.PipedInputStream;
import java.io.PipedOutputStream;

public abstract class FilterThread extends Thread {
    private InputStream unfiltered;
    public void setUnfilteredStream(InputStream unfiltered) {
        this.unfiltered = unfiltered;
    }
    private OutputStream threadOutput;
    public void setThreadOutputStream(OutputStream threadOutput) {
        this.threadOutput = threadOutput;
    }    

    // read from unfiltered stream, filter and write to thread output stream
    public abstract void run();
}

...

public InputStream getFilteredStream(InputStream unfiltered, FilterThread filter) {
    PipedInputStream filteredInputStream = new PipedInputStream();
    PipedOutputStream threadOutputStream = new PipedOutputStream(filteredInputStream);

    filter.setUnfilteredStream(unfiltered);
    filter.setThreadOuptut(threadOutputStream);
    filter.start();

    return filteredInputStream;
}

...

public void clientCode() {
    ...
    ZipInputStream zis = ...;// get ZIP stream
    FilterThread filter = ...; // assign your implementation of FilterThread that transforms your ZipInputStream

    InputStream filteredZipInputStream = getFilteredStream(zis, filter);
    ...
}

Answer 6

郵政編碼相當簡單，但我在將 ZipInputStream 作為 Inputstream 返回時遇到了問題。 出於某種原因，zip 中包含的某些文件的字符被丟棄。 以下是我的解決方案，到目前為止它一直在工作。

private Map<String, InputStream> getFilesFromZip(final DataHandler dhZ,
        String operation) throws ServiceFault
{
    Map<String, InputStream> fileEntries = new HashMap<String, InputStream>();
    try
    {

        DataSource dsZ = dhZ.getDataSource();

        ZipInputStream zipIsZ = new ZipInputStream(dhZ.getDataSource()
                .getInputStream());

        try
        {
            ZipEntry entry;
            while ((entry = zipIsZ.getNextEntry()) != null)
            {
                if (!entry.isDirectory())
                {
                    Path p = Paths.get(entry.toString());
                    fileEntries.put(p.getFileName().toString(),
                            convertZipInputStreamToInputStream(zipIsZ));
                }

            }
        }
        finally
        {
            zipIsZ.close();
        }

    }
    catch (final Exception e)
    {
        faultLocal(LOGGER, e, operation);
    }

    return fileEntries;
}
private InputStream convertZipInputStreamToInputStream(
        final ZipInputStream in) throws IOException
{
    ByteArrayOutputStream out = new ByteArrayOutputStream();
    IOUtils.copy(in, out);
    InputStream is = new ByteArrayInputStream(out.toByteArray());
    return is;
}