如何将 ZipInputStream 转换为 InputStream？

Question

我有代码，其中 ZipInputSream 转换为 byte[]，但我不知道如何将其转换为输入流。

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 的子类。

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

Answer 2

这是我解决这个问题的方法。现在我可以将单个文件作为 InputStream 从 ZipInputStream 获取到内存。

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

请查看下面的函数示例，该函数将从 ZIP 存档中提取所有文件。此函数不适用于子文件夹中的文件：

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

以下适用于我的 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 允许直接读取 ZIP 内容：使用 getNextEntry() 进行迭代，直到找到要读取的条目，然后从 ZipInputStream 中读取。

如果您不想只读取 ZIP 内容，但需要在进入下一步之前对流应用额外的转换，则可以使用 PipedInputStream 和 PipedOutputStream。这个想法与此类似（从内存中编写，甚至可能无法编译）：

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

邮政编码相当简单，但我在将 ZipInputStream 作为输入流返回时遇到了问题。由于某种原因，zip 中包含的某些文件的字符被删除。以下是我的解决方案，到目前为止它一直有效。

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;
}