Question

我想将多个文件压缩为zip文件，我正在处理大文件，然后将它们下载到客户端，目前我正在使用此文件：

@RequestMapping(value = "/download", method = RequestMethod.GET, produces = "application/zip")
public ResponseEntity <StreamingResponseBody> getFile() throws Exception {
    File zippedFile = new File("test.zip");
    FileOutputStream fos = new FileOutputStream(zippedFile);
    ZipOutputStream zos = new ZipOutputStream(fos);
    InputStream[] streams = getStreamsFromAzure();
    for (InputStream stream: streams) {
        addToZipFile(zos, stream);
    }
    final InputStream fecFile = new FileInputStream(zippedFile);
    Long fileLength = zippedFile.length();
    StreamingResponseBody stream = outputStream - >
        readAndWrite(fecFile, outputStream);

    return ResponseEntity.ok()
        .header(HttpHeaders.ACCESS_CONTROL_EXPOSE_HEADERS, HttpHeaders.CONTENT_DISPOSITION)
        .header(HttpHeaders.CONTENT_DISPOSITION, "attachment;filename=" + "download.zip")
        .contentLength(fileLength)
        .contentType(MediaType.parseMediaType("application/zip"))
        .body(stream);
}

private void addToZipFile(ZipOutputStream zos, InputStream fis) throws IOException {
    ZipEntry zipEntry = new ZipEntry(generateFileName());
    zos.putNextEntry(zipEntry);
    byte[] bytes = new byte[1024];
    int length;
    while ((length = fis.read(bytes)) >= 0) {
        zos.write(bytes, 0, length);
    }
    zos.closeEntry();
    fis.close();
}

在压缩所有文件然后开始下载之前，这需要花费很多时间，对于大型文件，此kan需要花费很多时间，这是造成延迟的原因：

while ((length = fis.read(bytes)) >= 0) {
    zos.write(bytes, 0, length);
}

有没有办法在压缩文件的同时立即下载文件？

Answer 1

试试看。而不是使用ZipOutputStream来包装FileOutputStream，而是将zip写入文件，然后将其复制到客户端输出流，而不是使用ZipOutputStream来包装客户端输出流，这样当您添加zip条目和数据时，它将直接发送到客户端。如果您还想也将其存储到服务器上的文件中，则可以将ZipOutputStream写入拆分输出流，以一次写入两个位置。

@RequestMapping(value = "/download", method = RequestMethod.GET, produces = "application/zip")
public ResponseEntity<StreamingResponseBody> getFile() throws Exception {

    InputStream[] streamsToZip = getStreamsFromAzure();

    // You could cache already created zip files, maybe something like this:
    //   String[] pathsOfResourcesToZip = getPathsFromAzure();
    //   String zipId = getZipId(pathsOfResourcesToZip);
    //   if(isZipExist(zipId))
    //     // return that zip file
    //   else do the following

    StreamingResponseBody streamResponse = clientOut -> {
        FileOutputStream zipFileOut = new FileOutputStream("test.zip");

        ZipOutputStream zos = new ZipOutputStream(new SplitOutputStream(clientOut, zipFileOut));
        for (InputStream in : streamsToZip) {
            addToZipFile(zos, in);
        }
    };

    return ResponseEntity.ok()
            .header(HttpHeaders.ACCESS_CONTROL_EXPOSE_HEADERS, HttpHeaders.CONTENT_DISPOSITION)
            .header(HttpHeaders.CONTENT_DISPOSITION, "attachment;filename=" + "download.zip")
            .contentType(MediaType.parseMediaType("application/zip")).body(streamResponse);
}


private void addToZipFile(ZipOutputStream zos, InputStream fis) throws IOException {
    ZipEntry zipEntry = new ZipEntry(generateFileName());
    zos.putNextEntry(zipEntry);
    byte[] bytes = new byte[1024];
    int length;
    while ((length = fis.read(bytes)) >= 0) {
        zos.write(bytes, 0, length);
    }
    zos.closeEntry();
    fis.close();
}

public static class SplitOutputStream extends OutputStream {
    private final OutputStream out1;
    private final OutputStream out2;

    public SplitOutputStream(OutputStream out1, OutputStream out2) {
        this.out1 = out1;
        this.out2 = out2;
    }

    @Override public void write(int b) throws IOException {
        out1.write(b);
        out2.write(b);
    }

    @Override public void write(byte b[]) throws IOException {
        out1.write(b);
        out2.write(b);
    }

    @Override public void write(byte b[], int off, int len) throws IOException {
        out1.write(b, off, len);
        out2.write(b, off, len);
    }

    @Override public void flush() throws IOException {
        out1.flush();
        out2.flush();
    }

    /** Closes all the streams. If there was an IOException this throws the first one. */
    @Override public void close() throws IOException {
        IOException ioException = null;
        for (OutputStream o : new OutputStream[] {
                out1,
                out2 }) {
            try {
                o.close();
            } catch (IOException e) {
                if (ioException == null) {
                    ioException = e;
                }
            }
        }
        if (ioException != null) {
            throw ioException;
        }
    }
}

对于第一个要压缩的资源集的请求，您将不知道生成的zip文件的大小，因此您无法将长度和响应一起发送，因为您在压缩文件时正在流式传输文件。

但是，如果您希望重复请求压缩同一组资源，则可以缓存zip文件，并在随后的任何请求中简单地将它们返回；您还将知道缓存的zip文件的长度，因此也可以在响应中发送该文件。

如果要执行此操作，则必须能够为要压缩的资源的每种组合一致地创建相同的标识符，以便可以检查这些资源是否已压缩，并返回缓存的文件（如果它们已被压缩）。是。您也许可以对将要压缩的资源的ID（可能是完整路径）进行排序，然后将它们连接起来以创建zip文件的ID。

无需使用byte []即可立即从流中获取Java zip文件

1 个答案: