SpringRest-创造。压缩文件并将其发送到客户端

Question

我想创建. zip文件，其中包含我从后端收到的压缩文件，然后将此文件发送给用户。两天来，我一直在寻找答案，但找不到合适的解决方案，也许你可以帮我：）

目前，代码是这样的：（我知道我不应该在spring控制器中完成所有工作，但不关心这一点，它只是为了测试目的，以找到使其工作的方法）

    @RequestMapping(value = "/zip")
    public byte[] zipFiles(HttpServletResponse response) throws IOException{
        //setting headers
        response.setContentType("application/zip");
        response.setStatus(HttpServletResponse.SC_OK);
        response.addHeader("Content-Disposition", "attachment; filename=\"test.zip\"");

        //creating byteArray stream, make it bufforable and passing this buffor to ZipOutputStream
        ByteArrayOutputStream byteArrayOutputStream = new ByteArrayOutputStream();
        BufferedOutputStream bufferedOutputStream = new BufferedOutputStream(byteArrayOutputStream);
        ZipOutputStream zipOutputStream = new ZipOutputStream(bufferedOutputStream);

        //simple file list, just for tests
        ArrayList<File> files = new ArrayList<>(2);
        files.add(new File("README.md"));

        //packing files
        for (File file : files) {
            //new zip entry and copying inputstream with file to zipOutputStream, after all closing streams
            zipOutputStream.putNextEntry(new ZipEntry(file.getName()));
            FileInputStream fileInputStream = new FileInputStream(file);

            IOUtils.copy(fileInputStream, zipOutputStream);

            fileInputStream.close();
            zipOutputStream.closeEntry();
        }

        if (zipOutputStream != null) {
            zipOutputStream.finish();
            zipOutputStream.flush();
            IOUtils.closeQuietly(zipOutputStream);
        }
        IOUtils.closeQuietly(bufferedOutputStream);
        IOUtils.closeQuietly(byteArrayOutputStream);
        return byteArrayOutputStream.toByteArray();
    }

但问题是，使用的代码，当我输入URL：localhost:8080/zip我得到文件：test.zip.html而不是. zip文件

我不知道我还能做什么。我还试着用以下内容替换tearrayouputstream：

OutputStream outputStream = response.getOutputStream();

并设置为无效的方法，所以它什么也不返回，但它创建了. zip文件，这是...损坏？

打开测试包后在我的macbook上。zip我正在接受测试。拉链cpgz再次给我测试。压缩文件等等。。

在窗户上。正如我所说，zip文件已损坏，甚至无法打开。

我也认为，自动删除. html扩展名将是最好的选择，但如何呢？希望它没有那么难，因为它似乎是：）谢谢

Answer 1

@RequestMapping(value="/zip", produces="application/zip")
public ResponseEntity<StreamingResponseBody> zipFiles() {
    return ResponseEntity
            .ok()
            .header("Content-Disposition", "attachment; filename=\"test.zip\"")
            .body(out -> {
                var zipOutputStream = new ZipOutputStream(out);

                // create a list to add files to be zipped
                ArrayList<File> files = new ArrayList<>(2);
                files.add(new File("README.md"));

                // package files
                for (File file : files) {
                    //new zip entry and copying inputstream with file to zipOutputStream, after all closing streams
                    zipOutputStream.putNextEntry(new ZipEntry(file.getName()));
                    FileInputStream fileInputStream = new FileInputStream(file);

                    IOUtils.copy(fileInputStream, zipOutputStream);

                    fileInputStream.close();
                    zipOutputStream.closeEntry();
                }

                zipOutputStream.close();
            });
}

Answer 2

@RequestMapping(value="/zip", produces="application/zip")
public void zipFiles(HttpServletResponse response) throws IOException {

    //setting headers  
    response.setStatus(HttpServletResponse.SC_OK);
    response.addHeader("Content-Disposition", "attachment; filename=\"test.zip\"");

    ZipOutputStream zipOutputStream = new ZipOutputStream(response.getOutputStream());

    // create a list to add files to be zipped
    ArrayList<File> files = new ArrayList<>(2);
    files.add(new File("README.md"));

    // package files
    for (File file : files) {
        //new zip entry and copying inputstream with file to zipOutputStream, after all closing streams
        zipOutputStream.putNextEntry(new ZipEntry(file.getName()));
        FileInputStream fileInputStream = new FileInputStream(file);

        IOUtils.copy(fileInputStream, zipOutputStream);

        fileInputStream.close();
        zipOutputStream.closeEntry();
    }    

    zipOutputStream.close();
}

Answer 3

好像解决了。我替换了：

response.setContentType("application/zip");

与：

@RequestMapping(value = "/zip", produces="application/zip")

现在我明白了，很美。压缩文件：）

如果你们中有人有更好或更快的提议，或者只是想给一些建议，那就去吧，我很好奇。