Question

再次使用Java EE。

在示例中

在servlet上我有byte [] fileContent ----＆gt;（file.txt）
我尝试将其添加到zip文件

将servlet设置为

response.setContentType("Content-type: text/zip");
response.setHeader("Content-Disposition",
        "attachment; filename=mytest.zip");

// List of files to be downloaded
List files = new ArrayList();
files.add(new File("C:/first.txt"));

ServletOutputStream out = response.getOutputStream();
ZipOutputStream zos = new ZipOutputStream(new BufferedOutputStream(out));

for (File file : files) {

    System.out.println("Adding " + file.getName());
    zos.putNextEntry(new ZipEntry(file.getName()));

    // Get the file
    FileInputStream fis = null;
    try {
        fis = new FileInputStream(file);

    } catch (FileNotFoundException fnfe) {
        // If the file does not exists, write an error entry instead of
        // file
        // contents
        zos.write(("ERRORld not find file " + file.getName())
                .getBytes());
        zos.closeEntry();
        System.out.println("Couldfind file "
                + file.getAbsolutePath());
        continue;
    }

    BufferedInputStream fif = new BufferedInputStream(fis);

    // Write the contents of the file
    int data = 0;
    while ((data = fif.read()) != -1) {
        zos.write(data);
    }
    fif.close();

    zos.closeEntry();
    System.out.println("Finishedng file " + file.getName());
}

zos.close();

} }

我对第二点有疑问。我尝试了几种解决方案，但他们都很难使用我的servlet或者没有工作。

有人可以通过示例向我展示如何将文件作为字节数组添加到zip并在servlet上作为atachment返回？感谢您的任何提示和建议

- 发布编辑

我尝试过类似的东西，但是如何添加byte []而不是file。

Answer 1

以下是解决方案的示例：

public class ZipFileServlet extends HttpServlet {

private static final int DEFAULT_BUFFER_SIZE = 10240; // 10KB.
private YourFileDAO yourFileDAO = YourDAOFactory.getYourFileDAO();

@Override
protected void doPost(HttpServletRequest request, HttpServletResponse response)
    throws ServletException, IOException
{
    String[] fileIds = request.getParameterValues("fileId");
    response.setContentType("application/zip");
    response.setHeader("Content-Disposition", "attachment; filename=\"allfiles.zip\"");
    ZipOutputStream output = null;
    byte[] buffer = new byte[DEFAULT_BUFFER_SIZE];

    try {
        output = new ZipOutputStream(new BufferedOutputStream(response.getOutputStream(), DEFAULT_BUFFER_SIZE));

        for (String fileId : fileIds) {
            YourFileItem item = yourFileDAO.find(fileId);
            if (item == null) continue; // Handle yourself. The fileId may be wrong/spoofed.
            InputStream input = null;

            try {
                input = new BufferedInputStream(item.getInputStream(), DEFAULT_BUFFER_SIZE);
                output.putNextEntry(new ZipEntry(item.getName()));
                for (int length = 0; (length = input.read(buffer)) > 0;) {
                    output.write(buffer, 0, length);
                }
                output.closeEntry();
            } finally {
                if (input != null) try { input.close(); } catch (IOException logOrIgnore) { /**/ }
            }
        }
    } finally {
        if (output != null) try { output.close(); } catch (IOException logOrIgnore) { /**/ }
    }
}

 }

如何从Servlet上的bytearray返回zip

1 个答案: