6

I use Servlet 3 @MultiPartConfig annotation to implement file upload in my application. I need set multipart-config location parameter at the runtime (not hardcode in annotaion parameter). Is there any API for programmatic access to multipart-config of servlet?

Thanks

4

2 回答 2

5

@MultiPartConfig 实际上只是容器的标记接口。当 servlet 初始化时，提供的注释值通过代理对象映射到它。当传入的请求是 multipart/form-data 时，上传的部分将映射到请求，并且容器根据注释中的值和请求中的部分执行必要的工作。您无法拦截此过程，因为这一切都发生在容器的内部。但是，还有一种选择。它需要第二次执行文件系统操作。由于您拥有所有部分，因此您可以重建文件并将其“重新上传”到您选择的位置。它可能看起来像下面的方法。

@Override
protected void doPost(HttpServletRequest httpServletRequest, HttpServletResponse httpServletResponse)
        throws ServletException, IOException {

    httpServletResponse.setContentType("text/html");
    PrintWriter printWriter = httpServletResponse.getWriter();

    InputStream inputStream;
    FileOutputStream fileOutputStream;

    for (Part part : httpServletRequest.getParts()) {

        inputStream = httpServletRequest.getPart(part.getName()).getInputStream();
        int i = inputStream.available();
        byte[] b = new byte[i];
        inputStream.read(b);
        String fileName = "";

        for (String temp : part.getHeader("content-disposition").split(";")) {
            if (temp.trim().startsWith("filename")) {
                fileName = temp.substring(temp.indexOf('=') + 1).trim().replace("\"", "");
            }
        }

        String uploadDir = "/temp";
        fileOutputStream = new FileOutputStream(uploadDir + "/" + fileName);
        fileOutputStream.write(b);
        inputStream.close();
        fileOutputStream.close();

        printWriter.write("Uploaded file " + uploadDir + "/" + fileName + ".");
    }
}
于 2013-10-13T02:02:18.980 回答
1

我也遇到了同样的问题，解决方案很简单：标准 Servlet 3.0 文件上传是不够的：只需从 Apache FileUpload Commons 获取 jar 即可

看看这个带有 Streaming API 的清晰示例

   ServletFileUpload upload = new ServletFileUpload();

   // Parse the request
    FileItemIterator iter = upload.getItemIterator(request);
    while (iter.hasNext()) {
      FileItemStream item = iter.next();
      String name = item.getFieldName();
      InputStream stream = item.openStream();
      if ( item.isFormField() == false) 
          System.out.println("File field " + name + " with file name "
            + item.getName() + " detected."); 
          FileOutputStream fos = new FileOutputStream("your_location"); 
          Streams.copy ( stream,  fos, true);

       }
    }
于 2014-02-08T05:04:53.163 回答