Question

以下是我的servlet的get（）和post（）方法。我是servlet的新手。当我从我的客户端调用时，这个servlet创建一个新页面并覆盖我客户端中的所有html元素。我想做的就是留在我的html页面中，让servlet独自完成他的工作。

public void doPost(HttpServletRequest request, 
               HttpServletResponse response)
              throws ServletException, java.io.IOException {
      // Check that we have a file upload request
      isMultipart = ServletFileUpload.isMultipartContent(request);

      DiskFileItemFactory factory = new DiskFileItemFactory();
      // maximum size that will be stored in memory
      factory.setSizeThreshold(maxMemSize);
      // Location to save data that is larger than maxMemSize.
      factory.setRepository(new File("c:\\temp"));

      // Create a new file upload handler
      ServletFileUpload upload = new ServletFileUpload(factory);
      // maximum file size to be uploaded.
      upload.setSizeMax( maxFileSize );

      try{ 
      // Parse the request to get file items.
      List fileItems = upload.parseRequest(request);

      // Process the uploaded file items
      Iterator i = fileItems.iterator();

      while ( i.hasNext () ) 
      {
         FileItem fi = (FileItem)i.next();
         if ( !fi.isFormField () )  
         {
            // Get the uploaded file parameters
            String fieldName = fi.getFieldName();
            String fileName = fi.getName();
            String contentType = fi.getContentType();
            boolean isInMemory = fi.isInMemory();
            long sizeInBytes = fi.getSize();
            // Write the file
            if( fileName.lastIndexOf("\\") >= 0 ){
               file = new File( filePath + 
               fileName.substring( fileName.lastIndexOf("\\"))) ;
            }else{
               file = new File( filePath + 
               fileName.substring(fileName.lastIndexOf("\\")+1)) ;
            }
            fi.write( file ) ;

         }
      }

   }catch(Exception ex) {
       System.out.println(ex);
   }
   }
   public void doGet(HttpServletRequest request, 
                       HttpServletResponse response)
        throws ServletException, java.io.IOException {

        throw new ServletException("GET method used with " +
                getClass( ).getName( )+": POST method required.");
   }

Answer 1

我想要做的就是留在我的html页面中，让servlet独自完成他的工作。

这是使用Asynchronous call AKA Ajax的完美候选者。

学习AJAX。

http://api.jquery.com/jquery.ajax/

How to use Servlets and Ajax?

Answer 2

如果您希望servlet应该进行处理并且它应该显示html页面，那么您可以在servlet的try块末尾使用RequestDispatcher。因此，当点击你的html页面中的任何动作时，它将转到servlet并执行适当的操作，并且它将再次重定向到html。

RequestDispatcher rd = request.getRequestDispatcher("index.html");
            rd.include(request, response);

如何避免servlet响应新的HTML页面？

2 个答案: