Question

我试着找出servlet是如何工作的。我用这段代码来设计我的servlet

client！

formPanel.setAction（GWT.getModuleBaseURL（）+ “文件上传”）;

并点击

formPanel.Sumit（）;

server！

在Server中，我没有理解客户端如何调用这个doPost方法。

当我点击提交按钮时，我可以在开发模式中“选择test.doc”。

请有人帮忙。

源代码。客户端。

   final FormPanel formPanel = new FormPanel();
    formPanel.addFormHandler(new FormHandler() {

        public void onSubmitComplete(final FormSubmitCompleteEvent event) {
            // TODO Auto-generated method stub
            Window.alert(event.getResults());
        }

        public void onSubmit(final FormSubmitEvent event) {
            // TODO Auto-generated method stub
            event.setCancelled(true);
        }
    });
 final FileUpload upload = new FileUpload();
 formPanel.setMethod(FormPanel.METHOD_POST);
    formPanel.setEncoding(FormPanel.ENCODING_MULTIPART);
    formPanel.setAction(GWT.getModuleBaseURL()+"fileupload");
             formPanel.setWidget(upload);

      Button btnAdd = new Button("Add");

        btnAdd.addClickHandler(new ClickHandler() {
        public void onClick(ClickEvent event) {
            GWT.log("you selected " + upload.getFilename(), null);
            formPanel.submit();
        }
    });

服务器

    public class FileUpload extends HttpServlet {

public void dopost(HttpServletRequest request,HttpServletResponse response){
    ServletFileUpload upload = new ServletFileUpload();
    System.out.println("pratyush file upload");
    try {
        FileItemIterator iterator = upload.getItemIterator(request);

        while (iterator.hasNext()){
            FileItemStream itemStream = iterator.next();

            String name = itemStream.getFieldName();
            InputStream stream = itemStream.openStream();

            ByteArrayOutputStream outputStream = new ByteArrayOutputStream();
            int len;
            byte[] buffer = new byte[8192];
            while ((len = stream.read(buffer, 0, buffer.length)) != -1) {
                outputStream.write(buffer, 0, len);

            }

            int maxFileSize = 2*(1024*1024); 
               if (outputStream.size() > maxFileSize) { 
                   throw new RuntimeException("File is > than " + maxFileSize);
               }

        }
    } catch (FileUploadException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }catch(Exception e){
        throw new RuntimeException();
    }
}
 }

Answer 1

form.setMethod(FormPanel.METHOD_POST);  //will generate <form method="post"></form>
form.setAction(GWT.getModuleBaseURL()+"fileupload"); 
// and now <form method="post" action="domain/testapp/fileupload"></form>

因此，当您点击提交时，其路径将与fileUploaderServler网址格式匹配，因此com.testapp.server.FileUpload.doPost(HttpServletRequest request, HttpServletResponse response);将被执行。

servlet在GWT中的工作原理是什么？

1 个答案: