servlet在GWT中如何工作？ [英] How exactly servlet Work in GWT?

问题描述

我试图找出servlet如何工作。
我使用此代码设计我的servlet

客户端 ！

formPanel.setAction（GWT.getModuleBaseURL（）+fileupload）;

/ p>

formPanel.Sumit（）;

$

b $ b

当我点击o提交按钮，我可以你选择test.doc在开发模式。

请有人帮助。

源代码。
客户。

  final FormPanel formPanel = new FormPanel 
 formPanel.addFormHandler（new FormHandler（）{
 
 public void onSubmitComplete（final FormSubmitCompleteEvent event）{
 // TODO自动生成方法存根
 Window.alert event.getResults（））; 
} 
 
 public void onSubmit（final FormSubmitEvent event）{
 // TODO自动生成方法存根
 event.setCancelled ）; 
} 
}）; 
 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文件上传）; 
 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自动生成的catch块
 e.printStackTrace 
} catch（IOException e）{
 // TODO自动生成的catch块
 e.printStackTrace（）; 
} catch（Exception e）{
 throw new RuntimeException（）; 
} 
} 
} 
  

解决方案>

  form.setMethod（FormPanel.METHOD_POST）; //将生成< form method =post>< / form> 
 form.setAction（GWT.getModuleBaseURL（）+fileupload）; 
 //现在< form method =postaction =domain / testapp / fileupload>< / form> 
 

因此，当您单击提交时，其路径将匹配 fileUploaderServler url pattern，因此 com.testapp.server.FileUpload.doPost（HttpServletRequest请求，HttpServletResponse响应）;将执行。

I am try to find out How servlet work. I used this code to design my servlet

client!

formPanel.setAction(GWT.getModuleBaseURL()+"fileupload");

and on click

formPanel.Sumit();

server!

in Server, i didnt Understand how this doPost method will be called by the client.

When i click o submit button , i can "you selected test.doc" in development mode.

Please someone help.

Source Code. Client.

   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();
        }
    });

Server

    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();
    }
}
 }

解决方案

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>

So when you click submit its path will match the fileUploaderServler url pattern, consequently com.testapp.server.FileUpload.doPost(HttpServletRequest request, HttpServletResponse response); will be executed.

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