Question

我在前向方法调度程序中收到错误。这是我的示例代码，我收到错误消息。我需要使用s2.java文件显示名称。

它显示了html页面，我把名字放在第一位和最后一位，然后提交它显示错误。

＆＃13;

/*this is s1.java(first java file)*/
package rqdis;

import java.io.IOException;
import java.io.PrintWriter;

import javax.servlet.RequestDispatcher;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

public class S1 extends HttpServlet{
	public void doPost(HttpServletResponse rs,HttpServletRequest rq)
	throws ServletException,IOException{
		try{
			rs.setContentType("text/html");
			PrintWriter p=rs.getWriter();
			String s1=rq.getParameter("n1");
			String s2=rq.getParameter("n2");
			p.print(s1+" "+s2);
		RequestDispatcher rd=rq.getRequestDispatcher("/fff");
		rd.forward(rq, rs);
		p.print("am from server 1.!");
		p.close();
		}catch (Exception e){
			e.printStackTrace();
		}
	}
}
/*this is s2.java(second java file)*/
package rqdis;

import java.io.IOException;
import java.io.PrintWriter;

import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

public class s2 extends HttpServlet{
	public void doPost(HttpServletRequest rq,HttpServletResponse rs)
	throws ServletException,IOException{
		try{
		rs.setContentType("text/html");
		PrintWriter p=rs.getWriter();
		String s1=rq.getParameter("n1");
		String s2=rq.getParameter("n2");
		p.print(s1+" "+s2);
		p.print("am from server 2.!");
		p.close();
		}catch (Exception e){
			e.printStackTrace();
		}
	}
}

＆＃13;

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
  <display-name>rqdis</display-name>
  <welcome-file-list>
    <welcome-file>main.html</welcome-file>
    </welcome-file-list>
    <servlet>
    <servlet-name>firstservlet</servlet-name>
    <servlet-class>rqdis.S1</servlet-class>
    </servlet>
    <servlet-mapping>
    <servlet-name>firstservlet</servlet-name>
    <url-pattern>/ff</url-pattern>
    </servlet-mapping>
    <servlet>
    <servlet-name>secondservlet</servlet-name>
    <servlet-class>rqdis.s2</servlet-class>
    </servlet>
    <servlet-mapping>
    <servlet-name>secondservlet</servlet-name>
    <url-pattern>/fff</url-pattern>
    </servlet-mapping>
</web-app>

＆＃13;

<html>

<head>
<meta charset="ISO-8859-1">
<title>disfor</title>
</head>

<body>
<form action="forward metod" method="post">
    First name:
    <input type="text" name="n1"> Last name:
    <input type="text" name="n2">
    <input type="submit" value="Submit">
</form>
</body>

</html>

＆＃13;

Answer 1

当您转发到另一个servlet时，不应该在servlet中写入，因为响应将在另一个servlet中提交。

参考： Cause of Servlet's 'Response Already Committed'

Answer 2

正如我从代码中看到的那样，您将获得的错误是404（未找到）。因为您的表单操作是forward metod在web.xml文件中没有url映射。表单操作用于指定应在何处进行请求的URL。尝试将表单操作更改为action="/ff"。

在调度程序servlet的正向方法上获得错误

2 个答案: