Question

我在JSP中有3个按钮。我需要每个按钮来单击调用不同的Java方法，而不刷新页面。搜索一下，看起来我需要servlets / AJAX，但我从来没有使用过这些。如果我也可以在按下按钮时调用相应的JavaScript函数，这将会很有帮助。我现在的代码适用于每个按钮以调用正确的方法，但刷新页面是为了这样做。

我的JSP代码：

<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
    pageEncoding="ISO-8859-1"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Insert title here</title>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<script>
    $('.btn').click(function (e) {
        e.preventDefault();
        $.post("http://localhost:8080/PracticeProject/myservlet",
            {button: $(this).val()}).
            done(function( response ) {
                console.log(response);
            });
        });
    });
</script>
</head>
<body>
    <video id="myVideo" width="512" height="384" controls autoplay>
        <source id="videoPlayer" src="videos/Cold.mp4" type="video/mp4">
    </video>
    <form onsubmit="return false;" id="myForm">
        <button class="submit" value="button1">Button 1</button>
        <button class="submit" value="button2">Button 2</button>
        <button class="submit" value="button3">Button 3</button>
    </form>
    <button class="btn" value="button1">Button 1</button>
    <button class="btn" value="button2">Button 2</button>
    <button class="btn" value="button3">Button 3</button>
    <form action="${pageContext.request.contextPath}/myservlet" method="post">
        <button type="submit" name="button" value="button1" action="${pageContext.request.contextPath}/myservlet" method="post">Button 1</button>
        <button type="submit" name="button" value="button2" action="${pageContext.request.contextPath}/myservlet" method="post">Button 2</button>
        <button type="submit" name="button" value="button3" action="${pageContext.request.contextPath}/myservlet" method="post">Button 3</button>
    </form>

    <script>



</body>
</html>

我的Servlet代码：

import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

/**
 * Servlet implementation class MyServlet
 */
@WebServlet("/myservlet")
public class MyServlet extends HttpServlet {
    private static final long serialVersionUID = 1L;

    /**
     * @see HttpServlet#HttpServlet()
     */
    public MyServlet() {
        super();
        // TODO Auto-generated constructor stub
    }

    /**
     * @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
     */
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        // TODO Auto-generated method stub
        response.getWriter().append("Served at: ").append(request.getContextPath());
    }

    /**
     * @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
     */
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        MyClass myClass = new MyClass();
        String button = request.getParameter("button");
        System.out.println("In the servlet");

        if ("button1".equals(button)) {
            myClass.method1();
        } else if ("button2".equals(button)) {
            myClass.method2();
        } else if ("button3".equals(button)) {
            myClass.method3();
        } else {
            // ???
        }
    }

}

我的班级代码：

public class MyClass {

    public void method1(){
        // do method 1
        System.out.println("method 1");
    }
    public void method2(){
        // do method 2
        System.out.println("method 2");
    }
    public void method3(){
        // do method 3
        System.out.println("method 3");
    }
}

Answer 1

您可以使用纯AJAX或jQuery等框架。这是一个简单而非常基本的jQuery示例，它将数据发布到您的servlet，但在此之前，您需要在HTML中进行一些小修改：

<form method="post">
        <button class="submit" value="button1">Button 1</button>
        <button class="submit" value="button2">Button 2</button>
        <button class="submit" value="button3">Button 3</button>
</form>

然后：

$('.submit').click(function(){
    $.post("${pageContext.request.contextPath}/myservlet", {button: $(this).val()}, function( data ) {
        alert( "Done!" );
    });
    return false;
});

Answer 2

您可以使用Ajax的Post方法将请求发送到服务器。首先，您应该捕获表单的提交事件并阻止默认提交，因为页面被重新加载。其次删除 request.getRequestDispatcher("/theJSP.jsp").forward(request, response); 因为如果你正在做一个ajax请求，你的服务器将返回html代码。

下面是更新的ajax代码，使用单击处理程序，而不是表单提交

<!DOCTYPE html>
<html lang="en">
<head>
  <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
</head>
<body>
<button class="btn" value="button1">Button 1</button>
<button class="btn" value="button2">Button 2</button>
<button class="btn" value="button3">Button 3</button>
</body>
<script>
$('.btn').click(function (e) {
                e.preventDefault();
        $.post("http://localhost:8080/PracticeProject/myservlet",
                 {button: $(this).val()}).
                done(function( response ) {
                    console.log(response);
                });
            });
    </script>

这完全符合我的目的。请确保您的Web服务器正在运行。

在按钮单击时调用Java方法，而不是刷新JSP

2 个答案: