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");
}
}
答案 0 :(得分:0)
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;
});
答案 1 :(得分:0)
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。