Question

我花了10年时间编写ASP / Ajax并且必须切换到JAVA。我开始非常简单的脚本，以便在深入了解更复杂的东西之前清楚地理解这些概念。我正在使用Tomcat 7，Servlet 3

问题是。成功调用servlet但不传输表单数据

这是编辑过的servlet（Main.java）

import java.io.IOException;
import java.io.PrintWriter;

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

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

/**
 * @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
 */
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws    ServletException, IOException {
       // TODO Auto-generated method stub
    //int ids;
    //String ids;
    //response.setContentType("text/html;charset=UTF-8");

    //PrintWriter out = response.getWriter();
    String val = request.getParameter("id");
    String name = request.getParameter("name");
    //response.getWriter().write(val);
    //response.getWriter().write("::");
    //response.getWriter().write(name);

    response.setContentType("text/plain");
    response.setCharacterEncoding("UTF-8");
    response.getWriter().write(val +" "+ name);

    //System.out.print(val);
    //if(val != null){
     //   //ids = Integer.parseInt(val);
      //  //out.print(ids);
       // out.print(val);
    //}
}
}

这是编辑过的JSP（Main.jsp）：

<!DOCTYPE html>
<html lang="en">
<head>
    <title>SO question 4112686</title>
    <script src="http://code.jquery.com/jquery-latest.min.js"></script>

    <script>
$.ajaxSetup({ cache: false });
$(document).ready(function() {   
$('#submit').click(function(event) {
    var form = (event.target.form),
        url = "/Main"
            + "?lName=" + escape(form.elements.lName.text)
            + "&fName=" + escape(form.elements.fName.text);

    $.get(url, function(getData) { 
        $('#somediv').text(getData);         
    });
});
});
</script>

</head>
<body>
<form id="Main" action="Main" method="post">
<h4>AJAX Demo using Jquery in JSP and Servlet</h4>
Enter your Name:
<br/><input type="text" id="fName" name="fName"/>
<br/><input type="text" id="lName" name="lName"/>
<br/><input type="submit" id="submit" value="Ajax Submit"/>
<br/>
</form>

<div id="somediv">....</div>
</body>
</html>

Answer 1

编辑：如果未传输表单数据，那是因为您没有将其发送到服务器。 GET请求意味着您需要在网址中附加lName=abc&fName=def：

<script>
$.ajaxSetup({ cache: false });
    $(document).ready(function() {   
        $('#submit').click(function(event) {
            var form = (event.target.form),
                url = "/SomeServlet"
                    + "?lName=" + escape(form.elements.lName.text)
                    + "&fName=" + escape(form.elements.fName.text);

            $.get(url, function(getData) { 
                $('#somediv').text(getData);         
            });
        });
    });
</script>

使用ajax将表单数据从jsp发送到servlet

1 个答案: