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  • 前后端交互技术之servlet与form表单提交请求及ajax提交请求

    1.先来个简单的form表单 login.jsp,建在webcontent目录下(url写相对路径就可以了

    <!DOCTYPE html>
    <html>
    <head>
    <meta charset="UTF-8">
    <title>Insert title here</title>
    </head>
    <body>
    <form action="login">
    账号:<input type = "text" name = "username"><br>
    密码:<input type = "text" name = "psaaword"><br>
    <input type = "submit" value = "登录">
    </form>

    </body>
    </html>

    2.ajax提交数据(url写相对路径就可以了)

    $("#btn").on("click",function(){
        sqlStatement = document.getElementById("sqlStatement").value;
        $.ajax({
            type: "GET",
            url : "login",
            dataType:"text",
            async: false,
            data:{sql: sqlStatement, pCount:pageCount},
            success: function(resultData){
                jObject = JSON.parse(resultData);
                document.getElementById("result").innerHTML = jObject.table;
            },
            error:function(xhr, status, errMsg){
                alert("Data transmission failed!");
            }
        });

    3..建个login的servlet. eclipse创建的话直接右键项目,new-->servlet-->填写具体信息

     

    package com.tcc;

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

    /**
    * Servlet implementation class loginServlet
    */
    public class loginServlet extends HttpServlet {

    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    String name = request.getParameter("username");
    String password = request.getParameter("psaaword");
    System.out.println("username"+name);
    System.out.println("psaaword"+password);

    自己去写逻辑处理,爱干嘛干嘛

    }


    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

    }

    }

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  • 原文地址:https://www.cnblogs.com/tongcc/p/11439675.html
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