1.先來個簡單的form表單 login.jsp,建在webcontent目錄下（url寫相對路徑就可以了）
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title>Insert title here</title>
</head>
<body>
<form action="login">
賬號：<input type = "text" name = "username"><br>
密碼：<input type = "text" name = "psaaword"><br>
<input type = "submit" value = "登錄">
</form>
</body>
</html>
2.ajax提交數據（url寫相對路徑就可以了）
$("#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..建個login的servlet. eclipse創建的話直接右鍵項目，new-->servlet-->填寫具體信息
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);
自己去寫邏輯處理，愛干嘛干嘛
}
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
}
}