aJax get和post
阿新 • • 發佈：2018-11-01
<%@ page language="java" pageEncoding="utf-8"%> <!DOCTYPE html> <html> <head> <meta http-equiv="Content-Type" content="text/html;charset=utf-8"> <title>aJaxRequest!!!</title> <script type="text/javascript"> function ajaxReq(){//獲取使用者請求資料 var uname=document.getElementById("uname").value; var data="uname="+uname; //建立ajax引擎物件 var ajax; if(window.XMLHttpRequest){ ajax=new XMLHttpRequest(); }else if(window.ActiveXObject){ ajax=new ActiveXObject("Msxml2.XMLHTTP"); } //複寫onreadystatechange函式 ajax.onreadystatechange=function(){ //判斷ajax狀態碼 if(ajax.readyState==4){ //判斷響應狀態碼 if(ajax.status==200){ //獲取響應內容 varresult=ajax.responseText; //處理響應內容 alert(result); } } } //get方法 //ajax.open("get","ajaxServlet",true); //ajax.send(null); //pos方法 ajax.open("post","ajaxServlet",true); ajax.setRequestHeader("Content-Type","application/x-www-form-urlencoded"); ajax.send(data); } </script> <style type="text/css"> #myDiv{ border:solid 4px; width:100px; height:100px; } </style> </head> <body> <h3>aJax!!</h3> <input id="uname" type="text" name="uname" value=""> <button id="myButton" onclick="ajaxReq()" />show!!!</button> <div id="myDiv"></div> </body> </html>