Ajax提交表单并接收json实例代码
需求：
实现点击按钮后，数据以表单形式提交至服务器，并接收来自服务器的返回数据。过程中页面不刷新。
html代码
<html xmlns="http://www.w3.org/1999/xhtml"> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <script src="https://cdn.bootcss.com/jquery/1.12.4/jquery.min.js"></script> <script src="./testajaxjs.js"></script> <head> </head> <body> <form id="form1"> <p>xingming:<input type="text" name="xingming"/></p> <p>nianling:<input type="text" name="nianling"/></p> </form> <button type="button" id="mybt" onclick="mysubmmit()"> ajax提交 </button> </body> </html>
js代码
function mysubmmit(){
$.ajax({
type: "POST",
url: "testajaxend.php",
data: $('#form1').serialize(),
async: false,
success: function(databack){
//console.log("chenggong");
console.log(databack);
},
error: function(request){
console.log("shibaile");
}
});
}后端php代码
<?php
$name = $_POST['xingming'];
$age = $_POST['nianling'];
$myarray = array("name"=>$name, "age"=>$age);
$myjson = json_encode($myarray);
echo $myjson;
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