Question

jquery ajax为什么总是返回undefined ajax代码 function GetU

jquery ajax为什么总是返回undefined

ajax代码
function GetUserInfo() {
$.ajax({
type: "POST",
cache: false,
data: "name=aaa" ,
async: false,
url:“1.php”,
success: function(userinfo) {
alert(userinfo);
},
});
}
php代码
<?php
$user=$_POST["AAA"]
echo $user;
?>

Answer 1

把php改为$_POST["name"]

Answer 2

name=aaa改成name:aaa

Answer 3

name=aaa改成｛name:"aaa"}