jquery的ajax用get步骤无法将值传到后台,why?

WBOY
Release: 2016-06-13 12:51:30
Original
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jquery的ajax用get方法无法将值传到后台,why??????
如题:
//以下是js的方法
function login() {
$.ajax({
     type: "get",
     url:'valid.php', 
     data:"username=admin", 
     dataType:"json", 
     success:ajaxComplete(result), 
     error:ajaxError(response)});
   });
}

//以下是valid.php
echo $_GET['username'];
?>

触发login()方法后,页面报错 Notice: Undefined index: username,意思是找不到这个参数,这是为什么呢,求解!!

jquery ajax
------解决方案--------------------
你连 js 都写错了,怎么可以得到结果?
function login() {<br />
  $.ajax({<br />
    type: "get",<br />
    url:'valid.php', <br />
    data:"username=admin", <br />
    dataType:"json", <br />
    success:ajaxComplete, <br />
    error:ajaxError<br />
  });<br />
}<br />
Copy after login

success 和 error 需要的都是回调函数
你要么赋予函数名,要么直接定义匿名函数
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