Home > Backend Development > PHP Tutorial > ajax-我这段代码本页面可显示出传递的信息,但是在被传递页面中却无法接收到信息呢?谢谢!

ajax-我这段代码本页面可显示出传递的信息,但是在被传递页面中却无法接收到信息呢?谢谢!

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Release: 2016-06-02 11:34:01
Original
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ajaxjavascriptpostjqueryphp

<code> <meta http-equiv="Content-Type" content="text/html; charset=gb2312">
<title>Untitled Document</title>
<script type="text/javascript" src="jquery-1.3.2.js"></script><script type="text/javascript" src="json2.js"></script><script language="javascript">function checkemail(){  if($('#email').val() == ""){    $('#msg').html("please enter the email!");    $('#email').focus;    return false;  }  if($('#address').val() == ""){    $('#msg').html("please enter the address!");    $('#address').focus;    return false;  }  ajax_post(); }function ajax_post(){  $.post("http://localhost:8086/a1.1.php",{email:$('#email').val(),address:$('#address').val()},  function(data){    //$('#msg').html("please enter the email!");    //alert(data);    $('#msg').html(data);  },  "json");//杩</script></code>
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