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ajax返回处理

WBOY
WBOYOriginal
2016-06-23 13:48:521015browse



$.get("pai.php",{bbid:bbid,pid:pid,},function(data){
alert("a:"+data);
                                alert("b:"+data.status);
},"json");
返回值:{"status":1,"info":"\u7ed3\u675f","data":[{"wjc_title":"2\u53a2\u5c0f\u8f7f\u8f66"},{"wjc_title":"3\u53a2\u5c0f\u8f7f\u8f66"}],"jieg":11}
返回值不会处理了,还是对jq不太了解。
我怎么把返回值data里的wjc_title的值插入到div里的li里,有几个wjc_title就循环几个

  • 插入值




  • 插入值1



  • 插入值2





回复讨论(解决方案)

<script type="text/javascript">    var data = {"status":1, "info":"\u7ed3\u675f", "data":[        {"wjc_title":"2\u53a2\u5c0f\u8f7f\u8f66"},        {"wjc_title":"3\u53a2\u5c0f\u8f7f\u8f66"}    ], "jieg":11};    function insert(data) {        if (!jQuery.isArray(data.data)) return false;        var $target = jQuery('div.row').empty();        for (var i = 0; i < data.data.length; ++i) {            $target.append("<ul><li>" + data.data[i].wjc_title + "</li></ul>");        }        return true;    }    insert(data);    </script>

<script type="text/javascript">    var data = {"status":1, "info":"\u7ed3\u675f", "data":[        {"wjc_title":"2\u53a2\u5c0f\u8f7f\u8f66"},        {"wjc_title":"3\u53a2\u5c0f\u8f7f\u8f66"}    ], "jieg":11};    function insert(data) {        if (!jQuery.isArray(data.data)) return false;        var $target = jQuery('div.row').empty();        for (var i = 0; i < data.data.length; ++i) {            $target.append("<ul><li>" + data.data[i].wjc_title + "</li></ul>");        }        return true;    }    insert(data);    </script>


效果没起作用

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"><html> <head>  <meta http-equiv="content-type" content="text/html; charset=utf-8">  <title> New Document </title>  <script src="//code.jquery.com/jquery-1.11.0.min.js"></script> </head> <body>	<div class="row">		<ul>			<li>插入值1</li>		</ul>		<ul>			<li>插入值2</li>		</ul>	<div>  <script type="text/javascript">    var bbid = 1;    var pid = 1;		$.get("pai.php",{bbid:bbid,pid:pid,},function(data){		for(var i=0; i<data.data.length; i++){			$($('.row').find('li').get(i)).html(data.data[i].wjc_title);		}	},"json");  </script> </body></html>


pai.php
<?phpecho '{"status":1,"info":"\u7ed3\u675f","data":[{"wjc_title":"2\u53a2\u5c0f\u8f7f\u8f66"},{"wjc_title":"3\u53a2\u5c0f\u8f7f\u8f66"}],"jieg":11}';?>

本末倒置!
php 作为服务端,理应按客户端的要求返回数据
你连客户端代码都不会写,那还 ajax 做什么?

本末倒置!
php 作为服务端,理应按客户端的要求返回数据
你连客户端代码都不会写,那还 ajax 做什么?


不是把客户端需求传给服务端,服务处理结果在返给前端显示?
客户端代码和JQ不一样吧!
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