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toJSONF步骤的使用

发布时间: 2012-10-11 10:16:10 作者: rapoo

toJSONF方法的使用

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd"><html>    <head>        <meta http-equiv="Content-Type" content="text/html; charset=utf-8">        <script type="text/javascript" src="lib/jquery/jquery-1.4.2.js"></script><!--<script type="text/javascript" src="lib/jquery/json_parse.js"></script>--><script type="text/javascript" src="lib/jquery/json2.js"></script>        <title>toJSON方法研究参考http://www.popo4j.com/article/Javscript-parsing-JSON.html</title>        <script type='text/javascript'>            $(function(){                        })        </script>        <script type="text/javascript">                                           </script><script>var jsonstring = '{"name":"axis", "org":"ph4nt0m", "blog":"http://hi.baidu.com/aullik5"}';//格式上是有要求的http://www.cnblogs.com/kenn/archive/2010/07/08/1773470.html//var ok = json_parse(jsonstring);   // 在ie里必须引入包json_parse.jsvar ok=JSON.parse(jsonstring);// 浏览器内置的APIalert(ok.blog);</script><script type="text/javascript">  var t="{'firstName': 'cyra', 'lastName': 'richardson', 'address': { 'streetAddress': '1 Microsoft way', 'city': 'Redmond', 'state': 'WA', 'postalCode': 98052 },'phoneNumbers': [ '425-777-7777','206-777-7777' ] }";  //这种表示方法是错误的  var t = '{"firstName": "cyra", "lastName": "richardson", "address": { "streetAddress": "1 Microsoft way", "city": "Redmond", "state": "WA", "postalCode": 98052 },"phoneNumbers": [ "425-777-7777","206-777-7777" ] }';  //var obj=json_parse(t);  var obj=JSON.parse(t);    alert(obj.firstName);  alert(obj.lastName);  </script>    </head>    <body>    有两个类库可以使用,一个是json2.js,另一个是json_parse.js参考文献有http://www.popo4j.com/article/Javscript-parsing-JSON.htmlhttp://blog.csdn.net/loseone/archive/2009/05/20/4203300.aspxhttp://www.cnblogs.com/kenn/archive/2010/07/08/1773470.html    </body></html>

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