用jquery ajax获取网站Alexa排名的代码

  

复制代码 代码如下:

  <html>

  <head>

  <title></title>

  <script type="text/javascript" src="http://dl.glzy8.com/img/jslib/jquery/jquery-1.3.2.min.js"></script>

  <script type="text/javascript">

  $(function(){

  $("#siteName").blur(function(){

  $.ajax({

  type: "GET",

  url: "http://data.alexa.com/data/?cli=10&dat=snba&ver=7.0&url="+$("#siteName").val(),

  dataType: "xml",

  success: function(xml)

  {

  $("#count").text($(xml).find("ALEXA SD POPULARITY").attr("TEXT"));

  }

  })

  });

  });

  </script>

  </head>

  <body>

  <input type="text" id="siteName" />

  <span id="count"></span>

  </body>

  </html>