2011-12-19 64 views
0

我试图将jquery导入codeigniter 2.X而不退出。我把链接放到库中,我在调试器中看到它加载它没有错误,但是,当我使用Docuemtn就绪功能时,它不起作用。有谁能够帮助我?将Jquery导入codeigniter 2.X

这是代码:

<html> 
<head> 
    <script src="//ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js" type="text/javascript"></script> 
    <script type="javascript"> 
     $(document).ready(function() { 
      $("a").click(function() { 
       alert("Hello world!"); 
      }); 
     }); 

    </script> 
    <title>Hardware Deal</title> 
</head> 
<body> 
    <div id='header'> 
     <h1> Hardware deal </h1> 
    </div> 
    <div id='body'> 
     <form action='' method = 'post'> 
      Customer Name:<input type='text' name='Cname' id='CName'/><br> 
      Adress:<input type='text' name='CAddress'/><br> 
      Existing Customer? yes<input type='radio' name='existing' value='yes'> no<input type='radio' name='existing' value='no'><br> 

      Printer Contition: <select name='PCondition'/> 
           <option value=''></option> 
           <option value='New Press'>New Press</option> 
           <option value='Second Hand'>Second Hand</option> 
           </select> 
     </form> 
     <a href="">Link</a> 

    </div> 
</body> 

回答

0

你错过了从第一个脚本源的http:。将第一个脚本标记更改为:

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js" type="text/javascript"></script> 

第二个脚本标记的类型错误。将第二个脚本更改为:

<script type="text/javascript"> 
    $(document).ready(function() { 
     $("a").click(function() { 
      alert("Hello world!"); 
     }); 
    }); 
</script>