关于Ajax的问题。

Question

function Load() {

  var el,li;
  
      el = document.querySelector("#wrapper ul li");
     
      $.ajax('').done(function(data){
           
           for( var i in data)
           
            li=document.getElementById("li");
            
            li.appendChild(document.createTextnode(data[i].name));
            
            el.appendChild(li,el.childNodes[0]); 
           
    })
    
     wrapper.refresh();
}    
    
    

如果后台传入[{name:'item1'},{name:'item2'},{name:'item3'}]可以使用上面的方法，但是如果后台传入多个例如：
[{"book_company_cover":"http://www.100gao.com/data/upload/2016/01/25/book_1453716846.jpg","book_company_id"
:"66","book_company_name":"u5f20u5bb6u53e3u8d44u6e90u4e66u5e97","add_time":"1453716846","sns_app_id"
:"59","type_id":"3","level_id":"0","book_company_address":"u5f20u5bb6u53e3u6668u5149u82b1u57ce
u5c0fu533a","level_name":null},{"book_company_cover":"http://www.100gao.com/data/upload/2016\
/01/25/book_1453714084.jpg","book_company_id":"65","book_company_name":"u5f20u5bb6u53e3u65b0u534e
u4e66u5e97","add_time":"1453714084","sns_app_id":"59","type_id":"3","level_id":"0","book_company_address"
:"u5f20u5bb6u53e3u6668u5149u82b1u57ceu5c0fu533a","level_name":null}]，请问这时改怎么写呢，谢谢！

Answer 1

data[i].book_company_cover
data[i].book_company_id
....

另外构造dom的方法我不建议你这么写，用js模板引擎

https://github.com/aui/artTemplate

这个是他的一个例子

<script id="test" type="text/html">
<h1>{{title}}</h1>
<ul>
    {{each list as value i}}
        <li>索引 {{i + 1}} ：{{value}}</li>
    {{/each}}
</ul>
</script>