jquery-ajax中的parseJSON toJSON响应不起作用

问题描述：

I am new to AJAX using jquery. I have a json response as below:

[{"customer_name":"Customer A"},{"customer_name":"Customer B"},{"customer_name":"Customer C"}]

And my ajax file is :

 function(result){
    $('#resdiv').html(result);
    console.log(result);
    var json_obj = $.parseJSON(result);//parse JSON
    alert(json_obj);

    var output="<ul>";
    for (var i in json_obj)
    {
    output+="<li>" + json_obj[i].customer_name + "</li>";
    }
    output+="</ul>";
    $('#resdiv1').html(output);
}

Though I can view the JSON response in div id resdiv, the div id resdiv1 is empty ! Also alert(json_obj); does not alerts anything ! Whats wrong with the file ?

NB: I am learning from Zuch Tutorial

我是使用jquery的AJAX新手。我有一个json响应如下： p>

  [{“customer_name”：“Customer A”}，{“customer_name”：“Customer B”}，{“customer_name”：  “客户C”}] 
  code>  pre> 
 
 我的ajax文件是： p> 
 
 
  function（result）{
  $（'＃resdiv'）。html（result）; 
 console.log（result）; 
 var json_obj = $ .parseJSON（result）; //解析JSON 
 alert（json_obj）; 
 
 var  output =“＆lt; ul＆gt;”; 
 for（var i in json_obj）
 {
 output + =“＆lt; li＆gt;”  + json_obj [i] .customer_name +“＆lt; / li＆gt;”; 
} 
输出+ =“＆lt; / ul＆gt;”; 
 $（'＃resdiv1'）。html（输出）; 
} \  n  code>  pre> 
 
 虽然我可以在div id  resdiv  code>中查看JSON响应，但div id  resdiv1  code>是空的！ 此外 alert（json_obj）;  code>不会发出任何警报！ 该文件有什么问题？ p> 
 
 
注意：我正在学习 Zuch教程   p> 
  div>

答

You dont need to parse the json again.Simply do a iteration and try like this

var json = [{"customer_name":"Customer A"},{"customer_name":"Customer B"},{"customer_name":"Customer C"}];

var output="<ul>";
$.each(json,function(key,val){
  output+="<li>" + val.customer_name + "</li>";
});
output+="</ul>";

console.log(output);

See DEMO

答

Can you try this

   function(result){ 
      $('#resdiv').html(result); 
      console.log(result); 
      var json_obj = $.parseJSON(result);
      //parse JSON alert(json_obj);

      var output="<ul>"; 
       $.each(json_obj,function(key,val){ 
           output+="<li>" + val.customer_name + "</li>"; 
           console.log(key+":::"+val);
      });
        output+="</ul>";

     $('#resdiv1').html(output); 
  }

答

check this out

// Need to parse if string.
var result = '[{"customer_name":"Customer A"},{"customer_name":"Customer B"},{"customer_name":"Customer C"}]';
var json_obj = $.parseJSON(result);//parse JSON

// No need to parse
var json_obj = [{"customer_name":"Customer A"},{"customer_name":"Customer B"},{"customer_name":"Customer C"}];

Also check this

// if undefined, you don't have resdiv1 div or you have call function before your div render.
alert($('#resdiv1').html());

jquery-ajax中的parseJSON toJSON响应不起作用

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