Question

我是codeigniter框架和json的新手。我想在用户点击网址时从数据库中获取数据，但是我收到了错误。

我的剧本：

function abc(i)
{
   if(i == 1)
   {
     $.ajax({
       type: "post",
       url: "<?php echo base_url('welcome/liststd');?>",
       data: 'postData='+ i ,
       success: function(json)
       {
         var abc = json;
         document.getElementById("fname").innerHTML=abc[0].fname; 
       }
    });
  }
 }

Firebug回复：

[
   {
     "id":"31",
     "fname":"Darshan",
      "mname":"D",
      "lname":"Dave",
      "std":"1",
      "marks":"12000",
      "image":"image31.jpg",
      "id1":"1",
      "in_date":"0000-00-00",
      "upd_date":"2013-07-10"
   },
   {
      "id":"34",
      "fname":"Darshan",
      "mname":"D",
      "lname":"Dave",
      "std":"1",
      "marks":"12000",
      "image":"image34.jpg",
      "id1":"1",
      "in_date":"2013-07-06",
      "upd_date":"2013-07-09"
     }
   ]

当我在json函数中提醒success时，它会向我显示：

[
   {
     "id":"31",
     "fname":"Darshan",
      "mname":"D",
      "lname":"Dave",
      "std":"1",
      "marks":"12000",
      "image":"image31.jpg",
      "id1":"1",
      "in_date":"0000-00-00",
      "upd_date":"2013-07-10"
   },

我的错误是什么？

Answer 1

您没有解析JSON。将dataType : 'json'添加到ajax调用中。这将告诉jQuery自动为您解析JSON。

 $.ajax({
     type: "post",
     url: "<?php echo base_url('welcome/liststd');?>",
     data: 'postData=' + i,
     dataType: 'json', // <----- tell jQuery to parse the JSON
     success: function (json) {
         var abc = json;
         document.getElementById("fname").innerHTML = abc[0].fname;
     }
 });

Answer 2

try this in your response...
also in your $.ajax tell jquery to parse json
 dataType: 'json'
in your response..
var response = JSON.parse(json);
response.fname;

codeigniter中的JSON响应错误

2 个答案: