如何显示从jQuery返回的JSON

Question

How to get display return values from the JSON values.I need to get value the user id

 $('User_id').observe('blur', function(e) {
   var txt = $('User_id').value;
   jQuery.ajax({
     type: 'get',
     url: BASE_URL + 'admin/index/user_id',
     data: {
       user_id: txt
     },
     dataType: 'json',
     success: function(data) {
       console.log('success' + data.success);
       if (data.success) {
         var Value = data.location.user_id;
         alert(Value);
       }
     }
   });
 });

These values are getting in html form. In that I need to store user id in Value varable. But I receive successundefined as a output..

[{
  "user_id": "139",
  "mobile": "9042843911",
  "gender": "male",
  "hashcode": "DfAbMqLApAV6nVa1z940",
  "username": "anandhsp21",
  "password": "74bcff7d1199012e154f364e3f65e31d:8I",
  "authorized_person": "Anandh",
  "created": "2015-06-08 13:46:55",
  "modified": "2015-06-08 06:43:35",
  "logdate": "2015-06-08 08:16:55",
  "lognum": "12",
  "reload_acl_flag": "0",
  "is_active": "1",
  "extra": "N;",
  "rp_token": null,
  "rp_token_created_at": null,
  "app_name": "",
  "api_key": ""
}] 

Please some one help. Thanks in Advance

Answer 1

Your get the data in array so use loop in success data

for (var i=0; i<data.length; i++) {
    console.log('success' + data[i].user_id );
}

If you know the record length is 1 then use directly

 console.log('success' + data[0].user_id );

Answer 2

Your data is an array that contains one object. So you can access this object using :

success: function(data){
    console.log('success' + data[0].user_id );

Trying to log success is pointless, because there is no success key whatsoever in the received data.

Answer 3

Make sure that you get the response in proper json format,and as harshad pointed String male should be wrapped in double quotes.

After you get that fixed,you can access the user_id as:

data[0].user_id

Answer 4

data.success is undefined, because the received data is stored directly in data . That's the first argument of the success block. You can access the received data directly by data[0] to get the object inside of the array, or if you have a larger array you can do a for each loop over it, etc..

Answer 5

Try this, simply use json.parse()

 $(document).ready(function() { var v = ['{"user_id":"139","mobile":"9042843911"}']; var obj = JSON.parse(v); alert(obj.user_id); }); 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> 

Answer 6

To get userid please follow below code I edited,

$('User_id').observe('blur', function(e) {
   var txt = $('User_id').value;
   jQuery.ajax({
   type: 'get',
   url: BASE_URL + 'admin/index/user_id',
   data: {
     user_id: txt
   },
   dataType: 'json',
   success: function(data) {
     // this below userid is the value user_id you want.
     var userid = data[0].user_id;
  }
 });
});

Answer 7

There is a json error

"gender":male

Strings male should be wrapped in double quotes.

you need to make sure that your response is formatted appropriately and according JSON.org standards.