CODEWITHSUNDEEP
javascriptjqueryjsonajax
Matarishvan
Matarishvan

Reputation: 2422

Cannot display JSON result in AJAX succcess

My ajax code looks like this

$.ajax({
    type : "POST",
    dataType : "JSON",
    url : "index.php",
    success : function(data){           
        for(i=0;i<data.length;i++)
        {
            console.log(data[i].1_assigned_accepted);
        }
    }
})

Now console.log(data[i].1_assigned_accepted); is giving me error, even though i have 1_assigned_accepted as column name in my table. jQuery doesn't consider 1_. How to resolve this?

Upvotes: 0

Views: 71

Answers (3)

Rhys Bradbury
Rhys Bradbury

Reputation: 1707

To fix your code:

$.ajax({
    type: "POST",
    dataType: "JSON",
    url: "index.php",
    success: function(data) {
        console.log(data); //test to see if we get something back
        for (var i = 0; i < data.length; i++) {
            console.log(data[i]['1_assigned_accepted']);
        }
    }
});

The difference OP and mine is assignment of i=0; -> var i = 0; and using brackets for getting the value.

Some Notes:

When doing a post you usually send data using the data: option. (remember to Json.stringify(input)). When posting you may need more options:

function postJson(url, data, successCallback, errorCallback) {
  $.ajax({
      url: url,
      type: 'POST',
      data: JSON.stringify(data),
      contentType: 'application/json',
      dataType: 'json',
      success: successCallback,
      error: errorCallback
  });
}

Notice there's contentType, dataType and data being used here. You usually want all of these when doing a post.

Invoke like so:

var data = { name: 'Rhys' };
postJson(
  'index.php',
   data,
   function(responseData) {
     console.log(responseData);
   },
   function(err) {
     console.error(err);
   }
);

Another point:

You are assuming that your response object is an array, you could test using instanceof.

I hope this helps,

Rhys

Upvotes: -1

Yurii Kramarenko
Yurii Kramarenko

Reputation: 1054

Try to use bracket form.

console.log(data[i]['1_assigned_accepted']);

Upvotes: 1

zerohero
zerohero

Reputation: 583

Try:

console.log(data[i].["1_assigned_accepted"]);

Upvotes: 1

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