JSON.parse is an “undefined” object

Question

I created json with php. Data coming with ajax. But JSON.parse is giving an “undefined” object. Why ?

Php CODE

 $emparray = array();
 while($row =mysqli_fetch_assoc($result))
 {
     $emparray[] = $row;
 }

 echo json_encode($emparray);

Ajax CODE

   $.ajax({
        type: "GET",
        url: "http://localhost:8080/xxx.php/",
        success: function (msg, result, status, xhr) {
            var obj= JSON.parse(msg);
            alert(obj.name);// giving undefined
        },
        error: function (error) {
        }
    });

json

[{"name":"eng","a":"sdf"}]

because you have an array that has an object in the first index. `obj[0].name` — epascarello, Dec 06 '17 at 13:45
obj is an array, try accessing the first element. `obj[0].name` — ztadic91, Dec 06 '17 at 13:45
Please, [quit using `alert()` for troubleshooting.](http://stravid.com/en/stop-the-javascript-alert-madness/), use `console.log()` instead. — Jay Blanchard, Dec 06 '17 at 13:45
If an answer solved your problem, consider accepting the answer. Here's how http://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work then return here and do the same with the tick/checkmark till it turns green. This informs the community, a solution was found. Otherwise, others may think the question is still open and may want to post (more) answers. You'll earn points and others will be encouraged to help you. *Welcome to Stack!* — Jay Blanchard, Dec 06 '17 at 14:10

score 1 · Answer 1 · answered Dec 06 '17 at 13:46

1

You should obj[0].name

Because you are accessing the name property of the of the first element of the array.

answered Dec 06 '17 at 13:46

Eddie

26,593
6
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chrisv · Accepted Answer · 2017-12-06T13:52:32.420

1

Your JSON is an array, meaning you'll have to point to the index of the object before accessing the property.

This code should work:

console.log(obj[0].name); //Returns "eng"

If your JSON array was something like this:

[{"name":"eng","a":"sdf"}, {"name":"esp", "a":"abc"}]

Then obj[1].name would return "esp".

edited Dec 06 '17 at 13:52

answered Dec 06 '17 at 13:47

chrisv

724
6
18

Meenesh Jain · Answer 3 · 2017-12-06T14:07:17.133

A better way to get data from server

$emparray = array();
 while($row =mysqli_fetch_assoc($result))
 {
     $emparray[] = $row;
 }

 echo json_encode(array("data"=>$emparray));

Put all the json response on a key which is data here then at front end define that the server response is in JSON by dataType and then there is no need to parse data by JSON.parse()

msg.data.length will provide you the validaton on if the data received from server is empty or not which will prevent the undefined error

 $.ajax({
        type: "GET",
        dataType: "JSON",
        url: "http://localhost:8080/xxx.php/",
        success: function (msg, result, status, xhr) {
            var obj= msg.data;
           if(msg.data.length){       
                  alert(msg.data[0].name);// wll give name at 0 index
              }
        },
        error: function (error) {
        }
    });

Why is this better? A ***good answer*** will always have an explanation of what was done and why it was done in such a manner, not only for the OP but for future visitors to SO. — Jay Blanchard, Dec 06 '17 at 13:50

JSON.parse is an “undefined” object

3 Answers3