CODEWITHSUNDEEP
jqueryajaxjsonundefined
Ryan
Ryan

Reputation: 787

AJAX call returning 'undefined'

AJAX call as follows:

$.get('file.php')
.done(function(data) {
    console.log(data);
});

Console returns:

[{"0":"PS001","code":"PS001"}]

However if amend the log to console.log(data.code) it returns undefined. Also, traversing JSON using each as such:

.done(function, data) {
    $.each(data, function(key, value) {
        console.log(value.code);
    });
)};

results in the following error:

[Error] TypeError: [{"0":"PS001","code":"PS001"}] is not a valid argument for
'in' (evaluating 'b-1 in a')

JSON looks well-formed, so not sure what the issue is?

Upvotes: 0

Views: 814

Answers (4)

Mox Shah
Mox Shah

Reputation: 3015

JSON_ENCODE returns string.

Learn more about JSON_ENCODE

You should convert that JsonArrayString to JsonArray

$.get('file.php')
.done(function(data) {
    var arrayObj = JSON.parse(data);
     $.each(arrayObj , function(key, value) {
            console.log(value.code);
        });
});

Upvotes: 2

Sadikhasan
Sadikhasan

Reputation: 18600

Try this, because your code return json array.

console.log(data[0].code);  //If you getting JSON format

OR

console.log(JSON.parse(data)[0].code);  //If you getting JSON as String

Demo

Upvotes: 1

Sougata Bose
Sougata Bose

Reputation: 31739

It is a json string. You need to parse it first. Try with - 

data = $.parseJSON(data);
$.each(data, function(key, value) {
    console.log(value.code);
});

Upvotes: 1

Vano Maisuradze
Vano Maisuradze

Reputation: 5899

Try this:

$.get('file.php')
.done(function(data) {
    var json = JSON.parse(data)[0];
    console.log(json .code);
});

Upvotes: 1

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