JSON data can't get back via jquery each function

Question

 $(function () {
    $.ajax({
        url: 'image/show',
        dataType: 'json',
        success: function (data) {
            $.each(function (i, item) {
                alert(item.ratio);
            });
        }
    });
 });

get data back like these:

{"id":2,"ratio":0.66666666666667,"src":"public\/media\/flickr\/13842755234_c8e89150b1_m.jpg"}{"id":3,"ratio":1.0491803278689,"src":"public\/media\/flickr\/13845062983_00fd2bebe2_n.jpg"}{"id":4,"ratio":1.5015015015015,"src":"public\/media\/flickr\/13845347184_24ca8d4e5d.jpg"}

QUESTION: I CAN'T ALERT DATA VIA MY JQUERY CODE.IS ANYTHING GOES WRONG IN MY CODES?

I think maybe these data format goes error.

public function show() {

    $dirname = "public/media/flickr/";
    $images = scandir($dirname);
    $ignore = Array('.',"..");

    foreach ($images as $key=>$curimg) {
        if (!in_array($curimg, $ignore)) {
            $imagefile = $dirname.$curimg;
            list($width,$height) = getimagesize($imagefile);    
            $ratio = $width/$height;
            $data = array(
                    'id' =>$key,
                    'ratio' => $ratio,
                    'src' => $imagefile);
            echo json_encode($data);
        }

    }
}

my php function error. I made a mistake in foreach.

Does the call enter `success` at all? Try the `error` callback as well. — Gaurav Bhor, Apr 15 '14 at 09:57

score 1 · Answer 1 · answered Apr 15 '14 at 09:32

1

Try,

 alert(data.ratio);

There is no need of .each() here.

Full code,

 $(function () {
    $.ajax({
        url: 'image/show',
        dataType: 'json',
        success: function (data) {
            alert(data.ratio);
        }
    });
 });

answered Apr 15 '14 at 09:32

Rajaprabhu Aravindasamy

66,513
17
101
130

He would need to use `alert(data["ratio"]);` I guess. – Gaurav Bhor Apr 15 '14 at 09:34
@GauravBhor why you are guessing like so..? both are same.. you are accessing the value using bracket notation.. that's all..\ – Rajaprabhu Aravindasamy Apr 15 '14 at 09:35
1

@Gaurav, how would that be different? – Frédéric Hamidi Apr 15 '14 at 09:35
This should clear the air http://stackoverflow.com/questions/2904131/what-is-the-difference-between-json-and-object-literal-notation – Gaurav Bhor Apr 15 '14 at 09:39
1

@SanChen First make an empty alert inside the success call back.. There may be a problem in the server side code.. – Rajaprabhu Aravindasamy Apr 15 '14 at 09:46

score 0 · Answer 2 · answered Apr 15 '14 at 09:33

0

Your $.each it wrong, it should be:

$.each(data, function(index, item) {
 alert(item.ratio)
})

answered Apr 15 '14 at 09:33

Slaven Tomac

1,552
2
26
38

JSON data can't get back via jquery each function

2 Answers2