Getting {object Object} in jquery JSON php

Question

I'm trying to get two dates based on a selection from a previous dropdown.

Functionally, the process is working. However, I am getting a result of {object Object} instead of the dates.

PHP

$data=array();

while($row=mysql_fetch_array($result)) {

    $data[]=array('start_date'=>date("m/d/Y",strtotime($row['start_date'])));
    //log_msg(print_r($data, true));
}

echo (json_encode($data));

jQuery

 .ajax({
        type: 'POST',
        url: 'admin_workshop_update_retrieve.php',
        dataType: 'json',
        data: 'value=' + choice,
        success: function(data) {
            $.each(data, function(k, v) {
            $("#workshop_dates").append($('<option value ="' + v + '" >' + v + '</option><br>'));
            });
        }
    });

It seems like I'm missing something easy to turn the JSON object into a string or text. Would greatly appreciate any help.

tip : use JSON_FORCE_OBJECT parameter with json_encode — xkeshav
– xkeshav, Commented Aug 24, 2012 at 5:30

Mihai Iorga · Accepted Answer · 2012-08-24 05:28:41Z

4

You should treat them as objects:

$("#workshop_dates").append($('<option value ="' + v.start_date + '" >' + v.start_date + '</option><br>'));

answered Aug 24, 2012 at 5:28

Mihai Iorga

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3 Comments

xkeshav Over a year ago

may i know what is the usage of k in function(k,v) ?

nnnnnn Over a year ago

@diEcho - data seems to be an array of objects, so in this case k would be the numeric index of the current array element where v is the actual array element at that index. See the $.each() doco.

Mihai Iorga Over a year ago

k is the array index, eg: 0, 1

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