Jquery and ajax with json data parse

Question

What is wrong with my code that i can not show anything in resultTXT

txtfld shows the array

[{
    "user_id": "2790",
    "freelancer_name": "",
    "order_id": "9121",
    "orderamount": "0.00",
    "payment_method": " ....... "
}]

I want the user ID to be in resultTXT

ajaxRequest.onreadystatechange = function(){
    if(ajaxRequest.readyState == 4){
        $.post('userfind.php', function(data) {
            $("#txtfld").val(data);

            var json = data,
            obj = JSON.parse(json);
            alert(""+obj.user_id);
            $("#resultTXT").val(obj.user_id);
        },'json');
    }
};
ajaxRequest.open("POST", "userfind.php", true);
ajaxRequest.send(null);

Please help me what should be changed.

Why do you have jQuery inside of a ajaxRequest.open? Use either jQuery AJAX or vanilla JS AJAX, not both. — gen_Eric
– gen_Eric, Commented Jul 17, 2012 at 14:26

gen_Eric · Accepted Answer · 2012-07-17 14:30:18Z

3

I see a few things wrong with this. First, why do you POST to userfind.php twice? Why are you using vanilla JS AJAX and jQuery AJAX? Just use one.

Second, the ,'json' in $.post means jQuery will parse the JSON for you, you don't need JSON.parse.

Third, your JSON is an array (of objects), so you need to get the array element first, then the user_id property.

$.post('userfind.php', function(data) {
    $("#txtfld").val(data);  // data is an object,
                             // so this will just put [object Object] in the field,
                             // probably not what you want

    alert(data[0].user_id);  // data is an array of (one) object(s)
    $("#resultTXT").val(data[0].user_id);
},'json');

answered Jul 17, 2012 at 14:30

gen_Eric

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1 Comment

Shahab Maboud Over a year ago

$("#txtfld").val(data); shows the array i can see the array elements in the box. and alert(data[0].user_id); // data is an array of (one) object(s) $("#resultTXT").val(data[0].user_id); },'json'); doesn't show anything

gen_Eric · Accepted Answer · 2012-07-17 14:26:42Z

0

I think that is because the json is an array so maybe try obj[0].user_id

edited Jul 17, 2012 at 14:26

gen_Eric

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answered Jul 17, 2012 at 14:25

4NDR01D3

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