copying json object to javascript array

Question

I want to copy json object to javascript array,

{
"profiledata1":{"id":0,"music":0,"image_x":130,"image_y":155,"mouth_x":0,"mouth_y":-28.125,"active":true,"default":false},
"profiledata2":{"id":1,"music":0,"image_x":130,"image_y":155,"mouth_x":0,"mouth_y":0,"active":true,"default":false},
"profiledata3":{"id":2,"music":0,"image_x":0,"image_y":0,"mouth_x":0,"mouth_y":0,"active":false,"default":false},
"profiledata4":{"id":3,"music":0,"image_x":0,"image_y":0,"mouth_x":0,"mouth_y":0,"active":false,"default":false},
"profiledata5":{"id":4,"music":0,"image_x":0,"image_y":0,"mouth_x":0,"mouth_y":0,"active":false,"default":false},
"upload":"http:\/\/localshost\/",
"format":"jpeg","status":1 }

This is my json object returned when i call some.php through ajax, I want to copy profiledata1 to userdata_arr[0],profiledata2 to userdata_arr[1],profiledata3 to userdata_arr[2],profile4data to userdata_arr[3],profiledata5 to userdata_arr[5] in java script. My java script is as follows,

$.ajax({
    type: "POST",
    url: "some.php",
    data: {action:'load',
            id:7}

}).done(function(o) {
    var data = $.parseJSON(o);
    if (!data || data === null) {
        someError(true);

    }else{ 
        if(data.status==true){
            userdata_arr[0] = data.profiledata1[0];
            userdata_arr[1] = data.profiledata2[0];
            userdata_arr[2] = data.profiledata3[0];
            userdata_arr[3] = data.profiledata4[0];
            userdata_arr[4] = data.profiledata5[0];
            uploadDir = data.upload;
            imgFormat = data.format;


            somefunction();
        }else{
            someError(true);
        }
    }
});

when i execute this script i'm getting userdata_arr as undefined! Please help me to rectify this problem. I'm also attaching the some.php here,

<?php
if ($_POST['action']=='load') {
$uid=$_POST['id'];

header("content-type:application/json");
// fetch contents from db with $uid;
$query = mysqli_query($link,$sql);
while ($row = mysqli_fetch_array($query)) {
    $prof1 = $row['prof1'];
    $prof2 = $row['prof2'];
    $prof3 = $row['prof3'];
    $prof4 = $row['prof4'];
    $prof5 = $row['prof5'];
}
$jp1 = json_decode($prof1, 1);
$jp2 = json_decode($prof2, 1);
$jp3 = json_decode($prof3, 1);
$jp4 = json_decode($prof4, 1);
$jp5 = json_decode($prof5, 1);

echo json_encode($dta = array('profile1data' =>json_decode($prof1),'profile2data' =>json_decode($prof2),'profile3data' =>json_decode($prof3),'profile4data' =>json_decode($prof4),'profile5data' =>json_decode($prof5) ,'upload' =>'http://localhost/img/', 'format' =>'jpeg', 'status' =>1 )); ?>

Thanks in advance!

NaijaProgrammer · Accepted Answer · 2015-10-27 13:21:39Z

2

That's because you haven't declared your userdata_arr. To fix it, declare your array/object variable before using it. In your else code-block, do this:

else{ 

    var userdata_arr = {}// declare your object

    if(data.status==true){ //proceed to use your already-declared object, also notice the quote marks surrounding the object members/indexes
        userdata_arr["0"] = data.profiledata1[0];
        userdata_arr["1"] = data.profiledata2[0];
        userdata_arr["2"] = data.profiledata3[0];
        userdata_arr["3"] = data.profiledata4[0];
        userdata_arr["4"] = data.profiledata5[0];
        uploadDir = data.upload;
        imgFormat = data.format;


        somefunction();
    }else{
        someError(true);
    }
}

answered Oct 27, 2015 at 13:21

NaijaProgrammer

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1 Comment

Adarsh Raveendra Over a year ago

Thanks @NaijaProgrammer

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