Sum the values of an array if their id's are the same

Question

Please see table below:

items

id  title   fields
1   Dog     [{"id":"11","value":"50"},{"id":"19","value":"100"},{"id":"22","value":"150"}]
2   Cat     [{"id":"19","value":"50"},{"id":"22","value":"100"},{"id":"37","value":"150"}]
3   Bird    [{"id":"14","value":"50"},{"id":"19","value":"100"},{"id":"22","value":"150"}]
4   Mouse   [{"id":"16","value":"50"},{"id":"22","value":"100"},{"id":"34","value":"150"}]
5   Horse   [{"id":"6","value":"50"},{"id":"22","value":"100"},{"id":"47","value":"150"}]

I want to echo the sum of all the values that equals to id 22 I am a complete novice to php and mysql and will appreciate any assistance, so please explain the code to me clearly.

This is what I have found, but does not work:

<?php
    if ($fields->id == 22) {
        $result += $fields['value'];
    }
    return $result;
?>

please help.... thank you

Do "var_dump($array);" and post the result in your question above please. — jgroenen
– jgroenen, Commented Jan 9, 2014 at 10:56
can you show what is the result of var_dump($fields);Also i guess you need to loop throught he items and find the sum. — 웃웃웃웃웃
– 웃웃웃웃웃, Commented Jan 9, 2014 at 10:57
are the values json_encoded? What does your actual array look like? — Nouphal.M
– Nouphal.M, Commented Jan 9, 2014 at 10:59
after investigation i can confirm that it is json data and that i've tried Sankalp's answer, but with no success. You refer to var_dump.... I have no idea what you are talking about or even where i need to enter that code. I did however add it to the php output file and the only result i received was NULL, if that makes any sense. I tried both answers below, but not working. — WDCT
– WDCT, Commented Jan 9, 2014 at 12:24

William Buttlicker · Accepted Answer · 2014-01-09 11:02:00Z

2

Looks like $fields is json data. If it is, then

Try this:

    $data = json_decode($fields);
    foreach($data as $element) {
      if ($fields->id == 22) {
         $result += intval($fields->value);
      }
    }
    return $result;

answered Jan 9, 2014 at 11:02

William Buttlicker

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