How to convert mysql array result into new array for json encoding

Question

I can’t seem to get the result I need. Here is the array as it is provided by the sql query:

Array (
 [0] => Array ( 
     [id] => 7
     [description] => Accepted )
 [1] => Array ( 
     [id] => 8
     [description] => Declined )
 [2] => Array (
     [id] => 11 
     [description] => Deferred ) 
     )

Here is the format for how I need to have it foreach of the objects listed above:

['7'][‘7’] = “Accepted”;
['7'][‘8’] = “Declined ”;
['7'][‘11’] = “Deferred”;

...where the first array[‘7’] is an added value and needed for each object.

Seems easy enough, but the foreach statements I created return an error “cannot use scalar value as an array”

DiverseAndRemote.com · Accepted Answer · 2012-09-18 22:23:06Z

2

$rows = /*Your Data above*/
$data = array('7'=>array());
foreach($rows as $row)
    $data['7'][$row['id']] = $row['description'];

$jsonData = json_encode($data);

Might get 'er done

answered Sep 18, 2012 at 22:23

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5 Comments

Patrick Over a year ago

Close, but not quite: {"7":{"7":"Accepted","8":"Declined","11":"Deferred"}}

DiverseAndRemote.com Over a year ago

That looks correct to me. You wont get it showing up as {{"7":{"7":"Accepted"}}, {"7":{"8":"Declined"}}, {"7":{"11":"Deferred, but accepted"}}} because the index 7 is common among the elements. the '7' index will only showup once when json_encoded

Patrick Over a year ago

Gotcha. You're right. I was just contemplating that very thing. Thanks, man... :)

Patrick Over a year ago

Sorry for the follow up, I'll open a new question if needed; but now when I array_merge($rowsA, $rowsB, and $rowsC) (not provided in the above example) with the same format you provided, I lose the leading array (in my case '7', '6', and '5'). Does array_merge not take that into account?

Patrick Over a year ago

array_merge() flattens an array. duh. Sorry.

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