Make json array from mysql query

Question

I am trying the following code:

$order = array(); $imageURL = array(); $name = array();
        while($row = mysql_fetch_array($mysql->result)) {
            $order[] = $row["order"]; 
            $imageURL[] = $row["imageURL"];
            $name[] = $row["name"];
        }
        $res = array($order, $imageURL,$name);
        return json_encode($res);

But it is not outputting in json format, any ideas?

Output:

[["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30"],["previews\/en-1-1.gif","previews\/en-1-2.gif","previews\/en-1-3.gif","previews\/en-1-4.gif","previews\/en-1-5.gif","previews\/en-1-6.gif","previews\/en-1-7.gif","previews\/en-1-8.gif","previews\/en-1-9.gif","previews\/en-1-10.gif","previews\/en-1-11.gif","previews\/en-1-12.gif","previews\/en-1-13.gif","previews\/en-1-14.gif","previews\/en-1-15.gif","previews\/en-1-16.gif","previews\/en-1-17.gif","previews\/en-1-18.gif","previews\/en-1-19.gif","previews\/en-1-20.gif","previews\/en-1-21.gif","previews\/en-1-22.gif","previews\/en-1-23.gif","previews\/en-1-24.gif","previews\/en-1-25.gif","previews\/en-1-26.gif","previews\/en-1-27.gif","previews\/en-1-28.gif","previews\/en-1-29.gif","previews\/en-1-30.gif"],["Helasd you?","Where sasaddsdam?","Weasdd!","Tasasdther","AtsaddsaBeach","Cheasd Hotel","At the Hotel","aaaaaaaaat?","At the Market","Aasdt's","Mesadasd th","Shdsdsssg","sssss","On aaaa","Do you work or study?","aaaaaaa","At tadstation","aaaaaae Gym","How doasdto\u2026?","Planning a Trip","At adsk","At the asdurant","At the Inads00e9","My Taaaog","A Meetiaaaaay adher","Tourist sdsadn Centre","Saaaaing","Aaaaaa Match","Lookaaasd","At tasda"]]

Are you returning from a function? If not, try outputting it instead using echo or print — scrowler
– scrowler, Commented Jun 26, 2014 at 22:13
What are you getting? are you sure the query result is not empty, and why split the array? and change return for echo — Emilio Gort
– Emilio Gort, Commented Jun 26, 2014 at 22:13
You're also calling mysql_fetch_array() proceduarlly while using an object as your query resource... While this is not necessarily incorrect it's definitely weird and most likely isn't what you're meaning to do... — scrowler
– scrowler, Commented Jun 26, 2014 at 22:14

scrowler · Accepted Answer · 2014-06-26 23:01:24Z

I think you're just after a better format to work with in your output, because you've currently set it up with three arrays and you'll have to reference each record's properties across these arrays by a numeric key.

I've reconstructed an example of what your original database output would look like:

$array = json_decode($json, true);

$row = array();
list($orders, $imageURLS, $names) = $array;
foreach($orders as $key => $val) {
  $row[] = array(
    'order' => $val,
    'imageURL' => $imageURLS[$key],
    'name' => $names[$key]
  );
}

So in your code, you should try this:

$output = array();
while($row = mysql_fetch_array($mysql->result)) {
    $output[] = $row;
}
return json_encode($output);

And you'll get a much easier data structure to work with. Example:

[
  {
    "order": "1",
    "imageURL": "previews\/en-1-1.gif",
    "name": "Helasd you?"
  },
  {
    "order": "2",
    "imageURL": "previews\/en-1-2.gif",
    "name": "Where sasaddsdam?"
  }
]

Christian Bonato · Accepted Answer · 2014-06-26 23:02:17Z

0

Judging by the comments to your original question, it looks like you're looking for :

 $main_array = array() ;    


    while($row = mysql_fetch_array($mysql->result)) {

        $res = array($row["order"], $row["imageURL"],$row["name"]);   
        $main_array[] = $res; 
    }

    return json_encode($main_array);

answered Jun 26, 2014 at 23:02

Christian Bonato

1,30612 silver badges27 bronze badges

Comments

Zack · Accepted Answer · 2014-06-26 23:03:06Z

0

The output is correct. A json array, containing three json arrays. But I guess you want each element to be a separate array/object, containing order, imageURL, name. If that is correct, try something like:

$result = array();

while($row = mysql_fetch_array($mysql->result)) {
$result[] = array("order"=>$row["order"], "imageURL" => $row["imageURL"],"name"=> $row["name"])
}
$res = array($order, $imageURL, $name);
return json_encode($result);

answered Jun 26, 2014 at 23:03

Zack

1891 gold badge4 silver badges18 bronze badges

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