Using array_shift on while loop with fetchAll() on SELECT * from mysql

Question

I have a question on understanding how using *array_shift* with a fetchAll on a while loop in this example works? The intended result is just to fetch the data from the db and use it for filling out a CRUD app.

I understand *array_shift* drops off the first result of an array but am not sure why this does not impact the results returned in this example. Just trying to wrap my head around this process.

$pdo = Database::connect();
$sql = $pdo->prepare("SELECT * FROM contacts ORDER BY name ASC");
$sql->execute();
$result = $sql->fetchAll(PDO::FETCH_ASSOC);

$contacts = array();
while ($obj = array_shift($result)) {
    $contacts[] = $obj;
}
return $contacts;

So why not just `$contacts = $sql->fetchAll(PDO::FETCH_ASSOC)`? That would be equivalent to what you are doing. — Explosion Pills, Jan 02 '14 at 19:22
Right, that is an alternative but my question is on how array_shift works in this example. — webdev, Jan 02 '14 at 19:25
It removes the first element from `$result` and sets it to `$obj`. This is done in a loop. When `$result` is empty, `array_shift` will return a falsy value and break the loop — Explosion Pills, Jan 02 '14 at 19:32
Hmm interesting, can you expand more on array_shift returning a falsey value and breaking the loop? — webdev, Jan 02 '14 at 19:37

score 1 · Accepted Answer · answered Jan 02 '14 at 20:03

array_shift will remove the first element from an array and return it. So if you have an array like

array("one", "two", "three")

Using array_shift will return "one" and the source array will be updated to array("two", "three"). When done in a loop, this means that eventually the source array will be pruned down to array().

According to the documentation, array_shift on an empty array returns NULL. You can think of the code as running:

if ($obj = NULL)

This is falsy, so the statement (while in your case) is not evaluated.

Cheers mate on the in-depth it makes more sense now on how array_shift works in this example. Appreciate you taking the time to explain it. — webdev, Jan 02 '14 at 20:18

Using array_shift on while loop with fetchAll() on SELECT * from mysql

1 Answers1