Populating MySQLi result into a PHP indexed array

Question

I need to load result of a mysqli select statement (single field with multiple rows) into an array like:

$post = [318,310,323]

As you can see I used the fetch_array():

$posts = [];
 .....
$stmt->bind_param("s", $sessien);
$stmt->execute();
$result = $stmt->get_result();
while ($row = $result->fetch_array(MYSQLI_NUM))
       {
            array_push($posts, $row);
       }
print_r($posts );

$result->free();
$stmt->close();
$customconn->close();

but on printing r the $posts is creating something like this (looks like an array in array):

Array
(
    [0] => Array
        (
            [0] => 318
        )

    [1] => Array
        (
            [0] => 310
        )

    [2] => Array
        (
            [0] => 323
        )

)

How can I fix this to have something like:

$post = [318,310,323]

hakre · Accepted Answer · 2021-10-01 23:07:31Z

1

As mysqli_result::fetch_array() will always return an array - even for 1 field, you need to add that field rather than the entire result to your overall result array...

$row = $result->fetch_array(MYSQLI_NUM)
array_push($posts, $row[0]);

edited Oct 1, 2021 at 23:07

hakre

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answered Dec 12, 2018 at 6:12

Nigel Ren

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Comments

shawn · Accepted Answer · 2018-12-12 06:15:09Z

1

You should call array_push($posts, $row[0]);

Or you can call array_column on your result.

array_column($posts, 0)

answered Dec 12, 2018 at 6:15

shawn

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Comments

Dharman · Accepted Answer · 2021-10-03 13:57:52Z

0

Since PHP 8.1, this became much easier. All you have to do is call fetch_column() and collect the elements into an array.

$result = $stmt->get_result();
while($id = $result->fetch_column()) {
    $posts[] = $id;
}

answered Oct 3, 2021 at 13:57

Dharman♦

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