php is not returning an array

Question

i'm probably make a really stupid mistake, but i can't find it.

i'm trying to return an array, and it just isn't working.

the count of $temp_array is 33 which is the correct number it should be, the count of $dropdown_array is 1. what am i doing wrong?

<?php
function dropmaker($stuff) {

require '../connect.php';

$dropdown_query = "SELECT $stuff FROM maps_sku_groups GROUP BY $stuff;";
$dropdown_result = mysqli_query($link, $dropdown_query);
while ($data = mysqli_fetch_array($dropdown_result)) {
    $temp_array[] = $data[$stuff];
}
echo count($temp_array);
return array($temp_array);
}?>

<?php 
$dropdown_array[] = dropmaker('cyclecount');
echo count($dropdown_array);
?>

I wouldn't include a file in a function! or at least use: require_once — Rizier123
– Rizier123, Commented Dec 2, 2014 at 16:28
did you try using var_dump($arrayname) or print_r($arrayname) to see the results? Also, as @Rizier123 mentioned, don't include a connection inside a function. Require it once at the top of your script, this can cause bugs. — alda1234
– alda1234, Commented Dec 2, 2014 at 16:29
i can't seem to get the function to work at all without having the file in it. which also makes no sense to me... — dehuman8
– dehuman8, Commented Dec 2, 2014 at 16:31

Jost · Accepted Answer · 2014-12-02 16:35:35Z

6

You are returning an array, which contains the array $temp_array as its only element. Just return $temp_array directly, and it should work.

EDIT (Thanks @Rizier123): Additionally, you are assigning the result of the call to dropmaker() to $dropdown_array[], which inserts the result of the function call into the (previously non-existing) array $dropdown_array, so $dropdown_array becomes an array with the function result as its only element.

edited Dec 2, 2014 at 16:35

answered Dec 2, 2014 at 16:29

Jost

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2 Comments

Rizier123 Over a year ago

Also he don't have to write this: $dropdown_array[] only $dropdown_array

Marc B Over a year ago

e. g a simple var_dump($dropdown_array) would've shown this problem.

Owen Kavanagh · Accepted Answer · 2014-12-02 16:34:55Z

1

The problem is this line $dropdown_array[] = dropmaker('cyclecount'); you're assigning the return value of dropmaker(), an array, to the first element of the array $dropdown_array[] rather than assigning $dropdown_array to the array itself.

Try this code

$dropdown_array = dropmaker('cyclecount');

Removing the [] is fixing this issue.

answered Dec 2, 2014 at 16:34

Owen Kavanagh

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Comments

VijayS91 · Accepted Answer · 2014-12-02 16:35:55Z

1

Try This Code :

return array($temp_array) instead of return $temp_array

Also $dropdown_array[] instead of $dropdown_array

answered Dec 2, 2014 at 16:35

VijayS91

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Comments

Bhavya Shaktawat · Accepted Answer · 2014-12-02 16:39:01Z

Try this

<?php
    function dropmaker($stuff) {

    require '../connect.php';

    $dropdown_query = "SELECT $stuff FROM maps_sku_groups GROUP BY $stuff;";
    $dropdown_result = mysqli_query($link, $dropdown_query);
    while ($data = mysqli_fetch_array($dropdown_result)) {
        $temp_array[] = $data[$stuff];
    }
    echo count($temp_array);
    return $temp_array;
    }?>

    <?php 
    $dropdown_array = dropmaker('cyclecount');
    echo count($dropdown_array);
    ?>

Your function returns array of array since you have written return array($temp_array);

And also you are assigning this return value to first index of array $dropdown_array so the count of this variable remains 1

when doing this it still is 33 & 1 the var_dump gives this: (which is one level less than my previous iteration.) array (size=1) 0 => array (size=33)

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