Failing to assign values to array from a loop

Question

I have got this code:

function retrieve_answers($array = array(), $id = null)
{
    include(root_path . '\config.php');
    if($id == null)
    {
        $id = $this->question_id;
    }
    $query = mysqli_query($link, "SELECT * FROM `answers` WHERE `question_id`='$id'");
    if(!mysqli_num_rows($query))
    {
        throw new Exception('Question not found.');
    }
    /* - Retrieves the answer rows
    - Loops through the array
    - Indexes the array and assigns the answerID to the index */
    else
    {
        while($result = mysqli_fetch_array($query))
        {
            for($i=0;$i<mysqli_num_rows($query);$i++) 
            {
                $array[$i] = $result["id"]; 
            }
        }
    }
}

Which is a part of a class.

What am I trying to do? I am trying to accept an array as a parameter, and assign values to the array, the values are to be answerIDs which are linked to the question.

The test.php file is here:

<?php
define('root_path', realpath(dirname(__FILE__)));
include(root_path . '\config.php');
require_once(root_path . '\includes\question.class.php');

$q = new Question(3);

$array = array();

$q->retrieve_answers($array);

var_dump($array);
?>

What happens? When I try to debug by dumping the array, it shows that the array contains nothing:

array(0) { }

I tried to execute the MySQL result through the class to debug, and it does succeed to retrieve the answer IDs, so I'm pretty positive the issue is in the array.

I would happy to get assistance, thanks in advance.

just return that array and assign it during the function execution — Zgr3doo
– Zgr3doo, Commented Jun 2, 2015 at 11:09

score 1 · Accepted Answer · 2015-06-02 11:52:26Z

1

return value in a function like this

function retrieve_answers($array = array(), $id = null)
{
    include(root_path . '\config.php');
    if($id == null)
    {
        $id = $this->question_id;
    }
    $query = mysqli_query($link, "SELECT * FROM `answers` WHERE `question_id`='$id'");
    if(!mysqli_num_rows($query))
    {
        throw new Exception('Question not found.');
    }
    /* - Retrieves the answer rows
    - Loops through the array
    - Indexes the array and assigns the answerID to the index */
    else
    {
        $i = 0;
        while($result = mysqli_fetch_array($query))
        {
            $array[$i] = $result["id"];                 
            $i++;
        }
    return $array;
    }


}

and then get it as

$arr = $q->retrieve_answers($array);

var_dump($arr);

edited Jun 2, 2015 at 11:52

answered Jun 2, 2015 at 11:22

user4767190

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7 Comments

Everett Over a year ago

I don't have 15 reputation so I can't, sorry! I accepted it though.

Everett Over a year ago

I have encountered another issue, I've added another answer to link to the same question, and the answer IDs that are executed via var_dump are the same, aka: 2,2 instead of 1,2 array(2) { [0]=> string(1) "2" [1]=> string(1) "2" }

user4767190 Over a year ago

Can you please explain that

Everett Over a year ago

I've added another answer which is linked to the question, basically it looks like this in the MySQL table: Table: id question_id answer_text Data: 1 3 Blabla 2 3 Blabla And instead of showing 1,2 as a result of the array, it shows 2,2 -> a.k.a the last result of the mysqli_fetch_array.

Everett Over a year ago

Thanks :) I did $array[] = $result["id"]; and it indexed automatically, works now.

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