Mysqli_fetch_assoc does not work with a function

Question

If I put my code in a function it does not work. If I get rid of the function it is responding correctly. What I'm doing wrong?

function dayClosure() {
$qClosure = 'SELECT * FROM timeRegistration WHERE department IN ("4")';
$rClosure = mysqli_query($conn, $qClosure);

    while($row = mysqli_fetch_assoc($rClosure)) {
        if ($row['status'] == '3' && $row['enddate'] == '23-10-2017') {
            $totalWorkedTime += $row['worktime'];
            return $totalWorkedTime;
        }
    }
}

echo dayClosure();

`$conn` is not defined in the function, so PHP has no idea what it is. You need to pass it as a variable or define it as a global([Not recommended](https://stackoverflow.com/questions/12445972/stop-using-global-in-php)). I suggest you make a class. — GrumpyCrouton, Oct 23 '17 at 13:58
Try to pass $conn as a parameter to function. or define it as global — Ravinder Reddy, Oct 23 '17 at 13:58
No error. Just blank output. If I make sure the function () { } is deleted. I get normal response from my database. — nhatimme, Oct 23 '17 at 13:58
@PankajMakwana No, [that is not recommended](https://stackoverflow.com/questions/12445972/stop-using-global-in-php). — GrumpyCrouton, Oct 23 '17 at 13:59
The wierd thing is I'm loading my external database.php (require_once). Global does not help neither. — nhatimme, Oct 23 '17 at 14:02
If any of the provided solution was helpful please be sure to accept it. — mega6382, Oct 23 '17 at 14:07

score 1 · Accepted Answer · answered Oct 23 '17 at 13:59

This is because the function cannot access $conn variable, You need to provide the $conn variable to the function as a parameter:

function dayClosure($conn) {
$qClosure = 'SELECT * FROM timeRegistration WHERE department IN ("4")';
$rClosure = mysqli_query($conn, $qClosure);

    while($row = mysqli_fetch_assoc($rClosure)) {
        if ($row['status'] == '3' && $row['enddate'] == '23-10-2017') {
            $totalWorkedTime += $row['worktime'];
            return $totalWorkedTime;
        }
    }
}

echo dayClosure($conn);

The wierd thing is I'm loading my external database.php (require_once). Global does not help neither. — nhatimme, Oct 23 '17 at 14:02

score 0 · Answer 2 · answered Oct 23 '17 at 13:59

0

It's because the function in unable to access $conn

The best practice is to declare this $conn as global.

global $conn;
$conn = mysqli_connect(......);

Another way is to pass $conn as function parameter:

function dayClosure($conn) {

answered Oct 23 '17 at 13:59

Indrasis Datta

8,692
2
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2

That is certainly not the best practice. Global variables are [not recommended](https://stackoverflow.com/questions/12445972/stop-using-global-in-php). – GrumpyCrouton Oct 23 '17 at 14:00

score 0 · Answer 3 · answered Oct 23 '17 at 14:05

Solution:

<?php

require_once('config.php');

function dayClosure($conn) {
$qClosure = 'SELECT * FROM timeRegistration WHERE department IN ("4")';
$rClosure = mysqli_query($conn, $qClosure);

    while($row = mysqli_fetch_assoc($rClosure)) {
        if ($row['status'] == '3' && $row['enddate'] == '23-10-2017') {
            $totalWorkedTime += $row['worktime'];
            return $totalWorkedTime;
        }
    }
}

echo dayClosure($conn); 

?>

I required my file and did sent the $conn variable to the function. Thanks guys!

Mysqli_fetch_assoc does not work with a function

3 Answers3