Mysqli_fetch_assoc不适用于函数[重复]

Question

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• Reference: What is variable scope, which variables are accessible from where and what are “undefined variable” errors? 3 answers

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(); 

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Answer 1

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);