PHP无法重新声明功能

Question

as far as I can tell the function is only declared once . but I get this error Cannot redeclare calcPercentages() and it tells me it was declared on line 17 which is correct

the function that contains the errant function...

function auditStats($audit){

    $mysqli = dbConnect();
    $audit_id = $audit;
    $result = $mysqli->query("SELECT imageGrade, SUM(imageGrade=1) AS grade1, SUM(imageGrade=2) AS grade2, SUM(imageGrade=3) AS grade3, COUNT(*) AS total_imgs FROM image WHERE type != 'standard' and auditID ='$audit_id'")or exit("Error code ({$mysqli->errno}): {$mysqli->error}");
    $row = $result->fetch_assoc();

    $grade1 = $row['grade1'];
    $grade2 = $row['grade2'];
    $grade3 = $row['grade3'];
    $totalImgs = $row['total_imgs'];

    function calcPercentages($grade, $total){
        $percent = round(($grade / $total) * 100,2);
        return $percent;
    }
    if ($totalImgs != 0){
        $percent_1 = calcPercentages($grade1, $totalImgs);
    }
    $return_array = [
        'total'=>$totalImgs,
        'percent_1'=>$percent_1
    ];
    return $return_array;
}

The function isn't being called anywhere except within the auditStats function and the results are called further down the page using

$percent_1 = auditStats($auditID)[percent_1];

Please excuse me if I have made on obvious newbie error, I am moving from procedural to OOP and am just starting out with it.

Answer 1

You declare calcPercentages inside the auditStats function, so it gets re-declared every time you call auditStats .

Move calcPercentages outside of auditStats .

Answer 2

In PHP, unlike say JS, a function is never local to another function, so when the line function calcPercentages... is reached, you're creating a global function called calcPercentages . When it's run again, you're trying to create another global function with the same name, and you get the error.

Either move the declaration outside globally, put it in a class or object or make an anonymous function .