MySQLi Prepared不工作UPDATE查询

Question

I have here my codes the should change the status of the specific crew once I click the submit button. but when I click the button, it gives me an FALSE result. can someone tell me what the problem is?

below is my code:

if(isset($_POST['submit'])) {
$crew = 'NEW';
$query_update = "UPDATE `crew_info` SET `crew_status` = ? WHERE `id` = ?";
$stmt2 = mysqli_prepare($conn, $query_update);
mysqli_stmt_bind_param($stmt2, 'ss', $_GET['id'],$crew);
mysqli_stmt_execute($stmt2);

    if (mysqli_stmt_affected_rows($stmt2) !== 0) {
        echo "Crew update accepted";
    } else {
        echo 'Error';
    }

}

Answer 1

mysqli_stmt_affected_rows shouldn't be used to detect errors.

Change your code to this:

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$query_update = "UPDATE `crew_info` SET `crew_status` = ? WHERE `id` = ?";
$stmt2 = mysqli_prepare($conn, $query_update);
mysqli_stmt_bind_param($stmt2, 'ss', $_GET['id'],$crew);
mysqli_stmt_execute($stmt2);
echo "Crew update accepted";