開捕致命錯誤：類的DateTime的對象無法轉換爲字符串

Question

我收到上述錯誤的代碼的最後一行，我已經試過的答案，但，這並不似乎工作

if (isset($_GET['logout'])) { 
    $name = $_SESSION["username"]; 
    date_default_timezone_set('Asia/Kolkata'); 
    $today = date('Y-m-d'); 
    $time = new DateTime(date('H:i:s')); 

    $statement = $db->prepare("SELECT `logintime` FROM `attendance` WHERE empid=? AND date_t=?"); 
    $statement->bind_param("ss", $name, $today); 
    $statement->execute(); 
    $statement->bind_result($logintime); 
    while ($statement->fetch()) { 

    } 
    $logintime = new DateTime($logintime); 

    $interval = $logintime->diff($time); 

    $hours = $interval->format('%h'); 
    $minutes = $interval->format('%i'); 
    $workinghours = $hours + $minutes/60; 


    $stmt = $db->prepare("UPDATE `attendance` SET `logouttime`=? ,`workinghours`=? WHERE empid=? AND date_t=?"); 
    $stmt->bind_param("ssss", $time, $workinghours, $name, $today); 
    $run = $stmt->execute(); 
} 

Answer 1

你必須使用format方法將DateTime對象轉換爲MySQL可以使用的東西。

date_format($time,'Y-m-d H:i:s') 

或

$time->format('Y-m-d H:i:s') 

所以......

$timeAsString = $time->format('Y-m-d H:i:s'); 
$stmt->bind_param("ssss",$timeAsString,$workinghours,$name,$today);