我試圖將一些mysqli代碼轉換爲PDO。Mysqli to PDO - If else statement to try catch
之前,我有這樣的:
$updateTaskQuery = "UPDATE `task` SET `user_id` = {$query['user_id']}, `status_id` = {$query['status_id']} WHERE `id` = {$query['task_id']}";
$updateTask = mysqli_query($mysqli, $updateTaskQuery);
//### Check for error
if($mysqliError = mysqli_error($mysqli)) {
echo json_encode(array('error' => 'Update Task MySQLi Error: '.$mysqliError));
exit;
} else {
echo json_encode(array('success' => true));
exit;
}
到目前爲止,我已經是convertet這樣:
$sql = $db->prepare ("UPDATE `task` SET `user_id` = {$query['user_id']}, `status_id` = {$query['status_id']} WHERE `id` = {$query['task_id']}");
$sql->execute();
$updateTask = $sql->fetchAll (PDO::FETCH_ASSOC);
try {
$updateTask;
} catch (PDOException $ex) {
//handle
}
我的問題在這裏,我怎麼能包括我的新代碼else語句?
UPDATE:錯了,但工作代碼
try {
// Update task
$query = $db->prepare ("UPDATE `task` SET `user_id` = {$query['user_id']}, `status_id` = {$query['status_id']} WHERE `id` = {$query['task_id']}");
$query->execute();
echo json_encode (array (
'success' => true
));
} catch (PDOException $e) {
// catch a pdo error
echo json_encode (array (
'error' => 'Update Task PDO Error: ' . $e->getMessage(),
'error_trace' => $e->getTraceAsString()
));
}
你正在擊敗準備好的陳述點.. – 2014-11-15 00:01:11
@DarylGill並嘗試捕捉太。 – ArtisticPhoenix 2014-11-15 00:03:31
@ArtisiticPhoenix我沒有注意到這一點。我的大腦被炸第一個缺陷 – 2014-11-15 00:04:15