我有這樣的代碼:如何處理PHP的SELECT 1的結果?
$db
->prepare("SELECT 1
FROM (SELECT count(*) AS num_week,
sum(date_time > unix_timestamp(DATE_SUB(now(), INTERVAL 1 day))) as num_day,
sum(date_time > unix_timestamp(DATE_SUB(now(), INTERVAL 1 hour))) as num_hour,
sum(date_time > unix_timestamp(DATE_SUB(now(), INTERVAL 1 minute))) as num_1min
FROM resend_pass
WHERE user_id = ?
AND date_time > unix_timestamp(DATE_SUB(now(), INTERVAL 1 WEEK))
) a
WHERE num_week < 12 AND num_day < 6 AND num_hour < 4 AND num_1min < 1;")
->execute(array($id));
那麼我該如何使用我查詢的結果?當WHERE
條款的所有條件爲時,返回1
真或者在出現問題時爲空。那麼我需要知道如何處理以SELECT 1
開頭的查詢結果?
注意,如果'date_time'是'DATETIME'然後查詢不應該使用'UNIX_TIMESTAMP()'這需要一個日期作爲參數,但在Unix紀元時間返回一個數值。 'SELECT 1 AS over_limit FROM ...'會給你的輸出列起一個名字。 –