如何处理PHP的SELECT 1结果？

Question

我有以下代码：

$db
->prepare("SELECT 1
           FROM ( SELECT count(*) AS num_week,
                         sum(date_time > unix_timestamp(DATE_SUB(now(), INTERVAL 1  day))) as num_day,
                         sum(date_time > unix_timestamp(DATE_SUB(now(), INTERVAL 1  hour))) as num_hour,
                         sum(date_time > unix_timestamp(DATE_SUB(now(), INTERVAL 1 minute))) as num_1min
                   FROM resend_pass
                  WHERE user_id   = ?  
                    AND date_time > unix_timestamp(DATE_SUB(now(), INTERVAL 1 WEEK))
                ) a 
           WHERE num_week < 12 AND num_day < 6 AND num_hour < 4 AND num_1min < 1;")
->execute(array($id));

那么如何使用查询结果？ 当WHERE子句上的所有条件都为true时，它返回1 ；否则，则返回WHERE 。 好吧，我需要知道如何处理以SELECT 1开头的查询的结果？

Answer 1

您甚至不需要知道结果是什么 ，因为无论如何您只会看到1 ，所以您要做的就是查看是否有结果（以及有多少个）： PDOStatement :: rowCount

$stmt = $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_day < 4 AND num_1min < 1;");
$stmt->execute(array($id));
if($stmt->rowCount() > 0) {
    // Your code
}

Answer 2

当仅选择包含单个值的单个行时，必须使用fetchColumn（）。
请注意，对于Vanila PDO，不能对SELECT查询使用方法链接

$stmt = $db->prepare("SELECT 1 whatever");
$stmt->execute(array($id));
$result = $stmt->fetchColumn();