Question

我有这段代码：

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

那我怎么能用我的查询结果呢？当1子句中的所有条件都为真时，它会返回WHERE，如果出现错误，则返回任何内容。好吧，我需要知道如何处理以SELECT 1开头的查询结果？

Answer 1

您甚至不需要知道结果，因为您无论如何只会看到1，所以您需要做的就是查看是否有结果（以及有多少）：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

如果只选择包含单个值的单行，则必须使用fetchColumn（）。
请注意，您不能对使用vanila PDO的SELECT查询使用方法链接

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

如何通过PHP处理SELECT 1的结果？

2 个答案: