Question

我想要一个SQL或PHP查询，在六个月后停用一个帐户。

简单的解决方案是

Select datediff(curdate(),last_date) as diff

我进入了复杂的情况，我不得不排除星期六和星期六。周日。

我被困在将来必须记录状态= 0

的日期或日期

我在这里找到的答案基于两个日期或日期范围但我必须计算第二次约会。

Answer 1

相当肯定必须有一种比这更有效的方法，但值得一试（使用NOW（）作为开始日期）： -

SELECT 
    CASE 
        WHEN DAYOFWEEK(NOW()) = 7
            THEN 
                CASE 
                    WHEN DAYOFWEEK(DATE_ADD(DATE(NOW()), INTERVAL (160 DIV 5) * 7 + (160 % 5) + 2 DAY)) = 7
                        THEN DATE_ADD(DATE(NOW()), INTERVAL (160 DIV 5) * 7 + (160 % 5) + 4 DAY)
                    WHEN DAYOFWEEK(DATE_ADD(DATE(NOW()), INTERVAL (160 DIV 5) * 7 + (160 % 5) + 2 DAY)) = 1
                        THEN DATE_ADD(DATE(NOW()), INTERVAL (160 DIV 5) * 7 + (160 % 5) + 3 DAY)
                    ELSE 
                        DATE_ADD(DATE(NOW()), INTERVAL (160 DIV 5) * 7 + (160 % 5) + 2 DAY)
                END
        WHEN DAYOFWEEK(NOW()) = 1
            THEN
                CASE 
                    WHEN DAYOFWEEK(DATE_ADD(DATE(NOW()), INTERVAL (160 DIV 5) * 7 + (160 % 5) + 1 DAY)) = 7
                        THEN DATE_ADD(DATE(NOW()), INTERVAL (160 DIV 5) * 7 + (160 % 5) + 3 DAY)
                    WHEN DAYOFWEEK(DATE_ADD(DATE(NOW()), INTERVAL (160 DIV 5) * 7 + (160 % 5) + 1 DAY)) = 1
                        THEN DATE_ADD(DATE(NOW()), INTERVAL (160 DIV 5) * 7 + (160 % 5) + 2 DAY)
                    ELSE 
                        DATE_ADD(DATE(NOW()), INTERVAL (160 DIV 5) * 7 + (160 % 5) + 1 DAY)
                END
        ELSE
            CASE 
                WHEN DAYOFWEEK(DATE_ADD(DATE(NOW()), INTERVAL (160 DIV 5) * 7 + (160 % 5) DAY)) = 7
                    THEN DATE_ADD(DATE(NOW()), INTERVAL (160 DIV 5) * 7 + (160 % 5) + 2 DAY)
                WHEN DAYOFWEEK(DATE_ADD(DATE(NOW()), INTERVAL (160 DIV 5) * 7 + (160 % 5) DAY)) = 1
                    THEN DATE_ADD(DATE(NOW()), INTERVAL (160 DIV 5) * 7 + (160 % 5) + 1 DAY)
                ELSE 
                    DATE_ADD(DATE(NOW()), INTERVAL (160 DIV 5) * 7 + (160 % 5) DAY)
            END

    END

需要天数并除以5才能获得周数，并将mod除以5以获得额外天数。如果开始日期是星期六，则会增加2天，如果星期日增加1天，则会增加7倍星期加上额外天数。然后，如果结果日是星期六，则又增加2天，如果是星期日又增加1天。

或者有点奇怪的做法： -

SELECT MAX(EndDate)
FROM
(
    SELECT @StartDate:=DATE_ADD(@StartDate, INTERVAL CASE WHEN DAYOFWEEK(@StartDate) = 6 THEN 3 WHEN DAYOFWEEK(@StartDate) = 7 THEN 3 WHEN DAYOFWEEK(@StartDate) = 1 THEN 2 ELSE 1 END DAY) AS EndDate
    FROM
    (
        SELECT units.i + tens.i * 10 + hundreds.i * 100 AS daysAdd
        FROM (SELECT 0 AS i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) units,
        (SELECT 0 AS i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) tens,
        (SELECT 0 AS i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) hundreds
        HAVING daysAdd < 160
    ) Sub1
    CROSS JOIN
    (
        SELECT @aCnt:=0, @StartDate:=DATE(NOW())
    ) sub2
) Sub0

160天后的未来日期或日期，不包括周六周日

1 个答案: