Question

昨天获得的纯粹MySQL方式是什么，但仅限于工作日？

现在我昨天使用SUBDATE(CURRENT_DATE, 1))，但周一它将在周日返回。

我希望它能够返回上周五的日期结果，换句话说，也就是前一个工作日的结果。

我应该澄清我试图在查询的WHERE部分（以及子查询）中使用它。

SELECT ... WHERE DATE(timestamp) = SUBDATE(CURRENT_DATE, 1))

以下是整个查询：

SELECT r.drivername, l.branchcity as location, COUNT(o.ordernum) as deliveries, COUNT(x.pictures) as pics, CONCAT(ROUND((COUNT(x.pictures) / COUNT(o.ordernum))*100,2),'%') as percentage_having_images 
FROM deliveries d, drivers r, locations l, staging s, stations t, orders o LEFT OUTER JOIN 
(SELECT a.ordernum AS pictures
FROM orders a WHERE a.stationID = '16' AND DATE(a.scantime) = SUBDATE(CURRENT_DATE, 1)) x 
ON x.pictures = o.ordernum
WHERE o.deliveryID = d.ID AND d.driverID = r.ID AND s.locationID = l.ID AND o.stationID = t.ID AND o.stagingID = s.ID AND t.ID IN ('11','12','13') AND DATE(o.scantime) = SUBDATE(CURRENT_DATE, 1) GROUP BY s.locationID, r.drivername ORDER BY s.locationID, percentage_having_images DESC

Answer 1

SELECT ..
.
.
AND DATE(a.scantime) = (CASE WEEKDAY(CURRENT_DATE) 
                             WHEN 0 THEN SUBDATE(CURRENT_DATE,3)
                             WHEN 6 THEN SUBDATE(CURRENT_DATE,2) 
                             WHEN 5 THEN SUBDATE(CURRENT_DATE,1)
                             ELSE SUBDATE(CURRENT_DATE,1) 
                        END)
..
..

Answer 2

试试这个来结束这一天：

     SELECT
CASE DAYOFWEEK(SUBDATE(CURRENT_DATE, 1))
      WHEN 1 THEN SELECT SUBDATE(CURRENT_DATE, 3);
      WHEN 7 THEN SELECT SUBDATE(CURRENT_DATE, 2);
      ELSE
        BEGIN
          SUBDATE(CURRENT_DATE, 1);
        END;
    END CASE;

然后使用子查询

SELECT ... WHERE DATE(timestamp) = (SELECT
    CASE DAYOFWEEK(SUBDATE(CURRENT_DATE, 1))
          WHEN 1 THEN SELECT SUBDATE(CURRENT_DATE, 3);
          WHEN 7 THEN SELECT SUBDATE(CURRENT_DATE, 2);
          ELSE
            BEGIN
              SUBDATE(CURRENT_DATE, 1);
            END;
        END CASE);

Answer 3

好的，所以不是最容易阅读的解决方案，而是更短更冷。 ; - ）

SELECT CURRENT-DATE - if((6+weekday(@dat))%7 > 4, (6+weekday(@dat))%7-3, 1) as  prev_workday

解释：至少在我的时区给出工作日（＆＃39;星期一 - 日期＆＃39;）= 0，在星期日6。用+6我将它转移到6 .. 12.使用modulu 7（ %7）我将其转移到星期六= 4，星期日= 5，星期一= 6。

因此，在大于4的那个班次之后的所有日子里，我减少了这个价值的日期 - 3，否则我会在＃39;之前的那一天。所以在周六我减少一天，周日减少两天，周一减少三天。

使用SQL选择上一个工作日

3 个答案: