MySQL

Question

所以我有 2 个表，分别是预订和办公室

预订：

办公室：

请注意，我使用 BITWISE 运算来计算 30 分钟迭代中的可用时间。

现在的问题是，当用户需要预订新预订时，我需要从每个办公室的办公室可用性 (offices.mon_hours) 中减去预订时间 (sum(bookings.hours_booked))，这样我可以获得可用办公室的列表。

到目前为止，这是我的查询，但出现了这个错误：子查询 returns 多于 1 行

select * from offices o  
where (mon_hrs -  (SELECT SUM(hours_booked)
 from bookings boo 
 where boo.week = 44
 AND boo.year = 2020
 AND CAST(from_timestamp as DATE) = '2020-10-26'
 GROUP BY office_id)) & 4 ;

Answer 1

测试

SELECT offices.* 
FROM offices   
JOIN (SELECT SUM(hours_booked) summ, office_id
      FROM bookings 
      WHERE week = 44
        AND year = 2020
-- next replacement avoids table scan and allows index usage
--      AND CAST(from_timestamp as DATE) = '2020-10-26'
        AND from_timestamp >= '2020-10-26' 
        AND from_timestamp < '2020-10-26' + INTERVAL 1 DAY
      GROUP BY office_id) sums ON offices.id = sums.office_id
WHERE (offices.mon_hrs - sums.summ) & 4 ;

MySQL - 子查询 Returns 多于 1 行 - 减法问题

MySQL - Subquery Returns more than 1 row - subtraction issue

bitwise-operators