sql查询到条件中的第一行

vyswwuz2  于 2021-06-21  发布在  Mysql
关注(0)|答案(2)|浏览(382)

我有一个sql查询:

SELECT (CASE WHEN count(_days) > 0 THEN 'yes' ELSE 'No' END) as availability 
  FROM ( 
        SELECT count(roomTypeDay2.room_type_id) as _days  
          FROM room_type_day as roomTypeDay2 
     LEFT JOIN room_type as roomType2 on roomTypeDay2.room_type_id = roomType2.id  
         WHERE roomType2.accommodation_id=3 
           AND roomTypeDay2.date IN ( '2018-06-09 00:00:00','2018-06-10 00:00:00','2018-06-11 00:00:00')  
      GROUP BY roomTypeDay2.room_type_id  
        HAVING COUNT(roomTypeDay2.room_type_id) = 3 
       ) as disponible

这很好,但是现在,我想过滤第一个日期行(2018-06-09 00:00:00)是否
“最短住宿时间<=3天,最长释放时间<=2天”
参数,但我不知道怎么做。
我尝试将以下行添加到查询:
和(roomtypeday2.date='2018-06-09 00:00:00'和roomtypeday2.min\u night\u stay<=3和roomtypeday2.release\u days<=2)
但这不是正确的查询。
乳突

SELECT (CASE WHEN count(_days) > 0 THEN 'yes' ELSE 'No' END) as availability 
   FROM ( 
        SELECT count(roomTypeDay2.room_type_id) as _days  
          FROM room_type_day as roomTypeDay2 
     LEFT JOIN room_type as roomType2 on roomTypeDay2.room_type_id = roomType2.id  
           AND roomType2.accommodation_id=3 
         WHERE roomTypeDay2.date IN ( '2018-06-09 00:00:00','2018-06-10 00:00:00','2018-06-11 00:00:00')  
      GROUP BY roomTypeDay2.room_type_id  
        HAVING COUNT(roomTypeDay2.room_type_id) = 3 
         ) as disponible

解决方案

SELECT (CASE WHEN count(_days) > 0 THEN 'yes' ELSE 'No' END) as availability 
    FROM ( 
        SELECT count(roomTypeDay2.room_type_id) as _days  
        FROM room_type_day as roomTypeDay2  
        LEFT JOIN room_type as roomType2 on roomTypeDay2.room_type_id = roomType2.id 
        WHERE roomType2.accommodation_id=3 
        AND roomTypeDay2.num_rooms_available > 0  
        AND (roomTypeDay2.date = '2018-06-12 00:00:00' AND roomTypeDay2.min_night_stay <= 2 AND roomTypeDay2.release_days <= 2 )
        OR roomTypeDay2.date IN ( '2018-06-13 00:00:00','2018-06-14 00:00:00')  
        GROUP BY roomTypeDay2.room_type_id  
        HAVING COUNT(roomTypeDay2.room_type_id) = 3 
    ) as disponible
v9tzhpje

v9tzhpje1#

这是你的 WHERE 条款:

WHERE roomTypeDay2.date IN ('2018-06-09 00:00:00',
                            '2018-06-10 00:00:00',
                            '2018-06-11 00:00:00')

现在,对于第一次约会,您需要附加条件。使用 AND 以及 OR 带圆括号:

WHERE 
(
  roomTypeDay2.date = '2018-06-09 00:00:00'
  AND
  roomTypeDay2.min_night_stay <= 3 
  AND
  roomTypeDay2.release_days <= 2
)
OR
roomTypeDay2.date IN ('2018-06-10 00:00:00','2018-06-11 00:00:00')
blmhpbnm

blmhpbnm2#

SELECT
  (
    CASE WHEN count(_days) > 0 THEN 'yes' ELSE 'No' END
  ) as availability
from
  (
    SELECT
      count(roomTypeDay2.room_type_id) as _days
    FROM
      room_type_day as roomTypeDay2
      LEFT JOIN room_type as roomType2 on roomTypeDay2.room_type_id = roomType2.id
      AND roomType2.accommodation_id = 3
    WHERE
      roomTypeDay2.date IN ('2018-06-10 00:00:00', '2018-06-11 00:00:00')
    GROUP BY
      roomTypeDay2.room_type_id
    HAVING
      COUNT(roomTypeDay2.room_type_id) = 3
    union
    SELECT
      count(roomTypeDay2.room_type_id) as _days
    FROM
      room_type_day as roomTypeDay2
      LEFT JOIN room_type as roomType2 on roomTypeDay2.room_type_id = roomType2.id
      AND roomType2.accommodation_id = 3
    WHERE
      roomTypeDay2.date = '2018-06-09 00:00:00'
      and roomTypeDay2.min_night_stay <= 3
      AND roomTypeDay2.release_days <= 2
    GROUP BY
      roomTypeDay2.room_type_id
    HAVING
      COUNT(roomTypeDay2.room_type_id) = 3
  ) disponible

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