Question

我想写一个脚本来检测我的数据库中的任何重复的guest虚拟机，以便它们可以合并。

我正在尝试选择我的预订表中预订次数最多的来宾ID，因为这是我要合并的来宾。因此，如果有任何重复，它将返回一行，而id将是具有最多预订的访客。

不确定我该怎么做？

select 
    count(*) as duplicates,
    guests.id as id,
    guest_name(null, null, guests.id) as guest_name
from
    guests
        left join
    bookings_guests_link
        left join
    bookings ON bookings.id = bookings_guests_link.booking_id ON guests.id = bookings_guests_link.guest_id
        and bookings_guests_link.primary_client = 1
group by title , firstname , lastname , street , city , postcode , state_id
having duplicates > 1
order by count(bookings.id) desc , duplicates desc

任何帮助表示赞赏！谢谢

Answer 1

这应该给你每位客人和预订的数量。

您可以使用此

创建临时表

SELECT g.id, g.firstname, g.lastname, street, city, postcode, state_id,
       COUNT(b.id) AS NoOfBookings
FROM guests g
LEFT JOIN bookings_guests_link bgl
  ON g.id = bgl.guest_id
LEFT JOIN bookings b
  ON b.id = bgl.booking_id
GROUP BY g.id, g.firstname, g.lastname

这将为您提供重复的客人列表

SELECT title, firstname, lastname, street, city, postcode, state_id
FROM guests
GROUP BY title, firstname, lastname, street, city, postcode, state_id
HAVING COUNT(id) > 1

所以你只需要找到两者之间的匹配，并决定你实际想要如何合并记录。

MYSQL在一个表中检索行，在另一个表中它发生的最多

1 个答案: