Question

嘿，我一直在试图弄清楚如何进行这些查询。有人可以帮我吗。这些是我目前的表格。

BOOKING

HOTEL_NO
GUEST_NO
DATE_FROM DATE_TO
ROOM_NO

GUEST

GUEST_NO
客人姓名 CITY
地址
ZIP_CODE

HOTEL

HOTEL_NO
HOTEL_NAME CITY
地址
ZIP_CODE
STAR

ROOM

ROOM_NO
HOTEL_NO
房型价格

这些是我需要做的查询。

- 在同一家酒店预订所有预订（过去和现在）的客人。

创建一个VIP-Guest视图，其中列出了仅预订4星级酒店的客人 4星级酒店

- 在VIP中找到总逗留时间最长的客人（按天数计算）。将此表达为具有视图且没有视图的查询

有人能帮助我吗？

Answer 1

这应该让你开始。要发布stackoverflow，您需要提出具体问题或错误或问题。就像你在评论中发布的查询一样....这可能是一个问题：“我有这些表，这个特定的目标（问题/结果集），我试过这个查询..它给了我这个结果，或者它给了我这个错误。“

BOOKING:  HOTEL_NO, GUEST_NO, DATE_FROM, DATE_TO, ROOM_NO

GUEST:    GUEST_NO, GUEST_NAME, CITY, ADDRESS, ZIP_CODE

HOTEL:    HOTEL_NO, HOTEL_NAME, CITY, ADDRESS, ZIP_CODE, STAR

ROOM:     ROOM_NO, HOTEL_NO, ROOM_TYPE, PRICE

所有客人和预订......

   -- all guests:  select * from guest;
   -- all bookings:  select * from booking; 

   select   * 
   from     guest
            join booking on guest.guest_no = booking.guest_no;

   -- which is the same as... 

   select   * 
   from     guest, booking 
   where    guest.guest_no = booking.guest_no;

   -- and... your comments query was missing a group by clause

   select   guest_no, guest_name, count(*) as booking_count
   from     guest
            join booking on guest.guest_no = booking.guest_no
   group    by guest_no, guest_name; 

   select   guest_no, guest_name, count(distinct hotel_no) as hotel_count
   from     guest
            join booking on guest.guest_no = booking.guest_no
   group    by guest_no, guest_name 
   having   count(distinct hotel_no) = 1;

和我count(distinct hotel_no)因为......他们可能在A酒店预订3次，在B酒店预订1次。基本的加入会为我提供4行预订。我不在乎预订多少。我在乎多少家酒店。所以我想计算每人hotel_no的不同出现次数（有那个分组）而不是每一行。

来自他们明星的客人......

   -- so we have to get guest and hotel joined.  bc hotel has stars.  
   -- booking has hotel_no.  so... we can use that last query and 
   -- join in HOTEL to get the star information.  in the WHERE you 
   -- will want to put your filter for the number of stars that you
   -- are looking for =4 or >=4 or something like that.
   -- you might want to check out DISTINCT to get just a list of names 
   -- instead of a row for each booking.

他们住的天数......

   -- use the second query.  
   -- datediff(date_to, date_from) as days_stay gives you the length of stay  
   -- i don't know what the view is. 
   -- to get the top length could go two ways... either ORDER BY and LIMIT if there is 
   -- only one person with the top length (let's say 10 days).  if there are many people
   -- who have stayed 10 days, you'll need to do a MAX on the days_stay and either join 
   -- that in or use it in the WHERE as a nested select.

这假设有一个最长的逗留时间。只有一个人住了10天。

SELECT  guest_no, guest_name, datediff(date_to, date_from) days_stayed

FROM    vip_guest 

        join booking on vip_guest.guest_no = booking.guest_no 

order   by datediff(date_to, date_from) desc
limit   1,1

这应该适用于许多人......（我不测试这些...只是看看它）

SELECT  distinct guest_no, guest_name, datediff(date_to, date_from) max_stay

FROM    vip_guest

        join booking on vip_guest.guest_no = booking.guest_no 

where   datediff(date_to, date_from) = (
                     select max(datediff(date_to, date_from)) as days_stayed 
                     from   booking )

嵌套查询获取每个人的最长逗留时间。 vip_guest和预订加在一起给我们客人和日期imfo。我们将为每个vip_guest获取所有预订。所以我们希望将其过滤到保持长度==最大停留长度的位置。如果一个人有10天多次停留（我的任意最长停留时间长度）...使用不同的。

现在......这是关于嵌套查询的一个好点。我不知道你的观点是什么。只要最长逗留时间，最大贵宾客人都不可能住宿。在这种情况下，此查询将不返回任何内容。

MySQL复杂查询

1 个答案: