Question

查询以获取入住率最高的房屋的房屋详细信息 有两个表房屋和租户历史记录用于计算最高入住率。
所以首先我要使用 datediff 函数来获取停留时间。然后创建视图以使其充当虚拟表，并可以访问该列以获取最大值。

create view [dbo].[vWHouseStay]
as
    select profile_id, 
    house_Id, 
    house_type,
    bhk_details,Bed_type,
    bed_count,
    furnishing_type ,
    DATEDIFF(MONTH,move_in_date,move_out_date) AS 'Total Length of stay'
from Tenancy_histories
join houses on  tenancy_histories.house_Id = houses.house_id

我需要的是住所信息最多的住所。我怎么做？基本上，它只应退还最多[入住总时间]的房子。

表结构：

**Tenancy_histories**
Field Type Null Key Default
id int(11) NO PRI auto_increment
profile_id int(11) NO FK
house_id int(11) NO FK
move_in_date date NO
move_out_date date YES
rent int(11) NO
Bed_type varchar(255) YES
move_out_reason varchar(255) YES

**Houses**
Field Type Null Key Default
house_id int(11) NO PRI auto_increment
house_type varchar(255) YES
bhk_details varchar(255) YES
bed_count int(11) NO
furnishing_type varchar(255) YES
Beds_vacant int(11) NO

样本数据

 house_id   house_type  bhk_details bed_count   furnishing_type OccupancyDays
5   Independent 4 BHK   4   fully-furnished 443
7   Apartment   3 BHK   3   semifurnished   417
4   Apartment   2 BHK   2   fully-furnished 397
18  Independent 2 BHK   2   fully-furnished 358
16  Apartment   3 BHK   3   fully-furnished 324
19  Independent 3 BHK   3   fully-furnished 290
3   Apartment   3 BHK   6   fully-furnished 226
1   Apartment   3 BHK   5   unfurnished NULL
2   Apartment   3 BHK   3   unfurnished NULL
17  Independent 3 BHK   3   fully-furnished NULL
6   Apartment   3 BHK   3   semifurnished   NULL
8   Apartment   2 BHK   4   fully-furnished NULL

样本数据[租赁历史]

    id  profile_id  house_id    move_in_date    move_out_date   rent    Bed_type    move_out_reason
242 1   5   2015-02-12  2016-04-30  7500    bed MOVE_OUT
243 2   2   2015-06-05  NULL    11000   room    
244 3   4   2015-10-28  2016-11-28  12000   room    RENT_CHANGE
245 4   1   2015-04-26  NULL    8000    bed 
246 5   3   2015-05-15  2015-12-27  9000    bed MOVE_OUT
247 6   8   2015-12-25  NULL    10200   room    
248 7   6   2015-11-20  NULL    6500    bed 
249 8   7   2015-11-10  2016-12-31  7200    bed MOVE_OUT
250 9   9   2015-10-15  NULL    7500    bed 
251 10  10  2015-06-20  NULL    7500    bed 
252 11  19  2015-08-29  2016-06-14  8000    bed INTERNAL_TRANSFER
253 12  15  2015-02-24  NULL    11000   room    
254 13  12  2015-02-25  NULL    12000   room    
255 14  18  2016-01-07  2016-12-30  13500   room    MOVE_OUT
256 15  13  2015-04-07  NULL    6500    bed 
257 16  17  2015-04-23  NULL    6500    bed 
258 17  14  2015-02-10  NULL    10500   room    
259 18  16  2015-10-16  2016-09-04  8000    bed MOVE_OUT
260 19  20  2015-09-26  NULL    7500    bed 
261 20  11  2015-09-30  NULL    9500    bed

Answer 1

好像您想汇总所有租赁历史记录。这将是这样。也可以在视图中使用它。

;with cte as(
select
    h.house_id
    ,h.house_type
    ,h.bhk_details
    ,h.bed_count
    ,h.furninshing_type
    ,OccupancyDays = sum(datediff(day,th.move_in_date,move_out_date))
from 
    Houses h
inner join
    Tenancy_histories th on th.house_id = h.house_id)
group by
    h.house_id
    ,h.house_type
    ,h.bhk_details
    ,h.bed_count
    ,h.furninshing_type

select TOP 1 --you can remove the TOP 1 to bring them all back 
*
from cte
order by OccupancyDays desc

使用View＆Max的最高详细信息的人？

1 个答案: