将最后一个外观添加到表中

Question

我有两张桌子（例如）
table1 将包含商店，有 2 种类型“糖果店”和“牙科店”。 每行包含有关客户在特定商店购买的信息

table1（购买）：

+----+---------+------------------+-------+
| id | store   | date of purchase | money |
| 1  | store 1 | 2016-01-01       | 10    |
| 1  | store 5 | 2018-01-01       | 50    |
| 2  | store 2 | 2017-01-20       | 10    |
| 2  | store 3 | 2019-02-20       | 15    |
| 3  | store 2 | 2017-02-02       | 20    |
| 3  | store 6 | 2019-01-01       | 60    |
| 1  | store 1 | 2015-01-01       | 20    |
+----+---------+------------------+-------+

table2（类型）：

+---------+--------+
| store   | type   |
| store 1 | candy  |
| store 2 | candy  |
| store 3 | candy  |
| store 4 | dental |
| store 5 | dental |
| store 6 | dental |
+---------+--------+

我希望我的查询返回这样的表：

+----+---------------+-----------------+---------------+
| id | the last place| the last date c |the last date d|
| 1  | store 5       | 2016-01-01      |  2018-01-01   |
| 2  | store 3       | 2019-02-20      |  -            |
| 3  | store 6       | 2017-02-02      |  2019-01-01   |
+----+---------------+-----------------+---------------+

在哪里

• [最后一个地方]是客户最后购买的店铺；


• [最后日期 c] - 在糖果店购买的最后日期


• [最后日期d]是最后在牙科店购买的日期

如果没有任何匹配项，则删除行（所有值均为空）

Answer 1

你可以使用条件聚合和join:

select t1.id,
       (array_agg(store order by t1.date_of_purchase desc))[1] as last_store,
       max(t1.date_of_purchase) filter (where t2.type = 'candy') as last_candy,
       max(t1.date_of_purchase) filter (where t2.type = 'dental') as last_dental
from table1 t1 join
     table2 t2
     using (store)
group by t1.id;

Answer 2

你可以在 postgresql 中使用 group by,max 和 window function :

select id
,max(case when rnb=1 then store else null end) the_last_place
,max(case when type='candy' then date else null end) the_last_date_c
,max(case when type='dental' then date else null end) the_last_date_d
 from(
    select t1.id,t1.store,t1.date,row_number() over(partition by t1.id order by t1.date desc) rnb
    from table1 t1 
    inner join table2 t2 
    on t1.store=t2.store
    ) t
group by id ;