[英]How to use SQL find first matching result from one table in another table?
假设我有两张桌子:
Customer -
ID | Name | Etc
1 | One |
2 | Two |
3 | Three |
4 | Four |
5 | Five |
... | ... |
Sales -
Customer ID | Date | Amount
5 | 1/20 | $45
5 | 3/19 | $145
3 | 8/19 | $453
7 | 3/20 | $4513
3 | 9/20 | ...
1 | 3/20 | ...
1 | 1/20 | ...
我想做的是编写一个查询,为每个客户找到第一笔销售。 我不确定该怎么做。 我觉得这是按问题分组的,但答案并没有出现。
编辑:我觉得我的第一个数据表并没有完全解释我的问题。 (老实说,在我编写解决方案之前,我什至没有意识到我的问题的这一方面) 注意:每次销售有多个客户。
Sales -
Sale ID | Customer ID | Date | Amount
1 | 5 | 1/20 | $45
5 | 5 | 3/19 | $145
8 | 3 | 8/19 | $453
7 | 7 | 3/20 | $4513
3 | 4 | 9/20 | ...
2 | 1 | 3/20 | ...
1 | 1 | 1/20 | ...
您可以使用子查询为每个客户的每次销售分配一个行号,按日期升序排序,然后 select 仅第一行:
SELECT "Customer ID", "Date", "Amount"
FROM (
SELECT "Customer ID", "Date", "Amount",
ROW_NUMBER() OVER (PARTITION BY "Customer ID" ORDER BY "Date") AS rn
FROM Sales) s
WHERE rn = 1
Nick 的解决方案可能是性能最高的,但如果你想在这里使用GROUP BY
,你可以这样做:
SELECT
c.ID,
c.Name,
s1.Date,
s1.Amount
FROM Customer c
INNER JOIN Sales s1 ON c.ID = s1."Customer ID"
INNER JOIN
(
SELECT "Customer ID", MIN(Date) AS FirstSaleDate
FROM Sales
GROUP BY "Customer ID"
) s2
ON s1."Customer ID" = s2."Customer ID" AND
s1.Date = s2.FirstSaleDate
ORDER BY
c.ID,
c.Name;
在 Oracle 中,您可以使用keep
:
select customer_id, min(date) as first_sales_date,
max(amount) keep (dense_rank first order by date asc) as first_amount
from sales
group by customer_id;
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