Question

我在SQL中遇到查询问题。我的查询是这样的：

SELECT t.id, t.attribute, t.name, t.date
FROM (
SELECT *
FROM table1
WHERE ...
UNION
SELECT *
FROM table2
WHERE
UNION
...) t
WHERE ...
GROUP BY t.attribute

我想要的是获取每个属性的最新日期的所有行（日期可以为null，然后是最新的）。我知道我可以创建一个临时表，但如果可能的话我想避免它。我也希望不再在WHERE和JOIN中再次嵌套FROM的相同查询。有什么办法吗？

谢谢！

Answer 1

这是使用CTE（公用表表达式）执行此操作的一种方法：

{{1}}

Answer 2

使用modern SQL，您可以使用窗口函数：

SELECT t2.id, t2.attribute, t2.name, t2.date
FROM (
    SELECT t1.id, t1.attribute, t1.name, t1.date, 
           row_number() over (partition by t1.attribute order by t1.date desc) as rn
    FROM (
      SELECT *
      FROM table1
      WHERE ...
      UNION
      SELECT *
      FROM table2
      WHERE ...
      UNION
      ...
   ) t1
) t2
where rn = 1;

或者使用公用表表达式来避免派生表的嵌套：

with data as (
  SELECT *
  FROM table1
  WHERE ...
  UNION
  SELECT *
  FROM table2
  WHERE ...
  UNION
), numbered as (
  SELECT id, attribute, name, date, 
         row_number() over (partition by t1.attribute order by t1.date desc) as rn
  FROM data
)
SELECT id, attribute, name, date
from numbered 
where rn = 1;

如何使用嵌套选择在查询列中获取具有最大值的整行？

2 个答案: