Question

我有一个查询从两个表中搜索最近的事务。查询工作正常并返回 2结果。

    SELECT *
FROM
  (SELECT ROW_NUMBER() over (partition BY lower(tb1.email) order by tb1.reserve_date DESC) AS ranking,
    tb1.id,
    tb1.reserve_date
  FROM TableTester1 tb1
  UNION
  SELECT ROW_NUMBER() over (partition BY lower(tb2.email) order by tb2.reserve_date DESC) AS ranking,
    tb2.id,
    tb2.reserve_date
  FROM TableTester2 tb2
  )
WHERE ranking = 1;

但是根据要求，我仍然需要减少它并获得更新的。因此，查询应该只返回 1结果。

我已阅读汇总功能并尝试使用 MAX（）和 GROUP BY 功能，如下所示：

SELECT MAX(reserve_date)
FROM
  (SELECT ROW_NUMBER() over (partition BY lower(tb1.email) order by tb1.reserve_date DESC) AS ranking,
    tb1.id,
    tb1.reserve_date
  FROM TableTester1 tb1
  UNION
  SELECT ROW_NUMBER() over (partition BY lower(tb2.email) order by tb2.reserve_date DESC) AS ranking,
    tb2.id,
    tb2.reserve_date
  FROM TableTester2 tb2
  )
WHERE ranking = 1;

但它只会导致 ORA-00979：不是GROUP BY表达式错误。

有关如何解决此问题的任何建议？

Answer 1

为什么不限制前两个：

SELECT id, reserve_date
FROM
(
    SELECT
        ROW_NUMBER() OVER (PARTITION BY LOWER(tb1.email)
            ORDER BY reserve_date DESC) AS rn,
        id,
        reserve_date
    FROM TableTester1
) t
WHERE rn <= 2;

如果你真的是指 rank 对应前两个记录的所有记录，那么你可以用ROW_NUMBER或RANK替换DENSE_RANK，具体取决于你要使用哪一个。

Answer 2

使用它：

 SELECT * FROM
 (SELECT ROW_NUMBER() over (partition BY lower(tb.email) order by tb.reserve_date DESC) AS ranking,
tb1.id,
tb1.reserve_date from 
(select * FROM TableTester1 tb1

UNION

select *  FROM TableTester2 tb2
)tb)B
WHERE ranking = 1;

Answer 3

在row_number()：

之后执行union all

SELECT tt.*
FROM (SELECT tt.*, ROW_NUMBER() OVER (ORDER BY reserve_date DESC) as seqnm
      FROM ((SELECT tb1.id, tb1.reserve_date
             FROM TableTester1 tb1
            ) UNION ALL
            (SELECT tb2.id, tb2.reserve_date
             FROM TableTester2 tb2
            )
           ) tt
     ) tt
WHERE seqnum = 1;

返回两个表的结果集中的最新结果

3 个答案: