Question

SELECT
   ass.assessmentAmount -- want to fetch assessmentAmount of min(ass.assessmentId)
   ass.assessmentId
FROM
   --bunch of joins  
WHERE
ass.assessmentId = (SELECT min(ass2.assessmentId) FROM Assessment ass2
   --same bunch of joins

它看起来很混乱，因为我有六个条件加入，我不想重复两次。还有另一种方法吗？

Answer 1

使用MIN( ass.assessmentId ) OVER ()分析函数：

SELECT *
FROM   (
  SELECT ass.assessmentAmount,
         ass.assessmentId,
         MIN( ass.assessmentId ) OVER () AS min_assessmentId
  FROM   --bunch of joins
)
WHERE assessmentId = min_assessmentId;

您还可以使用RANK()：

SELECT *
FROM   (
  SELECT ass.assessmentAmount,
         ass.assessmentId,
         RANK() OVER ( ORDER BY ass.assessmentId ) AS rnk
  FROM   --bunch of joins
)
WHERE rnk = 1;

如果assessmentId为UNIQUE，并且只能将一行作为最低限度，那么您可以将RANK替换为ROW_NUMBER;但是，您也可以使用ROWNUM伪列来获得所需的结果：

SELECT *
FROM   (
  SELECT ass.assessmentAmount,
         ass.assessmentId
  FROM   --bunch of joins
  ORDER BY ass.assessmentId ASC
)
WHERE ROWNUM = 1;

Answer 2

使用带有row_number的CTE

with CTE as
(
select assessmentId, 
       assessmentAmount , 
       row_number() over (order by assessmentid asc) as rn
from --bunch of joins
)
select *
from CTE
where rn = 1

Oracle：仅从具有min（id）的行中选择值

2 个答案: