Oracle：仅从具有 min(id) 的行中选择值

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 6 个带条件的连接，我不想重复两次。还有另一种方法可以做到这一点吗？

Answer 1

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

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是UNIQUE并且至少只能有一行，那么您可以替换RANK为ROW_NUMBER; 但是，您也可以使用伪列获得所需的结果ROWNUM：

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