Question

CREATE TABLE #TBLA (id int, numb int, value int)
INSERT INTO #TBLA 
VALUES (1,3,24),
       (1,4,14),
       (2,2,61),
       (2,1,12),       
       (2,3,54)



CREATE TABLE #TBLB (ID int, Value int)
INSERT INTO #TBLB (ID)
VALUES (1),(2)

我想更新#TBLB值列，以便该值与表A中的min（麻木）相关。预期结果： TBLB

id value 
1   24
2   12

谢谢！

Answer 1

UPDATE #TBLB
SET Value = A.value 
FROM #TBLA A
JOIN 
(select B.ID, min(A.numb)  as mi
from #TBLA A 
join #TBLB B ON a.id = B.id 
GROUP BY B.ID) res 
ON A.id = res.ID and a.numb = res.mi

Answer 2

;WITH CTE 
AS (
    SELECT *
      ,ROW_NUMBER() OVER (PARTITION BY ID ORDER BY Numb ASC) rn 
    FROM #TBLA
   )
UPDATE B
 SET B.Value = C.value
FROM #TBLB B 
INNER JOIN CTE C ON B.ID = C.id AND C.rn = 1

SQL Server 2012：根据另一个表的列中的最小值更新列

2 个答案: