SQL Server 2012中：基于另一个表中的列最小值更新列

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，这样的价值是eual从表A. 预期结果最小（麻木）： 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