如果我有下表:
CREATE TABLE #temp (
id int,
num int,
question varchar(50),
qversion int );
INSERT INTO #temp VALUES(1, 1, 'Question 1 v1', 1);
INSERT INTO #temp VALUES(2, 1, 'Question 1 v2', 2);
INSERT INTO #temp VALUES(3, 2, 'Question 2 v1', 1);
INSERT INTO #temp VALUES(4, 2, 'Question 2 v2', 2);
INSERT INTO #temp VALUES(5, 2, 'Question 2 v3', 3);
INSERT INTO #temp VALUES(6, 3, 'Question 3 v1', 1);
SELECT *
FROM #temp;
DROP TABLE #temp;
我想找一张桌子来展示他们最新版本中的三个问题?这是在SQL Server 2005
中答案 0 :(得分:3)
CREATE TABLE #temp (
id int,
num int,
question varchar(50),
qversion int );
INSERT INTO #temp VALUES(1, 1, 'Question 1 v1', 1);
INSERT INTO #temp VALUES(2, 1, 'Question 1 v2', 2);
INSERT INTO #temp VALUES(3, 2, 'Question 2 v1', 1);
INSERT INTO #temp VALUES(4, 2, 'Question 2 v2', 2);
INSERT INTO #temp VALUES(5, 2, 'Question 2 v3', 3);
INSERT INTO #temp VALUES(6, 3, 'Question 3 v1', 1);
WITH latest AS (
SELECT num, MAX(qversion) AS qversion
FROM #temp
GROUP BY num
)
SELECT #temp.*
FROM #temp
INNER JOIN latest
ON latest.num = #temp.num
AND latest.qversion = #temp.qversion;
DROP TABLE #temp;
答案 1 :(得分:1)
SELECT t1.id, t1.num, t1.question, t1.qversion
FROM #temp t1
LEFT OUTER JOIN #temp t2
ON (t1.num = t2.num AND t1.qversion < t2.qversion)
GROUP BY t1.id, t1.num, t1.question, t1.qversion
HAVING COUNT(*) < 3;
答案 2 :(得分:1)
您正在使用SQL Server 2005,因此至少可以探索over
子句:
select
*
from
(select *, max(qversion) over (partition by num) as maxVersion from #temp) s
where
s.qversion = s.maxVersion
答案 3 :(得分:1)
我想找个表来显示每个问题的三个最新版本。
desc
关键字。where qversion is not null
。如果qversion可以为null,并且需要在结果集中包含空值,则需要进行其他更改。CREATE TABLE #temp (
id int,
num int,
question varchar(50),
qversion int );
INSERT INTO #temp VALUES(1, 1, 'Question 1 v1', 1);
INSERT INTO #temp VALUES(2, 1, 'Question 1 v2', 2);
INSERT INTO #temp VALUES(3, 2, 'Question 2 v1', 1);
INSERT INTO #temp VALUES(4, 2, 'Question 2 v2', 2);
INSERT INTO #temp VALUES(5, 2, 'Question 2 v3', 3);
INSERT INTO #temp VALUES(7, 2, 'Question 2 v4', 4);
-- ^^ Added so at least one row would be excluded.
INSERT INTO #temp VALUES(6, 3, 'Question 3 v1', 1);
INSERT INTO #temp VALUES(8, 4, 'Question 4 v?', null);
select id, num, question, qversion
from (select *,
row_number() over (partition by num order by qversion desc) as RN
from #temp
where qversion is not null) T
where RN <= 3