3

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;

我想要一张表格来显示最新版本的三个问题？这是在 SQL Server 2005 中

4

4 回答 4

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;
于 2009-09-29T00:47:35.017 回答
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;
于 2009-09-29T00:55:36.070 回答
1

您使用的是 SQL Server 2005，因此至少值得探索该over子句：

select
    *
from
    (select *, max(qversion) over (partition by num) as maxVersion from #temp) s
where
    s.qversion = s.maxVersion
于 2009-09-29T01:52:46.800 回答
1

我想要一张表格来显示每个问题的三个最新版本。

  1. 我假设qversion 随着时间的推移而增加。如果这个假设是倒退的，请desc从答案中删除关键字。
  2. 表定义对 qversion 没有显式的非空约束。我认为应该排除空 qversion。（注意：根据设置，声明中缺少明确的 null/not null 可能会导致 not null 约束。）如果表确实有 not null 约束，where qversion is not null则应删除文本。如果 qversion 可以为空，并且结果集中需要包含空值，则需要进行额外的更改。
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
于 2009-09-29T02:28:27.617 回答