给定以下结构的表 foo (Oracle 11g)：
ID | GROUP_ID
1 | 100
2 | 100
3 | 100
4 | 200
5 | 300
6 | 300
7 | 400
我想选择前 n 行（按 ID 排序）或更多，这样我总是得到一个完整的组。
例子：
n = 2：我想至少获得前两行，但由于 ID 3 也属于组 100，我也想获得它。
n = 4：给我前四行，我很高兴 ;-)
n = 5：请求第 1-6 行。
非常感谢您的帮助！
问问题
2760 次
4 回答
7
解决方案使用rank()：
select id, group_id
from (select t.*, rank() over (order by group_id) as rnk
from t)
where rnk <= :n;
建筑测试数据：
SQL> create table t (id number not null primary key
2 , group_id number not null);
Table created.
SQL> insert into t values (1, 100);
1 row created.
SQL> insert into t values (2, 100);
1 row created.
SQL> insert into t values (3, 100);
1 row created.
SQL> insert into t values (4, 200);
1 row created.
SQL> insert into t values (5, 300);
1 row created.
SQL> insert into t values (6, 300);
1 row created.
SQL> insert into t values (7, 400);
1 row created.
SQL> commit;
Commit complete.
SQL>
跑步...
SQL> var n number
SQL> exec :n := 2;
PL/SQL procedure successfully completed.
SQL> select id, group_id
2 from (select t.*, rank() over (order by group_id) as rnk
3 from t)
4 where rnk <= :n;
ID GROUP_ID
---------- ----------
1 100
2 100
3 100
SQL> exec :n := 4;
PL/SQL procedure successfully completed.
SQL> select id, group_id
2 from (select t.*, rank() over (order by group_id) as rnk
3 from t)
4 where rnk <= :n;
ID GROUP_ID
---------- ----------
1 100
2 100
3 100
4 200
SQL> exec :n := 5;
PL/SQL procedure successfully completed.
SQL> select id, group_id
2 from (select t.*, rank() over (order by group_id) as rnk
3 from t)
4 where rnk <= :n;
ID GROUP_ID
---------- ----------
1 100
2 100
3 100
4 200
5 300
6 300
6 rows selected.
编辑这是包含for update子句 (:n = 2) 的版本：
SQL> select id, group_id
2 from T
3 where rowid in (select RID
4 from (select t.rowid as RID, t.*, rank() over (order by group_id) as rnk
5 from t)
6 where rnk <= :n)
7 for update;
ID GROUP_ID
---------- ----------
1 100
2 100
3 100
于 2011-07-25T16:14:23.597 回答
0
如果您ID的 s 始终从 1 开始是连续的（没有间隙）。并且如果您Group_ID的 s 从未在其他地方作为第二组出现。如果你Group_ID的 s 总是升值...
SELECT
*
FROM
foo
WHERE
Group_ID <= (SELECT Group_ID FROM foo WHERE ID = n)
ORDER BY
ID
您将受益于在ID和Group_ID
于 2011-07-25T16:09:19.630 回答
0
如果GROUP_ID连续和升序总是正确的，那么这很容易通过 SQL 使用分析ROW_NUMBER()函数解决：
SQL> select id
2 , group_id
3 from foo
4 where group_id <= ( select group_id
5 from (
6 select f.group_id
7 , row_number() over (order by f.id asc) rn
8 from foo f
9 )
10 where rn = &n )
11 order by id
12 /
Enter value for n: 2
old 10: where rn = &n )
new 10: where rn = 2 )
ID GROUP_ID
---------- ----------
1 100
2 100
3 100
SQL> r
1 select id
2 , group_id
3 from foo
4 where group_id <= ( select group_id
5 from (
6 select f.group_id
7 , row_number() over (order by f.id asc) rn
8 from foo f
9 )
10 where rn = &n )
11* order by id
Enter value for n: 4
old 10: where rn = &n )
new 10: where rn = 4 )
ID GROUP_ID
---------- ----------
1 100
2 100
3 100
4 200
SQL> r
1 select id
2 , group_id
3 from foo
4 where group_id <= ( select group_id
5 from (
6 select f.group_id
7 , row_number() over (order by f.id asc) rn
8 from foo f
9 )
10 where rn = &n )
11* order by id
Enter value for n: 5
old 10: where rn = &n )
new 10: where rn = 5 )
ID GROUP_ID
---------- ----------
1 100
2 100
3 100
4 200
5 300
6 300
6 rows selected.
SQL>
于 2011-07-25T16:11:01.080 回答
0
如果我们假设 group_id 是连续的并且是递增的，那么@Shannon 的答案就完美了。如果我们不做这个假设，并且我们有如下数据，例如：
SQL> select * from foo order by id;
ID GROUP_ID
-- --------
1 100
2 100
3 100
4 200
6 100
7 400
9 500
10 500
11 500
12 600
然后这是一个更棘手的问题。例如，如果 N = 3、4 或 5，那么我们需要通过 ID = 6 获取行。对于 N = 6，我们需要最多 ID = 7。对于 N = 7，我们需要通过 ID = 11。
我相信无论 group_id 的顺序如何，此查询都有效：
对于 N = 7：
WITH q AS (SELECT ID, group_id
, row_number() OVER (ORDER BY ID) rn
, MAX(id) OVER (PARTITION BY group_id) rn2
FROM foo)
SELECT ID, group_id FROM q
WHERE ID <= (SELECT max(rn2) FROM q WHERE rn <= :N)
ORDER BY ID;
ID GROUP_ID
-- --------
1 100
2 100
3 100
4 200
6 100
7 400
9 500
10 500
11 500
9 rows selected
对于 N = 6：
ID GROUP_ID
-- --------
1 100
2 100
3 100
4 200
6 100
7 400
对于 N = 1：
ID GROUP_ID
-- --------
1 100
2 100
3 100
4 200
6 100
于 2011-07-25T17:56:18.763 回答