Oracle SQL：至少选择前n行，直到列值与最后一行不同为止

Question

给出具有以下结构的表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，这样我总能得到一个完整的组。

例：

n = 2：我想至少获得前两行，但是由于ID 3也属于组100，所以我也希望得到它。

n = 4：给我前四行，我很高兴;-)

n = 5：请求1-6行。

非常感谢您的帮助！

Answer 1

使用rank()解决方案：

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>

正在运行...

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

Answer 2

如果您的ID始终从1开始是连续的（无间隔），并且您的Group_ID从未在其他地方作为第二个组出现。 如果您的Group_ID的价值始终在上升...

SELECT
  *
FROM
  foo
WHERE
  Group_ID <= (SELECT Group_ID FROM foo WHERE ID = n)
ORDER BY
  ID

您将从ID和Group_ID上具有单独的索引中Group_ID

Answer 3

如果始终确实GROUP_ID是连续且递增的，那么可以使用SQL的解析ROW_NUMBER()函数轻松解决此问题：

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>

Answer 4

如果我们假设group_id是连续的并且是递增的，那么@Shannon的答案将非常有效。 如果我们不做这个假设，并且我们有看起来像这样的数据，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

这是一个棘手的问题。 例如，如果N = 3、4或5，则需要通过ID = 6获取行。对于N = 6，我们需要高达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，该查询都有效：

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