Oracle SQL：至少選擇前n行，直到列值與最后一行不同為止

Question

給出具有以下結構的表foo（Oracle 11g）：

ID | GROUP_ID
 1 | 100
 2 | 100
 3 | 100
 4 | 200
 5 | 300
 6 | 300
 7 | 400

我想選擇前n行（按ID排序）或更多，這樣我總能得到一個完整的組。

例：

n = 2：我想至少獲得前兩行，但是由於ID 3也屬於組100，所以我也希望得到它。

n = 4：給我前四行，我很高興;-)

n = 5：請求1-6行。

非常感謝您的幫助！

Answer 1

使用rank()解決方案：

select id, group_id
from (select t.*, rank() over (order by group_id) as rnk
    from t)
where rnk <= :n;

建築測試數據：

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>

正在運行...

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.

編輯這是包含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始終從1開始是連續的（無間隔），並且您的Group_ID從未在其他地方作為第二個組出現。 如果您的Group_ID的價值始終在上升...

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

您將從ID和Group_ID上具有單獨的索引中Group_ID

Answer 3

如果始終確實GROUP_ID是連續且遞增的，那么可以使用SQL的解析ROW_NUMBER()函數輕松解決此問題：

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

如果我們假設group_id是連續的並且是遞增的，那么@Shannon的答案將非常有效。 如果我們不做這個假設，並且我們有看起來像這樣的數據，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

這是一個棘手的問題。 例如，如果N = 3、4或5，則需要通過ID = 6獲取行。對於N = 6，我們需要高達ID =7。對於N = 7，我們需要通過ID = 11。

我相信無論group_id的順序如何，該查詢都有效：

對於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

對於N = 6：

ID GROUP_ID
-- --------
 1      100
 2      100
 3      100
 4      200
 6      100
 7      400

對於N = 1：

ID GROUP_ID
-- --------
 1      100
 2      100
 3      100
 4      200
 6      100