这是我的桌子：
ID Name Department
1 Michael Marketing
2 Alex Marketing
3 Tom Marketing
4 John Sales
5 Brad Marketing
6 Leo Marketing
7 Kevin Production
我正在尝试查找ID 的范围，其中Department = 'Marketing'：
Range From To
Range1 1 3
Range2 5 6
任何帮助将不胜感激。
【问题讨论】：
这很容易通过称为Tabibitosan 的技术实现。
此技术的作用是将每个组的行的位置与整个行集进行比较，以确定同一组中的行是否彼此相邻。
例如，对于您的示例数据，如下所示：
WITH your_table AS (SELECT 1 ID, 'Michael' NAME, 'Marketing' department FROM dual UNION ALL
SELECT 2 ID, 'Alex' NAME, 'Marketing' department FROM dual UNION ALL
SELECT 3 ID, 'Tom' NAME, 'Marketing' department FROM dual UNION ALL
SELECT 4 ID, 'John' NAME, 'Sales' department FROM dual UNION ALL
SELECT 5 ID, 'Brad' NAME, 'Marketing' department FROM dual UNION ALL
SELECT 6 ID, 'Leo' NAME, 'Marketing' department FROM dual UNION ALL
SELECT 7 ID, 'Kevin' NAME, 'Production' department FROM dual)
-- end of mimicking your table with data in it. See the SQL below:
SELECT ID,
NAME,
department,
row_number() OVER (ORDER BY ID) overall_rn,
row_number() OVER (PARTITION BY department ORDER BY ID) department_rn,
row_number() OVER (ORDER BY ID) - row_number() OVER (PARTITION BY department ORDER BY ID) grp
FROM your_table;
ID NAME DEPARTMENT OVERALL_RN DEPARTMENT_RN GRP
---------- ------- ---------- ---------- ------------- ----------
1 Michael Marketing 1 1 0
2 Alex Marketing 2 2 0
3 Tom Marketing 3 3 0
4 John Sales 4 1 3
5 Brad Marketing 5 4 1
6 Leo Marketing 6 5 1
7 Kevin Production 7 1 6
在这里，我已经为整个数据集的所有行按 id 升序分配了一个行号（overall_rn 列），并为每个部门中的行分配了一个行号（@987654324 @column)，再次按 id 升序排列。
现在我已经完成了，我们可以从另一个中减去一个（grp 列）。
请注意 grp 列中的数字对于彼此相邻的部门行是如何保持不变的，但每次出现间隙时它都会改变。
例如对于营销部门，第 1-3 行彼此相邻并且 grp = 0，但第 4 行营销实际上位于整体结果集的第 5 行，因此它现在具有不同的 grp 编号。由于第 5 行营销在整个集合的第 6 行，它与第 4 行营销具有相同的 grp 编号，因此我们知道它们彼此相邻。
一旦我们有了这些 grp 信息，就可以简单地对部门和我们的新 grp 列进行聚合查询分组，使用 min 和 max 来查找开始和结束 id：
WITH your_table AS (SELECT 1 ID, 'Michael' NAME, 'Marketing' department FROM dual UNION ALL
SELECT 2 ID, 'Alex' NAME, 'Marketing' department FROM dual UNION ALL
SELECT 3 ID, 'Tom' NAME, 'Marketing' department FROM dual UNION ALL
SELECT 4 ID, 'John' NAME, 'Sales' department FROM dual UNION ALL
SELECT 5 ID, 'Brad' NAME, 'Marketing' department FROM dual UNION ALL
SELECT 6 ID, 'Leo' NAME, 'Marketing' department FROM dual UNION ALL
SELECT 7 ID, 'Kevin' NAME, 'Production' department FROM dual)
-- end of mimicking your table with data in it. See the SQL below:
SELECT department,
MIN(ID) start_id,
MAX(ID) end_id
FROM (SELECT ID,
NAME,
department,
row_number() OVER (ORDER BY ID) - row_number() OVER (PARTITION BY department ORDER BY ID) grp
FROM your_table)
GROUP BY department, grp;
DEPARTMENT START_ID END_ID
---------- ---------- ----------
Marketing 1 3
Marketing 5 6
Sales 4 4
Production 7 7
注意，我假设 id 列中的间隙并不重要（即如果没有 id = 6 的行（所以 Leo 和 Kevin 的 id 分别为 7 和 8），那么 Leo 和 Brad 仍然会出现在同一组中，起始 id = 5，结束 id = 7。
如果 id 列中的间隙算作表示一个新组，那么您可以只使用 id 来标记整个行集（即无需计算overall_rn；只需使用 id 列代替）。
这意味着您的查询将变为：
WITH your_table AS (SELECT 1 ID, 'Michael' NAME, 'Marketing' department FROM dual UNION ALL
SELECT 2 ID, 'Alex' NAME, 'Marketing' department FROM dual UNION ALL
SELECT 3 ID, 'Tom' NAME, 'Marketing' department FROM dual UNION ALL
SELECT 4 ID, 'John' NAME, 'Sales' department FROM dual UNION ALL
SELECT 5 ID, 'Brad' NAME, 'Marketing' department FROM dual UNION ALL
SELECT 7 ID, 'Leo' NAME, 'Marketing' department FROM dual UNION ALL
SELECT 8 ID, 'Kevin' NAME, 'Production' department FROM dual)
-- end of mimicking your table with data in it. See the SQL below:
SELECT department,
MIN(ID) start_id,
MAX(ID) end_id
FROM (SELECT ID,
NAME,
department,
ID - row_number() OVER (PARTITION BY department ORDER BY ID) grp
FROM your_table)
GROUP BY department, grp;
DEPARTMENT START_ID END_ID
---------- ---------- ----------
Marketing 1 3
Sales 4 4
Marketing 5 5
Marketing 7 7
Production 8 8
【讨论】：
我目前没有这个环境，但是你可以试试这样的
select * from tab1 where id in
(select min(id) from tab1 where Department = 'Marketing'
union
select max(id) from tab1 where Department = 'Marketing')
【讨论】：