如何在SQL Server中查询超过特定计数的连续值？

我有一张桌子：

id  dummy_data
1   a
2   a
3   a
4   b
5   b
6   c
7   b
8   c
9   c
10  c

我需要查询超过某个阈值（比如2）计数的所有连续伪_数据值，结果如下：

+----+------------+
| id | dummy_data |
+----+------------+
|  1 | a          |
|  2 | c          |
+----+------------+

我写了这个查询：

select 
  t1.dummy_data
from 
  data_table t1
    join data_table t2 on t1.id = t2.id + 1
    join data_table t3 on t1.id = t3.id + 2
where
    t1.dummy_data + ' ' + t2.dummy_data + ' ' + t3.dummy_data = 
    t1.dummy_data + ' ' + t1.dummy_data + ' ' + t1.dummy_data

…这让我：

+------------+
| dummy_data |
+------------+
| a          |
| c          |
+------------+

我的理解是，这是一种叫做缺口和孤岛的问题的变体

然而，我有两个问题： 1.我所做的显然是不可伸缩的，而且 2.我只是无法理解如何重置ID主键，以便在查询结果中重置它们

我怎样才能做到这一点？可在此处找到：

。。。但下面的模式创建只是以防万一：

create table data_table (
    id int primary key,
    dummy_data varchar(10)
);

insert into data_table (id, dummy_data) values
    (1, 'a'),
    (2, 'a'),
    (3, 'a'),
    (4, 'b'),
    (5, 'b'),
    (6, 'c'),
    (7, 'b'),
    (8, 'c'),
    (9, 'c'),
    (10, 'c');

---

谢谢大家!

这将容纳任意数量的连续匹配

with data as (
    select *,
        row_number() over (order by id) as rn,
        row_number() over (partition by dummy_data order by id) as rn2
    from T
)
select row_number() over (order by rn - rn2), min(dummy_data)
from data
group by rn - rn2
having count(*) >= X;

    

---

这将容纳任意数量的连续匹配

with data as (
    select *,
        row_number() over (order by id) as rn,
        row_number() over (partition by dummy_data order by id) as rn2
    from T
)
select row_number() over (order by rn - rn2), min(dummy_data)
from data
group by rn - rn2
having count(*) >= X;

    

---

调整您自己的解决方案，然后使用滞后而不是联接：

WITH SUB AS 
  (SELECT id, 
   dummy_data, 
   lag(dummy_data, 1) OVER (ORDER BY ID) as dd1, 
   lag(dummy_data, 2) OVER (ORDER BY ID) as dd2
  FROM data_table)
SELECT ROW_NUMBER() OVER (ORDER BY id) AS id, dummy_data 
FROM SUB
WHERE dummy_data = dd1 AND dd1 = dd2

---

调整您自己的解决方案，然后使用滞后而不是联接：

WITH SUB AS 
  (SELECT id, 
   dummy_data, 
   lag(dummy_data, 1) OVER (ORDER BY ID) as dd1, 
   lag(dummy_data, 2) OVER (ORDER BY ID) as dd2
  FROM data_table)
SELECT ROW_NUMBER() OVER (ORDER BY id) AS id, dummy_data 
FROM SUB
WHERE dummy_data = dd1 AND dd1 = dd2

您可以使用滞后或超前。对于连续三个：

select min(id), dummy_data
from (select t.*,
             lag(id, 2) over (order by id) as prev_id,
             lag(id, 2) over (partition by dummy_data order by id) as prev_id_dd
      from t
     ) t
where prev_id = prev_id_dd
group by dummy_data;

逻辑很简单。峰背n-1行。然后，峰值返回具有相同伪_数据值的n-1行。如果这些值相同，则所有中间行的值都相同。

您可以使用滞后或超前。对于连续三个：

select min(id), dummy_data
from (select t.*,
             lag(id, 2) over (order by id) as prev_id,
             lag(id, 2) over (partition by dummy_data order by id) as prev_id_dd
      from t
     ) t
where prev_id = prev_id_dd
group by dummy_data;

---

逻辑很简单。峰背n-1行。然后，峰值返回具有相同伪_数据值的n-1行。如果这些值相同，则所有中间行的值都相同。

谢谢。我收到错误列“data.dummy_data”在选择列表中无效，因为它不包含在聚合函数或GROUP BY子句中。这意味着什么？需要汇总。谢谢。我收到错误列“data.dummy_data”在选择列表中无效，因为它不包含在聚合函数或GROUP BY子句中。这意味着什么？它需要聚合。谢谢，我被from语法中的t变量弄糊涂了，这是代表表的吗？你能告诉我为什么需要它吗？@saktas001。这是问题中未指定的表名。注意：这比您目前接受的解决方案更通用。将其扩展到任意n值都很简单。谢谢，我被from语法中的t变量搞糊涂了，这是表的意思吗？你能告诉我为什么需要它吗？@saktas001。这是问题中未指定的表名。注意：这比您目前接受的解决方案更通用。将其扩展到n的任何值都很简单。