如何在 SQL 服务器中 select 顺序复制

Question

我想 select 从 SQL 服务器 table、 复制条目，但前提是 ID 是连续的。

我一直在努力 this answer 满足我的需要，但我无法让它发挥作用。

上面的答案是针对Oracle的，但是我看到SQL服务器也有lead和lag功能。

此外，我认为上面的答案在重复项旁边放置了 *，但我只想 select 重复项。

select 
    id, companyName, 
    case 
       when companyName in (prev, next) 
          then '*' 
    end match, 
    prev, 
    next 
from 
    (select
         id,
         companyName,
         lag(companyName, 1) over (order by id) prev,
         lead(companyName, 1) over (order by id) next
     from
         companies)
order by
    id;

示例：

所以从这个数据集中：

id      companyName
-------------------    
1       dogs ltd
2       cats ltd
3       pigs ltd
4       pigs ltd
5       cats ltd
6       cats ltd
7       dogs ltd
8       pigs ltd

我想select：

id      companyName
-------------------    
3       pigs ltd
4       pigs ltd
5       cats ltd
6       cats ltd

更新

时不时地，我对在 SO 上获得的答案的数量和质量感到吃惊。这是那些时代之一。我没有足够的专业知识来判断一个答案比另一个更好，所以我选择了 SqlZim，因为这是我看到的第一个有效答案。但很高兴看到不同的方法。尤其是一个小时前我还在想 "is this even possible?"。

Answer 1

您可以使用 Row_Number() 并根据子句

的分区获取重复项

;with cte as (
SELECT id, companyName,
    RowN = Row_Number() over (partition by id order by companynae) from #yourTable
    )
    Select * from cte where RowN > 1

您能否提供您的输入和预期输出来验证此查询

Answer 2

这是一个间隙和孤岛式的问题，但我们在最里面的子查询中使用了 id 和 row_number()，而不是使用两个 row_numbers()。接下来是 count() over() 以获取每个 grp 的计数，最后是 return 具有 cnt > 1.

的计数

select id, companyname 
from (
  select 
      id
    , companyName
    , grp
    , cnt = count(*) over (partition by companyname, grp)
  from (
    select *
      , grp = id - row_number() over (partition by companyname order by id)
    from
      companies
    ) islands
  ) d
where cnt  > 1
order by id

rextester 演示：http://rextester.com/ACP73683

returns:

+----+-------------+
| id | companyname |
+----+-------------+
|  3 | pigs ltd    |
|  4 | pigs ltd    |
|  5 | cats ltd    |
|  6 | cats ltd    |
+----+-------------+

Answer 3

在WHERE子句中你只需要限制companyName与prev或next相同的那些

select id, companyName
from (
   select id, companyName,
   lag(companyName, 1) over (order by id) as prev,
   lead(companyName, 1) over (order by id) as next
   from companies
 ) q
 where companyName in (prev, next)
 order by id;

为了确保 id 真的没有间隙，你可以这样做：

select id, companyName
from (
   select id, companyName,
   lag(concat(id+1,companyName), 1) over (order by id) as prev,
   lead(concat(id-1,companyName), 1) over (order by id) as next
   from companies
 ) q
 where concat(id,companyName) in (prev, next)
 order by id;

Answer 4

你非常接近你想要的：

select id, companyName
from (select c.*,
             lag(companyName, 1) over (order by id) prev,
             lead(companyName, 1) over (order by id) next
      from companies c
     ) a
where CompanyName in (prev, next)
order by id;

Answer 5

另一种形式，使用 LEAD() 和 LAG()（SQL 2012 年及以上）

SELECT id, CompanyName
FROM (
    SELECT *,
        LEAD(CompanyName, 1) OVER(ORDER BY id) as nc,
        LAG(CompanyName, 1) OVER(ORDER BY id) AS pc
    FROM #t t
    ) x
WHERE nc = companyName
    OR pc = companyName

这里是测试数据，大家可以自行查看

CREATE TABLE #T (id int not null PRIMARY KEY, companyName varchar(16) not null)

INSERT INTO #t Values 
(1,       'dogs ltd'),
(2,       'cats ltd'),
(3,       'pigs ltd'),
(4,       'pigs ltd'),
(5,       'cats ltd'),
(6,       'cats ltd'),
(7,       'dogs ltd'),
(8,       'pigs ltd')

如何在 SQL 服务器中 select 顺序复制

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