Question

我的数据库中有字符串，表示一周中的天数：

1234567 (all days of the week)
1230567 (all days but Thursday, EU standard - day 1 is Monday)
0000067 (no day apart from Saturday and Sunday)

我需要编写一个检查重叠的SQL问题。例如：

1234500 and 0000067 are NOT overlapping.
while 1234500 and 0030000 are overlapping (the 3).
and 1234500 and 0000567 IS overlapping (the 5).

每个条目都有一个ID，客户编号和此工作日表示。

我在想这样的事情：

SELECT
    *
FROM dbo.Customers c
JOIN dbo.Customers c2 ON c.CustomerNumber = c2.CustomerNumber
    AND c.Days <> c2.Days
WHERE 1 = 1
    AND ...?

要获得两个同一客户的条目，但当我来到WHERE语句时，我打了一个空白。在两个Days字段中查找子字符串（例如3）非常简单，但是当7个条目中的任何一个可以重叠并且我必须排除0（非活动日）时，我会感到困惑。

我需要一些帮助。

Answer 1

一种方法。每天匹配char字符串char并忽略0（通过替换非匹配值）。下面的查询将返回同一客户没有重叠天数（忽略0）的行。

SELECT
    *
FROM Customers c
JOIN Customers c2 ON c.CustomerNumber = c2.CustomerNumber
and c.days <> c2.days
where
(
REPLACE (substring (c.[days],1,1),'0','8') <> REPLACE (substring (c2.[days],1,1) ,'0','9')
AND 
REPLACE (substring (c.[days],2,1),'0','8') <> REPLACE (substring (c2.[days],2,1) ,'0','9')
AND
REPLACE (substring (c.[days],3,1),'0','8') <> REPLACE (substring (c2.[days],3,1) ,'0','9')
AND
REPLACE (substring (c.[days],4,1),'0','8') <> REPLACE (substring (c2.[days],4,1) ,'0','9')
AND 
REPLACE (substring (c.[days],5,1),'0','8') <> REPLACE (substring (c2.[days],5,1) ,'0','9')
AND
REPLACE (substring (c.[days],6,1),'0','8') <> REPLACE (substring (c2.[days],6,1) ,'0','9')
AND
REPLACE (substring (c.[days],7,1),'0','8') <> REPLACE (substring (c2.[days],7,1) ,'0','9')
)

Answer 2

使用两个common table expressions，一个用于微小的计数表，另一个用Days分割cross apply()和substring()每个位置使用微小的计数表以及count() over()窗口聚合函数，用Day计算每个CustomerNumber的出现次数。最终select显示每个重叠的Day：

;with n as (
  select i from (values (1),(2),(3),(4),(5),(6),(7)) t(i)
)
, expand as (
  select c.CustomerNumber, c.Days, d.Day
    , cnt = count(*) over (partition by c.CustomerNumber, d.Day)
  from Customers c
    cross apply (
      select Day = substring(c.Days,n.i,1)
      from n
      ) d
  where d.Day > 0
)
select * 
from expand
where cnt > 1

rextester演示：http://rextester.com/SZUANG12356

使用测试设置：

create table Customers (customernumber int, days char(7))
insert into Customers values 
 (1,'1234500')
,(1,'0000067') -- NOT overlapping
,(2,'1234500')
,(2,'0030000') -- IS overlapping (the 3).
,(3,'1234500')
,(3,'0000567') -- IS overlapping (the 5).
;

返回：

+----------------+---------+-----+-----+
| CustomerNumber |  Days   | Day | cnt |
+----------------+---------+-----+-----+
|              2 | 1234500 |   3 |   2 |
|              2 | 0030000 |   3 |   2 |
|              3 | 1234500 |   5 |   2 |
|              3 | 0000567 |   5 |   2 |
+----------------+---------+-----+-----+

<小时/> 参考：

Answer 3

没有任何DDL（基础表结构），就无法理解您将要比较的数据存在的位置。也就是说，使用ngrams8k，您要做的事情很简单。

请注意此查询：

declare @searchstring char(7) = '1234500';
select * from dbo.ngrams8k(@searchstring,1);

<强>返回

position    token 
----------- ------
1           1     
2           2     
3           3     
4           4     
5           5     
6           0     
7           0

考虑到这一点，这将对您有所帮助：

-- sample data
declare @days table (daystring char(7));
insert @days values ('0000067'),('0030000'),('0000567');

declare @searchstring char(7) = '1234500';

-- how to break down and compare the strings
select 
  searchstring    = @searchstring, 
  overlapstring   = OverlapCheck.daystring,
  overlapPosition = OverlapCheck.position,
    overlapValue    = OverlapCheck.token
from dbo.ngrams8k(@searchstring, 1) search
join 
(
  select * 
  from @days d
  cross apply dbo.ngrams8k(d.daystring,1)
  where token <> 0
) OverlapCheck on  search.position = OverlapCheck.position 
                  and search.token = OverlapCheck.token;

返回：

searchstring overlapstring overlapPosition      overlapValue
------------ ------------- -------------------- ---------------
1234500      0030000       3                    3
1234500      0000567       5                    5

在另一个字符串中查找字符串的任何部分transact sql

3 个答案: