查找字符串的任何部分在另一个字符串处理SQL

Question

我在我的数据库，它表示在这样的工作日字符串：

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) 

，我需要写一个SQL疑问，检查重叠。 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，客户号码，并且该平日表示。

我的想法是这样的：

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

要获得两个条目是相同的客户，但是当我来到WHERE语句我打了一个空白。在两个Days字段中查找子字符串（例如3）是非常容易的，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（不活跃的一天）时，我会感到困惑。

我需要一些帮助。

Answer 1

这样做的一种方法。每天通过char匹配字符串char并忽略0（通过替换为不匹配的值）。下面的查询将返回那些没有重叠日期（忽略0）的行的同一个客户。

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，一个用于一个微小的帐簿表，其他与cross apply()和substring()用于使用所述小帐簿表与count() over()窗口聚合函数沿每个位置分裂Days通过计数每Day的出现CustomerNumber。最终select表示各重叠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

与测试设置：

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). 
; 

回报：

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

---

参见ENCE：

• common table expression
• table value constructor (values (...),(...))
• cross apply()
• over()
• The "Numbers" or "Tally" Table: What it is and how it replaces a loop - Jeff Moden

Answer 3

没有任何DDL（基本表结构），这是不可能明白的地方数据存在你”我会比较。也就是说，你试图做的事情会很简单，使用ngrams8k。

注意此查询：

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

回报

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

考虑到这一点，这将帮助你：

-- 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