查找字符串的任何部分在另一個字符串處理SQL

Question

我在我的數據庫，它表示在這樣的工作日字符串：

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) 

，我需要寫一個SQL疑問，檢查重疊。 例如：

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

每個條目都有一個ID，客戶號碼，並且該平日表示。

我的想法是這樣的：

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

要獲得兩個條目是相同的客戶，但是當我來到WHERE語句我打了一個空白。在兩個Days字段中查找子字符串（例如3）是非常容易的，但是當7個條目中的任何一個可以重疊並且我必須排除0（不活躍的一天）時，我會感到困惑。

我需要一些幫助。

Answer 1

這樣做的一種方法。每天通過char匹配字符串char並忽略0（通過替換爲不匹配的值）。下面的查詢將返回那些沒有重疊日期（忽略0）的行的同一個客戶。

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

使用兩個common table expressions，一個用於一個微小的帳簿表，其他與cross apply()和substring()用於使用所述小帳簿表與count() over()窗口聚合函數沿每個位置分裂Days通過計數每Day的出現CustomerNumber。最終select表示各重疊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

與測試設置：

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). 
; 

回報：

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

---

參見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

沒有任何DDL（基本表結構），這是不可能明白的地方數據存在你」我會比較。也就是說，你試圖做的事情會很簡單，使用ngrams8k。

注意此查詢：

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

回報

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

考慮到這一點，這將幫助你：

-- 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