检查两个字符串中相同的位置字符是否相等

Question

我有一个如下表

如果相同位置的任何字符相等，我想获取表的结果

示例：

全名： ANURADHA KONGARI
名称： AK

没有缩写： AI

在此示例中，AK的首字母和AI的首字母相等。因此它应该返回结果。

我尝试使用substring，但是会得到所有记录。由于长度

我尝试过的例子

select fullname, names, withoutinitials from #localtable where substring(names,1,1)= substring(withoutinitials,1,1)
or
substring(names,2,1)= substring(withoutinitials,2,1)
or
substring(names,3,1)= substring(withoutinitials,3,1)

这是我尝试的一种方式。它可以正常工作，但是如果字符串长度大于4怎么办？

create table #lt2(fullname varchar(100))
insert into #lt2 select getName=case  
    when len(names)=1 and substring(names,1,1) = substring(withoutinitials,1,1) 
    then fullname 
    when len(names)=2 and ((substring(names,1,1) = substring(withoutinitials,1,1))or(substring(names,2,1) = substring(withoutinitials,2,1))) 
    then fullname 
    when len(names)=3 and ((substring(names,1,1) = substring(withoutinitials,1,1))or(substring(names,2,1) = substring(withoutinitials,2,1)) or (substring(names,2,1) = substring(withoutinitials,2,1))) 
    then fullname 
    when len(names)=4 and ((substring(names,1,1) = substring(withoutinitials,1,1))or(substring(names,2,1) = substring(withoutinitials,2,1)) or (substring(names,2,1) = substring(withoutinitials,2,1))
    or (substring(names,3,1) = substring(withoutinitials,3,1))) 
    then fullname else ''
    end
    from #localtable

select * from #lt2 where fullname!=''

但这将导致所有记录，因为某些记录的长度可能不为3，所以它甚至返回长度为1的名称。

如果两个字符串中的至少一个字符在相同位置相等，我想得到结果。

  

像字符串1的第一位置=字符串2或2的第一位置   字符串1的位置=字符串2的第二个位置。

谢谢。

Answer 1

您需要在位置上使用1和2，并且在没有两个字符的情况下还要添加一个空白检查。

declare @table table (fullname varchar(64), names varchar(2), withoutinititals varchar(2))
insert into @table
values
('GANGA RAJAM','GR','AM'),
('ANURADHA KONGARI','AK','AI'),
('PATEL SHIVAJI','R','H'),
('NEW NAME','X','X')

select
    *
    ,substring(names,1,1)
    ,substring(names,2,1)
    ,substring(withoutinititals,1,1)
    ,substring(withoutinititals,2,1)
from @table
where
    (substring(names,1,1) = substring(withoutinititals,1,1))
    or
    (substring(names,2,1) = substring(withoutinititals,2,1) and substring(withoutinititals,2,1) != '')

请注意，如果您删除where子句substring(withoutinititals,2,1) != ''的这一部分，则会得到误报

Answer 2

'0'和names的{​​{1}}字符不相等的右键盘：

'1'

带有示例数据：

withoutinitials

结果是：

select fullname, names, withoutinitials from #localtable where 
substring(names, 1, 1) = substring(withoutinitials, 1, 1)
or
substring(LEFT(CONCAT(names, '0'), 2), 2, 1) = substring(LEFT(CONCAT(withoutinitials, '1'), 2), 2, 1)
or
substring(LEFT(CONCAT(names, '00'), 3), 3, 1) = substring(LEFT(CONCAT(withoutinitials, '11'), 3), 3, 1)
or
substring(LEFT(CONCAT(names, '000'), 4), 4, 1) = substring(LEFT(CONCAT(withoutinitials, '111'), 4), 4, 1)

Answer 3

有点有趣，但是我尝试了递归CTE

--select distinct top 4000 employeeid, surname, ForeName1 forename into #test from isEmployeeMaster where Sequence  = 1 



;WITH S AS (SELECT 1 AS A, M.employeeid as E, M.Surname as sur, SUBSTRING(M.surname,1,1) ATOM FROM #test M 
            UNION ALL
            SELECT A + 1, E, sur, SUBSTRING(S.sur,A+1,1)  FROM S WHERE A < LEN(S.sur) 
            ),
      F AS (SELECT 1 AS A, M2.employeeid as E, M2.forename as fore, SUBSTRING(M2.forename,1,1) ATOM FROM #test M2
            UNION ALL
            SELECT A + 1, E, fore, SUBSTRING(f.fore,A+1,1)  FROM f WHERE A < LEN(f.fore) 
            )

            select t.* from #test t where t.EmployeeId  in 
            (select s.e  from S join F on S.E = F.E and S.Atom = F.Atom and S.A = F.a)



--drop table #test