I am working on a SQL Server database and would like to find a numbered column.
Suppose I have below values in a column
column_A
I have tried this:
select *
from Table1
where column_A like '^([0-9a-z])\1 '
but the regex doesn't seem to work in SQL Server.
Any help is very appreciated.
CodePudding user response:
You can simply try like following.
select * from Table1 where '0123456789' like '%' column_A '%'
The same thing can be done using recursive CTE like the following query.
declare @table table(column_A varchar(10))
insert into @table select '123456'
insert into @table select '345678'
insert into @table select '214562'
insert into @table select '457215'
insert into @table select '456789'
;with cte as (
select v.column_A os, v.column_A, 1 as m
from @table v
union all
select os,stuff(column_A, 1, 1, ''),
case when cast(left(column_A, 1) as int) 1 = cast(left(stuff(column_A, 1, 1, ''),1) as int) then 1 else 0 end as m
from cte
where column_A > ''
)
select os
from cte
where column_A <>'' and m=1
group by os
having count(m) =len(os)
Output
os
-----
123456
345678
456789
CodePudding user response:
Maybe this would work
SELECT column_A FROM TABLE1
where
column_A =
CONCAT((SELECT LEFT(column_A,1) FROM TABLE1)),(SELECT LEFT(column_A,1)
FROM TABLE1) 1,(SELECT LEFT(column_A,1) FROM TABLE1) 2,(SELECT
LEFT(column_A,1) FROM TABLE1) 3,(SELECT LEFT(column_A,1) FROM TABLE1) 4,
(SELECT LEFT(column_A,1) FROM TABLE1) 5)