Checking Palindrome - sql-server

Question

We have a string variable where we capture string listed below:
String-like >>
Temp Table Temp | Temp1 Table1 Temp1 | Temp2 Table2 Temp2 | ABD EFG
EFG
Now we need to check, in this particular string how many Palindromes exists.
So, can you help me with this, that how may I fetch the number of Palindrome counts exists.
Note: "|" this pipeline exists after every successful string completion.
Answer should be: 3
The query which I have written, I used Reverse() / Replace() functions but not able to understand how to split the string after every pipeline symbol.
So, please help me in doing that, I am a beginner in SQL Server.

Answer 1

It seems you are confusing your requirement with searching for palindromes, so I have put together a solution to your question as well as a few methods should anyone else come across this question looking for and answer relating to actual palindromes:
Answer to your question as it is here
To do this, you can split your string on the delimiter and then split the result again on the spaces (I have included the function I've used here at the end). With this ordered list of words, you can compare the words in order to the words in reverse order to see if they are the same:
declare #s nvarchar(100) = 'Temp Table Temp | Temp1 Table1 Temp1 | Temp2 Table2 Temp2 | ABD EFG EFG';
with w as
(
select s.item as s
,ss.rn
,row_number() over (partition by s.item order by ss.rn desc) as rrn
,ss.item as w
from dbo.fn_StringSplit4k(#s,'|',null) as s
cross apply dbo.fn_StringSplit4k(ltrim(rtrim(s.item)),' ',null) as ss
)
select w.s
,case when sum(case when w.w = wr.w then 1 else 0 end) = max(w.rn) then 1 else 0 end as p
from w
join w as wr
on w.s = wr.s
and w.rn = wr.rrn
group by w.s
order by w.s
Which outputs:
+----------------------+---+
| s | p |
+----------------------+---+
| ABD EFG EFG | 0 |
| Temp1 Table1 Temp1 | 1 |
| Temp2 Table2 Temp2 | 1 |
| Temp Table Temp | 1 |
+----------------------+---+
Solution for actual palindromes
Firstly to check if a string value is a proper palindrome (ie: spelled the same forwards and backwards) this is a trivial comparison of the original string with it's reverse value, which in the example below correctly outputs 1:
declare #p nvarchar(100) = 'Temp Table elbaT pmeT';
select case when #p = reverse(#p)
then 1
else 0
end as p
To do this across a set of delimited values within the same string, you should firstly feel bad for storing your data in a delimited string within your database and contemplate why you are doing this. Seriously, it's incredibly bad design and you should fix it as soon as possible. Once you have done that you can apply the above technique.
It that is genuinely unavoidable however, you can split your string using one of many set based table valued functions and then apply the above operation on the output:
declare #ps nvarchar(100) = 'Temp Table elbaT pmeT | Temp1 Table1 1elbaT 1pmeT | Temp2 Table2 Temp2 | ABD EFG EFG';
select ltrim(rtrim(s.item)) as s
,case when ltrim(rtrim(s.item)) = reverse(ltrim(rtrim(s.item))) then 1 else 0 end as p
from dbo.fn_StringSplit4k(#ps,'|',null) as s
Which outputs:
+---------------------------+---+
| s | p |
+---------------------------+---+
| Temp Table elbaT pmeT | 1 |
| Temp1 Table1 1elbaT 1pmeT | 1 |
| Temp2 Table2 Temp2 | 0 |
| ABD EFG EFG | 0 |
+---------------------------+---+
String split function
create function [dbo].[fn_StringSplit4k]
(
#str nvarchar(4000) = ' ' -- String to split.
,#delimiter as nvarchar(1) = ',' -- Delimiting value to split on.
,#num as int = null -- Which value to return, null returns all.
)
returns table
as
return
-- Start tally table with 10 rows.
with n(n) as (select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1)
-- Select the same number of rows as characters in #str as incremental row numbers.
-- Cross joins increase exponentially to a max possible 10,000 rows to cover largest #str length.
,t(t) as (select top (select len(isnull(#str,'')) a) row_number() over (order by (select null)) from n n1,n n2,n n3,n n4)
-- Return the position of every value that follows the specified delimiter.
,s(s) as (select 1 union all select t+1 from t where substring(isnull(#str,''),t,1) = #delimiter)
-- Return the start and length of every value, to use in the SUBSTRING function.
-- ISNULL/NULLIF combo handles the last value where there is no delimiter at the end of the string.
,l(s,l) as (select s,isnull(nullif(charindex(#delimiter,isnull(#str,''),s),0)-s,4000) from s)
select rn
,item
from(select row_number() over(order by s) as rn
,substring(#str,s,l) as item
from l
) a
where rn = #num
or #num is null;