Check palindrome without using string functions with condition - sql

Question

I have a table EmployeeTable.
If I want only that records where employeename have character of 1 to 5
will be palindrome and there also condition like total character is more then 10 then 4 to 8 if character less then 7 then 2 to 5 and if character less then 5 then all char will be checked and there that are palindrome then only display.
Examples :- neen will be display
neetan not selected
kiratitamara will be selected
I try this something on string function like FOR first case like name less then 5 character long
SELECT SUBSTRING(EmployeeName,1,5),* from EmaployeeTable where
REVERSE (SUBSTRING(EmployeeName,1,5))=SUBSTRING(EmployeeName,1,5)
I want to do that without string functions,
Can anyone help me on this?

Answer 1

You need at least SUBSTRING(), I have a solution like this:
(In SQL Server)
DECLARE #txt varchar(max) = 'abcba'
;WITH CTE (cNo, cChar) AS (
SELECT 1, SUBSTRING(#txt, 1, 1)
UNION ALL
SELECT cNo + 1, SUBSTRING(#txt, cNo + 1, 1)
FROM CTE
WHERE SUBSTRING(#txt, cNo + 1, 1) <> ''
)
SELECT COUNT(*)
FROM (
SELECT *, ROW_NUMBER() OVER (ORDER BY cNo DESC) as cRevNo
FROM CTE t1 CROSS JOIN
(SELECT Max(cNo) AS strLength FROM CTE) t2) dt
WHERE
dt.cNo <= dt.strLength / 2
AND
dt.cChar <> (SELECT dti.cChar FROM CTE dti WHERE dti.cNo = cRevNo)
The result will shows the count of differences and 0 means no differences.
Note :
Current solution is Non-Case-Sensitive for change it to a Case-Sensitive you need to check the strings in a case-sensitive collation like Latin1_General_BIN
You can use this solution as a SVF or something like that.

Answer 2

I dont realy understand why you dont want to use string functions in your query, but here is one solution. Compute everything beforehand:
Add Column:
ALTER TABLE EmployeeTable
ADD SubString AS
SUBSTRING(EmployeeName,
(
CASE WHEN LEN(EmployeeName)>10
THEN 4
WHEN LEN(EmployeeName)>7
THEN 2
ELSE 1 END
)
,
(
CASE WHEN LEN(EmployeeName)>10
THEN 8
WHEN LEN(EmployeeName)>7
THEN 5
ELSE 5 END
)
PERSISTED
GO
ALTER TABLE EmployeeTable
ADD Palindrome AS
REVERSE(SUBSTRING(EmployeeName,
(
CASE WHEN LEN(EmployeeName)>10
THEN 4
WHEN LEN(EmployeeName)>7
THEN 2
ELSE 1 END
)
,
(
CASE WHEN LEN(EmployeeName)>10
THEN 8
WHEN LEN(EmployeeName)>7
THEN 5
ELSE 5 END
)) PERSISTED
GO
Then your query will looks like:
SELECT * from EmaployeeTable
where Palindrome = SubString
BUT!
This is not a good idea. Please tell us, why you dont want to use string functios.

Answer 3

You could do it building a list of palindrome words using a recursive query that generates palindrome words till a length o n characters and then selects employees with the name matching a palindrome word. This may be a really inefficient way, but it does the trick
This is a sample query for Oracle, PostgreSQL should support this feature as well with little differences on syntax. I don't know about other RDBMS.
with EmployeeTable AS (
SELECT 'ADA' AS employeename
FROM DUAL
UNION ALL
SELECT 'IDA' AS employeename
FROM DUAL
UNION ALL
SELECT 'JACK' AS employeename
FROM DUAL
), letters as (
select chr(ascii('A') + rownum - 1) as letter
from dual
connect by ascii('A') + rownum - 1 <= ascii('Z')
), palindromes(word, len ) as (
SELECT WORD, LEN
FROM (
select CAST(NULL AS VARCHAR2(100)) as word, 0 as len
from DUAL
union all
select letter as word, 1 as len
from letters
)
union all
select l.letter||p.word||l.letter AS WORD, len + 1 AS LEN
from palindromes p
cross join letters l
where len <= 4
)
SEARCH BREADTH FIRST BY word SET order1
CYCLE word SET is_cycle TO 'Y' DEFAULT 'N'
select *
from EmployeeTable
WHERE employeename IN (
SELECT WORD
FROM palindromes
)

Answer 4

DECLARE #cPalindrome VARCHAR(100) = 'SUBI NO ONIBUS'
SET #cPalindrome = REPLACE(#cPalindrome, ' ', '')
;WITH tPalindromo (iNo) AS (
SELECT 1
WHERE SUBSTRING(#cPalindrome, 1, 1) = SUBSTRING(#cPalindrome, LEN(#cPalindrome), 1)
UNION ALL
SELECT iNo + 1
FROM tPalindromo
WHERE SUBSTRING(#cPalindrome, iNo + 1, 1) = SUBSTRING(#cPalindrome, LEN(#cPalindrome) - iNo, 1)
AND LEN(#cPalindrome) > iNo
)
SELECT IIF(MAX(iNo) = LEN(#cPalindrome), 'PALINDROME', 'NOT PALINDROME')
FROM tPalindromo