How to compare 2 version strings (of the format x.x.x.x where x : {0, 99})

Question

I am using Oracle Database 11g Enterprise Edition Release 11.2.0.2.0 - 64bit Production,
PL/SQL Release 11.2.0.2.0 - Production
I have a scenario like:
Table1:
Name Null Type
NAME NOT NULL VARCHAR2(64)
VERSION NOT NULL VARCHAR2(64)
Data:
Table1
Name Version
A 1
B 12.1.0.2
B 8.2.1.2
B 12.0.0
C 11.1.2
C 11.01.05
I want the output as:
Name Version
A 1
B 12.1.0.2
C 11.01.05
Basically ,I want to get rows for each name which having highest version. For this i am using following query:
SELECT t1.NAME,
t1.VERSION
FROM TABLE1 t1
LEFT OUTER JOIN TABLE1 t2
on (t1.NAME = t2.NAME and t1.VERSION < t2.VERSION)
where t2.NAME is null
Now 't1.VERSION < t2.VERSION' only works in normal version cases but in cases such as:
B 12.1.0.2
B 8.2.1.2
It fails, So i want a PL/SQL script to normalize the version strings and compare them for higher value.

Answer 1

Hi,
Here's one way:
ORDER BY  REGEXP_REPLACE ( REGEXP_REPLACE  ( version
                                       , '(\d+)'
                            , '0000\1'
                , '0+(\d{5})'
                , '\1'
                )This assumes that you never have more than 5 digits in a row. You can easily change that to any fixed number.
This will not cause an error if you have letters or other symbols in the string.
This makes no assumptions about how many numbers are in the string, or what characters delimit the numbers.
Here's how it works:
The inner REGEXP_REPLACE adds 4 '0's immediately before any substring of digits. This guarantees that all numbers are at least 5 digits long.
The outer REGEXP_REPLACE removes '0's from the beginning of numbers that are more than 5 digits long, which guaranteees that the numbers will be exactly 5 digits long. '00008' will come before '00012'