SQL Server - extract dates from strings in several formats - sql-server

Question

I've inherited quite a mess of a database table column called DOB, of type nvarchar - here is just a sample of the data in this column:
DOB: 1998-09-04US
Sex: M Race: White Year of Birth: 1950
12/31/00
January 5th, 1998
Date of Birth: 12/19/1938
AGE; 46
DOB: 11-24-1967
May 31, 1942, Split, Croatia
DOB:   12/28/1986
D.O.B.31-OCT-92
D.O.B.: January 8, 1973
31/07/1974 (44 years old)
Date Of Birth: 08/01/1979
78  (DOB: 12/09/1940)
1961 (56 years old)
12/31/1985 (PRIMARY)
DOB: 05/27/67
8-Jun-43
9/9/78
12/31/84 0:00
NA
Birth Year 2018
nacido el 29 de junio de 1959
I am trying to determine whether there is any way to extract the dates from these fields, with so many varying formats, without using something like RegEx patterns for every single possible variation in this column.
The resulting extracted data would look like this:
1998-09-04
1950
12/31/00
January 5th, 1998
12/19/1938
11-24-1967
May 31, 1942
12/28/1986
31-OCT-92
January 8, 1973
31/07/1974
08/01/1979
12/09/1940
1961
12/31/1985
05/27/67
8-Jun-43
9/9/78
12/31/84
NA
2018
29 de junio de 1959
While it may be a complete pipe dream, I was wondering if this could be accomplished with SQL, with some kind of "if it looks like a date, attempt to extract it" method. And if not out-of-the-box, perhaps with a helper extension or plugin?

Answer 1

It is possible, but there are potential pitfalls. This will certainly have to be expanded and maintained.
This is a brute-force pattern match where the longest matching pattern is selected
Example - See Full Working Demo
Select ID
,DOB
,Found
From (
Select *
,Found = substring(DOB,patindex(PatIdx,DOB),PatLen)
,RN = Row_Number() over (Partition By ID Order by PatLen Desc)
From #YourTable A
Left Join (
Select *
,PatIdx = '%'+replace(replace(Pattern, 'A', '[A-Z]'), '0', '[0-9]') +'%'
,PatLen = len(Pattern)
From #FindPattern
) B
on patindex(PatIdx,DOB)>0
) A
Where RN=1
Returns