Comparing string values within a table - sql

Question

Is there any way to compare two columns with strings to each other, and getting the matches?
I have two columns containing Names, once with the Full Name the other with (mostly) just the Surname.
I just tried it with soundex, but it will just return if the values are almost similar in both columns.
SELECT * FROM TABLE
WHERE soundex(FullName) = soundex(Surname)
1 John Doe Doe
2 Peter Parker Parker
3 Brian Griffin Brian Griffin
with soundex it will only match the 3rd line.

Answer 1

A simple option is to use instr, which shows whether surname exists in fullname:
SQL> with test (id, fullname, surname) as
2 (select 1, 'John Doe' , 'Doe' from dual union all
3 select 2, 'Peter Parker' , 'Parker' from dual union all
4 select 3, 'Brian Griffin', 'Brian Griffin' from dual
5 )
6 select *
7 from test
8 where instr(fullname, surname) > 0;
ID FULLNAME SURNAME
---------- ------------- -------------
1 John Doe Doe
2 Peter Parker Parker
3 Brian Griffin Brian Griffin
Another option is to use one of UTL_MATCH functions, e.g. Jaro-Winkler similarity which shows how well those strings match:
SQL> with test (id, fullname, surname) as
2 (select 1, 'John Doe' , 'Doe' from dual union all
3 select 2, 'Peter Parker' , 'Parker' from dual union all
4 select 3, 'Brian Griffin', 'Brian Griffin' from dual
5 )
6 select id, fullname, surname,
7 utl_match.jaro_winkler_similarity(fullname, surname) jws
8 from test
9 order by id;
ID FULLNAME SURNAME JWS
---------- ------------- ------------- ----------
1 John Doe Doe 48
2 Peter Parker Parker 62
3 Brian Griffin Brian Griffin 100
SQL>
Feel free to explore other function that package offers.
Also, note that I didn't pay attention to possible letter case differences (e.g. "DOE" vs. "Doe"). If you need that as well, compare e.g. upper(surname) to upper(fullname).

Answer 2

Please use instring function,
SELECT * FROM TABLE
WHERE instr(Surname, FullName) > 0;
SELECT * FROM TABLE
WHERE instr(upper(Surname), upper(FullName)) > 0;
SELECT * FROM TABLE
WHERE upper(FullName) > upper(Surname);

Answer 3

As far as I know there is nothing out of the box when matching becomes complicated. For the cases shown, however, the following expression would suffice:
where fullname like '%' || surname
Update
The main problem may be false positives:
The last name 'Park' appears in 'Peter Parker'. Above query solves this by looking at the full name's end.
Another problem may be upper / lower case as mentioned in the other answers (not shown in your sample data).
You want the last name 'PARKER' match 'Peter Parker'.
But when looking at the strings case insensitively, another problem arises:
The last name 'Strong' will suddenly match 'Louis Armstrong'.
A solution for this is to add a blank to make the difference:
where ' ' || upper(fullname) like '% ' || upper(surname)
' LOUIS ARMSTRONG' like '% STRONG' -> false
' LOUIS ARMSTRONG' like '% ARMSTRONG' -> true
' LOUIS ARMSTRONG' like '% LOUIS ARMSTRONG' -> true
Demo: https://dbfiddle.uk/?rdbms=oracle_18&fiddle=0ac5c80061b4aeac1153a8c5976e6e54