How to select the most matching subString to another String - sql

Question

Lets say the full String is
The following example examines the string, looking for the first substring bounded by comas
and the subString is
substing bounded
is there any way that I could check the full string if contains a 90% matching subString using sql
like the word substing bounded and substring bounded in my example
the subString could be a compound of more words so I can't split the full string into words .

Answer 1

First transform your text in a table of words. You'll find a lot as posts to this topic on SO, e.g. here
You'll have to adjust the list of delimiter characters to extract the words only.
This is a sample query
with t1 as (select 1 rn, 'The following example examines the string, looking for the first substring bounded by comas' col from dual ),
t2 as (select rownum colnum from dual connect by level < 16 /* (max) number of words */),
t3 as (select t1.rn, t2.colnum, rtrim(ltrim(regexp_substr(t1.col,'[^ ,]+', 1, t2.colnum))) col from t1, t2
where regexp_substr(t1.col, '[^ ,]+', 1, t2.colnum) is not null)
select * from t3;
COL
----------
The
following
example
examines
...
In the next step your the Levenshtein Distance to get the closes word.
with t1 as (select 1 rn, 'The following example examines the string, looking for the first substring bounded by comas' col from dual ),
t2 as (select rownum colnum from dual connect by level < 16 /* (max) number of words */),
t3 as (select t1.rn, t2.colnum, rtrim(ltrim(regexp_substr(t1.col,'[^ ,]+', 1, t2.colnum))) col from t1, t2
where regexp_substr(t1.col, '[^ ,]+', 1, t2.colnum) is not null)
select col, str, UTL_MATCH.EDIT_DISTANCE(col, str) distance
from t3
cross join (select 'commas' str from dual)
order by 3;
COL STR DISTANCE
---------- ------ ----------
comas commas 1
for commas 5
examines commas 6
...
Check the definition of the Levenshtein Distance and define a threshold on the distance to get your candidate words.
To match independent of the word boundary simple scan through your input and get all substring in a lenth of your match string adjusted for the diferentce e.g. adding some 10%.
You may limit the candidates by filtering such substrings only that start on the word boundary. The rest ist the same distance calculation.
with txt as (select 'The following example examines the string, looking for the first substring bounded by comas' txt from dual),
str as (select 'substing bounded' str from dual),
t1 as (select substr(txt, rownum, (select length(str) * 1.1 from str)) substr, /* add 10% length for the match */
(select str from str) str
from txt connect by level < (select length(txt) from txt) - (select length(str) from str))
select SUBSTR, STR,
UTL_MATCH.EDIT_DISTANCE(SUBSTR, STR) distance
from t1
order by 3;
SUBSTR STR DISTANCE
-------------------- ---------------- ----------
substring bounded substing bounded 1
ubstring bounded substing bounded 3
substring bounde substing bounded 3
t substring bound substing bounded 5
...

Answer 2

Experiment with the SOUNDEX function.
I haven't tested this but this might help you on your way:
WITH strings AS (
select regexp_substr('The following example examines the string, looking for the first substring bounded by comas','[ ]+', 1, level) ss
from dual
connect by regexp_substr('The following example examines the string, looking for the first substring bounded by comas', '[ ]+', 1, level) is not null
)
SELECT ss
FROM strings
WHERE SOUNDEX(ss) = SOUNDEX( 'commas' ) ;
The REGEXP_SUBSTR with CONNECT BY splits the long string into words (by space) - amend the delimited as required to include punctuation marks etc.
Here we are relying on the built-in SOUNDEX matching our expectations.