搜索模式并删除搜索字符串字符（如果找不到），然后在oracle sql中再次搜索

Question

I have to write a query which is having a where clause like below:

select x from table where col1 like '%6789%';

Now the requirement I have is if above query doesn't return anything then search with 678 or 789 or 67 or 89.

How can I achieve this in best optimised way?

Answer 1

Search once for any value containing 6, 7, 8 or 9, order by how well the value matches '6789' and take the top result.

(Edit: I've added an instr expression to prioritise occurrences of the exact string '6789' over the JW score, as otherwise '678 blah' would rank higher than 'blah 6789' .)

create table demo (col1) as
( select column_value
  from   table(sys.dbms_debug_vc2coll
         ( '6789','678X','X789','67XX','XX89','XXXX','6XXX','X7XX','11111167891111'
         , '167891','Germany 6, England 7 what a great match that was in 1989')) );

select col1
     , utl_match.jaro_winkler_similarity(col1,'6789') as match_rating
     , nullif(instr(col1,'6789'),0) as match_position
from   demo
where  regexp_like(col1,'[6789]')
order by match_position nulls last, match_rating desc;

COL1                                                     MATCH_RATING MATCH_POSITION
-------------------------------------------------------- ------------ --------------
6789                                                              100              1
167891                                                             88              2
11111167891111                                                     59              7
678X                                                               88 
X789                                                               83 
67XX                                                               73 
XX89                                                               66 
6XXX                                                               55 
X7XX                                                               50 
Germany 6, England 7 what a great match that was in 1989           34 

10 rows selected

Filter to get the first row (Oracle 12.1 onwards):

select col1
from   demo
where  regexp_like(col1,'[6789]')
order by
       nullif(instr(col1,'6789'),0) nulls last
     , utl_match.jaro_winkler_similarity(col1,'6789') desc
fetch first row only;

COL1
-----
6789

For older versions:

select col1
from   ( select col1
         from   demo
         where  regexp_like(col1,'[6789]')
         order by
                nullif(instr(col1, '6789'),0) nulls last
              , utl_match.jaro_winkler_similarity(col1,'6789') desc )
where rownum = 1;

http://sqlfiddle.com/#!4/a9d884/3

Note '6789' can be replaced with a variable in PL/SQL:

select col1 into :result
from   demo
where  regexp_like(col1,'['||p_pattern||']')
order by
       nullif(instr(col1, p_pattern),0) nulls last
     , utl_match.jaro_winkler_similarity(col1,p_pattern) desc
fetch first row only;

Answer 2

You can achieve executing different select with order using UNION ALL combined with NOT EXISTS , similar to this answer

select x from table where col1 like '%6789%'
UNION ALL
select x from table where col1 like '%678%' or col1 like '%789%' 
or col1 like '%67%' or col1 like '%89%'
AND NOT EXISTS (
select x from table where col1 like '%6789%'
)