LIKE on multiple values, all must match - sql

Question

I'm querying a legacy Oracle db to match a name using LIKE. It would be good if a multi-word pattern (could be 1 word, 3 words, 8 words, whatever) could be matched against names independently of word order. Eg, if the user searched for "rob jon", it would match
ROBERT JONES
JONATHON ADAM PROBE
but not match
ROB SMITH
PETER JONES
If the requirement was "at least one word should match" I could use REGEXP_LIKE(name, 'rob|jon", 'i').
But I can't work out how to make it match only when ALL words are found somewhere in the name. I could do this using Oracle Text, but I can't use that in this db. I also can't easily store the words in a table to join to either.
The query is currently static and uses bind variables, and so I would prefer not to add an extra LIKE predicate for each word in the search pattern.
Any ideas?

Answer 1

According to your question and the data presented in the post, I Used this table
create table name_pattern
(
name_1 varchar(200)
);
insert statements :
insert into name_pattern
select ( 'ROBERT JONES') from dual union all
select ('JONATHON ADAM PROBE') from dual union all
select ('ROB SMITH') from dual union all
select ('PETER JONES') from dual;
commit;
try this if it works for you if you are searching "rob" and "jon" separately.
select name from table_name where lower(name) like '%rob%' and lower(name) like '%jon%' ;
try this if you want to search "rob jon" combined and to get your expected result
select * from name_pattern where REGEXP_COUNT(name_1, 'rob|jon',1, 'i') =2;

Answer 2

You want your query to accept one bind variable containing a blank-separated list of name parts, e.g. 'rob jon'.
You can write a recursive query to get the single name parts from that string. Then use NOT EXISTS to only keep names for which not exists any mismatch.
with parameters as
(
select :namepartlist as namepartlist from dual
)
, nameparts (namepart, namepartlist, occurrence) as
(
select regexp_substr(namepartlist, '[^ ]+', 1, 1), namepartlist, 1
from parameters
union all
select regexp_substr(namepartlist, '[^ ]+', 1, occurrence + 1), namepartlist, occurrence + 1
from nameparts
where regexp_substr(namepartlist, '[^ ]+', 1, occurrence + 1) is not null
)
select *
from mytable
where not exists
(
select null
from nameparts
where lower(mytable.name) not like '%' || lower(nameparts.namepart) || '%'
)
order by name;
You can replace
where lower(mytable.name) not like '%' || lower(nameparts.namepart) || '%'
by
where not regexp_like(mytable.name, nameparts.namepart, 'i')
if you like that better.
Demo: https://dbfiddle.uk/?rdbms=oracle_18&fiddle=5e48caaa20e9397afe65516504c62acd

Answer 3

if you prefer not to add an extra LIKE predicate for each word in the search pattern (the best solution IMHO) you can create a PL/SQL function to do the matching
FUNCTION PATTERN_MATCHES(pattern in varchar2, name in varchar2) returns number
is
Begin *pseudocode*
*convert pattern to lowercase*
*split pattern by spaces*
*for each part, Loop*
if NOT lower(name) like %part% then
return 0; --early exit
end;
End Loop;
return 1; --all parts have a match
end;
select * from employee where PATTERN_MATCHES(:pattern, name)=1
Then you can have a static parameterized query.
On the bright side, you'll be able to extend the behavior (add advanced search for example) without touching the front-end app.
PS: The function code is intentionally left for you to write. Ask if you need help with that.
PS2: returning "number" and not a boolean will help you with legacy database drivers. Feel free to use integer or a boolean if that works for you.

Answer 4

Shamelessy building on #Jfrd's solution ... is this any good ? (Performance will be shocking on large tables).
Basically, replace all spaces with '%' (not catered here for successive spaces).
var search_text varchar2(30)
exec :search_text := 'Rob Jon';
select name
from table_name
where lower(name) like REPLACE('%' || :search_text || '%', ' ', '%' ) ;