PostgreSQL search lists of substrings in string column - sql

Question

I have the following table in a postreSQL database (simplified for clarity):
| serverdate | name | value
|-------------------------------------
0 | 2019-12-01 | A LOC 123 DISP | 1
1 | 2019-12-01 | B LOC 456 DISP | 2
2 | 2019-12-01 | C LOC 777 DISP | 0
3 | 2019-12-01 | D LOC 000 DISP | 10
4 | 2019-12-01 | A LOC 700 DISP | 123
5 | 2019-12-01 | F LOC 777 DISP | 8
name columns is of type string. The substrings LOC and DISP can have other values of different lengths but are not of interest in this question.
The problem: I want to SELECT the rows that only contain a certain substring. There are several substrings, passed as an ARRAY, in the following format:
['A_123', 'F_777'] # this is an example only
I would want to select all the rows that contain the first part of the substring (sepparating it by the underscore '_'), as well as the second. In this example, with the mentioned array, I should obtain rows 0 and 5 (as these are the only ones with exact matches in both parts of the):
| serverdate | name | value
|-------------------------------------
0 | 2019-12-01 | A LOC 123 DISP | 1
5 | 2019-12-01 | F LOC 777 DISP | 8
Row 4 has the first part of the substring correct, but not the other one, so it shouldn't be returned. Same thing with row 2 (only second part matches).
How could this query be done? I'm relatively new to SQL.
This query is part of process in Python, so I can adjust the input parameter (the substring array) if needed, but the behaviour must be the same as the one described.
Thanks!

Answer 1

Have you tried with regexp_replace and a subquery?
SELECT * FROM
(SELECT serverdate, substring(name from 1 for 1)||'_'||
regexp_replace(name, '\D*', '', 'g') AS name, value
FROM t) j
WHERE name IN('A_123', 'F_777');
Or using a CTE
WITH j AS (
SELECT serverdate, substring(name from 1 for 1)||'_'||
regexp_replace(name, '\D*', '', 'g') AS name2,
value,name
FROM t
) SELECT serverdate,name,value FROM j
WHERE name2 IN('A_123', 'F_777');
serverdate | name | value
------------+----------------+-------
2019-12-01 | A LOC 123 DISP | 1
2019-12-01 | F LOC 777 DISP | 8
(2 Zeilen)

Answer 2

Just unnest the array and join the table using a like clause
select
*
from
Table1
join
(
select
'%'||replace(unnest, '_', '%')||'%' pat
from
unnest(array['A_123', 'F_777'])
) pat_table on "name" like "pat"
Just replace unnest(array['A_123', 'F_777']) with unnest(string_to_array(str_variable, ','))

Answer 3

Thanks for your answers! Solution by Larry B got me an error, but it was caused by external factors (I run the queries using an internal tool developed by my company and it threw errors when using the % wildcard. Strange behaviour, I already contacted support team), so I could not test it properly.
Solution by Jim Jones seemed an alternative, but I found that, in some cases, the values in the name field would look like these (didn't notice it when writing the question, as it a rare case):
ABC LOC 123 DISP
So I modified the solution a little bit so as to grab the first part of the name when splitting it by the ' ' character.
(TLDR: 1st substring of name could be of arbitrary length, but is always at the start)
My solution is this one:
SELECT * FROM
(SELECT serverdate, split_part(name, ' ', 1)||'_'||
regexp_replace(name, '\D*', '', 'g') AS name, value
FROM t) j
WHERE name IN('A_123', 'F_777');

Answer 4

split_part(name,'_',1) + '_' + split_part(name,'_',3) as name
this is the break down of the query: A + _ + 123 = A_123