根据 Postgres 中的分隔符替换括号并将列拆分为多行

Question

I have a table with column with separated by ';'. The data looks like this:

row_id    col
1         p.[D389R;D393_W394delinsRD]
2         p.[D390R;D393_W394delinsRD]
3         p.D389R
4.        p.[D370R;D393_W394delinsRD]

I would like replace the '[]' brackets whereever they are and fetch the text. Later, I would like to split the string be ';' and concatenate 'p.' to the splitted text (if it is not there) and create a new row.

The expected output is:

row_id    new_col
1         p.D389R
2         p.D393_W394delinsRD
3         p.D390R
4         p.D393_W394delinsRD
5         p.D389R
6         p.D370R
7         p.D393_W394delinsRD

I have tried below query to get the desired output.

SELECT *,

        CASE        
            WHEN regexp_split_to_table(regexp_replace(col, '\[|\]', '', 'g'), E';') NOT LIKE 'p.[%' 
                THEN 'p.' || (regexp_split_to_table(regexp_replace(col, '\[|\]', '', 'g'), E';'))[1]
            ELSE regexp_split_to_table(regexp_replace(col, '\[|\]', '', 'g'), E';')[2]
        END AS new_col    


FROM table;

Any suggestions would be really helpful.

Answer 1

I would first remove the constant values ( p.[ and ] ) from the string and then unnest it.

with clean as (
  select row_id, regexp_replace(col, '^p\.(\[){0,1}|\]$', '', 'g') as col
  from the_table
)
select row_id, 'p.'|| t.c
from clean c
  cross join unnest(string_to_array(c.col, ';')) as t(c)

The CTE ( with ... ) isn't really necessary, but that way the unnest(...) stays readable.

Online example