Extracting specific part of column values in Oracle SQL

Question

I want to extract a specific part of column values. The target column and its values look like

TEMP_COL
---------------
DESCOL 10MG
TEGRAL 200MG 50S
COLOSPAS 135MG 30S

The resultant column should look like

RESULT_COL
---------------
10MG
200MG
135MG

Answer 1

This can be done using a regular expression:

SELECT regexp_substr(TEMP_COL, '[0-9]+MG')
FROM the_table;

Note that this is case sensitive and it always returns the first match.

Answer 2

I would probably approach this using REGEXP_SUBSTR() rather than base functions, because the structure of the prescription text varies from record to record.

SELECT TRIM(REGEXP_SUBSTR(TEMP_COL, '(\s)(\S*)', 1, 1))
FROM yourTable

The pattern (\\s)(\\S*) will match a single space followed by any number of non-space characters. This should match the second term in all cases. We use TRIM() to remove a leading space which is matched and returned.

Answer 3

how do you know what is the part you want to extract? how do you know where it begins and where it ends? using the white-spaces?

if so, you can use substr for cutting the data and instr for finding the white-spaces.

example:

select substr(tempcol, -- string
              instr(tempcol, ' ', 1), -- location of first white-space 
              instr(tempcol, ' ', 1, 2) - instr(tempcol, ' ', 1))  -- length until next space
from dual

another solution is using regexp_substr (but it might be harder on performance if you have a lot of rows):

SELECT REGEXP_SUBSTR (tempcol, '(\S*)(\s*)', 1, 2)
FROM dual;

edit: fixed the regular expression to include expressions that don't have space after the parsed text. sorry about that.. ;)