Take. a particular string from a TEXT in postgres - postgresql

Question

I have one question:
Below is an error message which is stored in a postgres table column, From this string i would like to extract only a part of the string, Is that possible to do in Postgres?
I would like to see odoo.exceptions.ValidationError: ('No MRP template was found for MO/10881!', None)' only this part.
In general all text starting with odoo.exceptions.ValidationError: until the end
How can i do it ? Any idea or suggestions?
'Traceback (most recent call last):
File "/opt/src/addons_OCA/queue/queue_job/controllers/main.py", line 101, in runjob
self._try_perform_job(env, job)
File "/opt/src/addons_OCA/queue/queue_job/controllers/main.py", line 61, in _try_perform_job
job.perform()
File "/opt/src/addons_OCA/queue/queue_job/job.py", line 466, in perform
self.result = self.func(*tuple(self.args), **self.kwargs)
File "/opt/src/addons/costs/models/mrp_production.py", line 163, in trigger_calculate_all_costs
self.calculate_all_costs()
File "/opt/src/addons/sucosts/models/costline_mixin.py", line 284, in calculate_all_costs
rec.generate_service_products()
File "/opt/src/addons/mrp_product_templates/models/mrp_production.py", line 660, in generate_service_products
MO=self.name)))
odoo.exceptions.ValidationError: ('No MRP template was found for MO/10881!', None)'

Answer 1

You can use regexp_replace function to search for particular text then select the text following.
The regexp_replace function provides substitution of new text for
substrings that match POSIX regular expression patterns. It has the
syntax regexp_replace(source, pattern, replacement [, flags ]). The
source string is returned unchanged if there is no match to the
pattern. ...
In this case it seems you want the text after ValidationError:. Something like: (see demo)
select regexp_replace (message, '.*ValidationError:(.*)','\1')
from test;