Regular Expressions - Snowflake [closed] - regex

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enter image description hereI am trying to get text till the fourth "\n \n" from the below text. Can you please help me to write the snowflake expression for this issue.
Hello Jeffrey,\n \nWe have not heard from you yet. I hope all is well with you.\n \nChecking in to gather your Goosehead approved office location address, so we can add you to our database here at ERGOS. Once added here, we can schedule your laptop setup.\n \nGoosehead requires all agents to be onboarded by ERGOS so that we can provide IT support as well as get your laptop in our database. \n \nDo you have a laptop ready for setup?

Answer 1

so every thing up to the first \n \n can be fetched with regexp_substr via:
select
regexp_substr(column1, '.*\n \n') as match
from values
('Hello Jeffrey,\n \nWe have not heard from you yet. I hope all is well with you.\n \nChecking in to gather your Goosehead approved office location address, so we can add you to our database here at ERGOS. Once added here, we can schedule your laptop setup.\n \nGoosehead requires all agents to be onboarded by ERGOS so that we can provide IT support as well as get your laptop in our database. \n \nDo you have a laptop ready for setup?');
MATCH
Hello Jeffrey,
now, if we add a group around that ( ) and ask for 4 matches {4}, and swap to a smaller sample text, to make things less ugly for the output
select
regexp_substr(column1, '(.*\n \n){4}') as match
from values
('1111\n \n222222222222222\n \n3333333333333333\n \n44444444444444444\n \n55555555555555555555555');
gives:
MATCH
1111 222222222222222 3333333333333333 44444444444444444
if you are expecting the \n in the output:
then
select
column1,
regexp_substr(column1, '[^\\\\]+\\\\n \\\\n') as match
from values
('1111\\n \\n22222\\n \\n33333333\\n \\n4444444\\n \\n55555\\n \\66666\\n \\n7777');
shows how they need to be encoded in the SQL to output, and thus how to encode the match.
these matches greedy and gives:
COLUMN1
MATCH
1111\n \n22222\n \n33333333\n \n4444444\n \n55555\n \66666\n \n7777
1111\n \n
thus putting the grouping back in:
select
column1,
regexp_substr(column1, '([^\\\\]+\\\\n \\\\n){4}') as match
from values
('1111\\n \\n22222\\n \\n33333333\\n \\n4444444\\n \\n55555\\n \\66666\\n \\n7777');
COLUMN1
MATCH
1111\n \n22222\n \n33333333\n \n4444444\n \n55555\n \66666\n \n7777
1111\n \n22222\n \n33333333\n \n4444444\n \n
Picture to example for escaped new lines: