AWS 雅典 regexp_extract

Question

I am trying to extract a part of the string 'c://abcd /abcdef/0012wetr_1234567890.csv' between the last '/' and '_' characters.

0012wetr

I am able to extract everything after the last '/' character

select regexp_extract('c://abcd /abcdef/0012wetr_1234567890.csv', '([^/]*)$');

0012wetr_1234567890.csv

Unfortunately I am stuck and don't know how to split it further.

Your help would be appreciated. Cheers, A.

Answer 1

Maybe it is overkill but I managed to get required result using next combination of lookaheads - (??\/)[^\/]+(?=_) :

select regexp_extract('c://abcd /abcdef/0012wetr_1234567890.csv', '(?!\/)[^\/]+(?=_)');

Output:

_col0
0012wetr

regex101.com

Answer 2

You can use a REGEXP_REPLACE approach:

REGEXP_REPLACE('c://abcd /abcdef/0012wetr_1234567890.csv', '.*/([^_]+).*', '$1')

See the regex demo .

If you need to keep the result blank if there is no match, add |.+ at the end of the pattern:

REGEXP_REPLACE('c://abcd /abcdef/0012wetr_1234567890.csv', '.*/([^_]+).*|.+', '$1')

Details :

• .* - any zero or more chars other than line break chars as many as possible
• / - a / char
• ([^_]+) - Group 1: any one or more chars other than _
• .* - the rest of the line
• | - or
• .+ - any one or more chars other than line break chars as many as possible.