Using RegEx to Extract Date
Hi all!
I am looking for advice on extracting a Maturity Date from a column containing a string of text, example:
string: 12months,MATURITY DATE:23-Jun-2022,BALANCE;$12345,PRODUCT
Return: 23-Jun-2022
I am able to successfully select the date by using: \d{2}-[A-z]{3}-\d{4}
But am unable to replace this so that it only returns the date. I always end up just getting the full text string.
Thanks in advance!
Best Answer
-
Try this:
REGEXP_REPLACE(`string`, '^.*(\d{2}-[A-z]{3}-\d{4}).*$', '$1')
You need to replace the entire string not just the substring. This is why there's ^.* and .*$ to match anything before and after the date value respectively.
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Answers
-
Try this:
REGEXP_REPLACE(`string`, '^.*(\d{2}-[A-z]{3}-\d{4}).*$', '$1')
You need to replace the entire string not just the substring. This is why there's ^.* and .*$ to match anything before and after the date value respectively.
**Was this post helpful? Click Agree or Like below**
**Did this solve your problem? Accept it as a solution!**2 -
Thank you @GrantSmith, worked perfectly. I appreciate it!
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