Regex Find (return pattern or null)

Matthew

hello everyone, i would like to find a Regex pattern within a string, and replace the entire string value with only a portion of that matched pattern

i know i can do this with the following:

REGEX_Replace([Dest], '.*TZ::([/\w]+?),.*', '$1')

But it only works if the pattern exists.. If the pattern isn't found, then the returned value is the entire string (i would prefer it to return null).. a complete solution requires me to first check for the pattern, and THEN replace it

if 
REGEX_Match([Dest], '.*TZ::([/\w]+?),.*') = -1 
then
REGEX_Replace([Dest], '.*TZ::([/\w]+?),.*', '$1')
else
null()
endif

Is there any way to accomplish this with a single statement without the "If" condition?

OllieClarke

@Matthew RegEx works left to right, so the following should work for you:

RegEx_Replace([Field],'.*TZ::([\/\w]+?),.*|.*','$1')

However, as this would return an empty string, you still need a two step process to get what you want.

Personally I think 

if 
REGEX_Match([Dest], '.*TZ::([/\w]+?),.*') = -1 
then
REGEX_Replace([Dest], '.*TZ::([/\w]+?),.*', '$1')
else
null()
endif

Is probably easier to understand than

SWITCH(REGEX_Replace([Field1],'.*TZ::([\/\w]+?),.*|.*','$1'),REGEX_Replace([Field1],'.*TZ::([\/\w]+?),.*|.*','$1'),'',NULL())

Or whatever approach you'd want

Ollie