Python regex search until specific word and exclude everything behind it - python-3.x

Question

I have a script that always have the "get the" and the "get" in a string.
The "ONE TWO THREE" can vary, like it also can be "THIRTEEN FORTY" or "SIX". After these variations there will always be a 2nd "get".
I have the following code:
variable = 'get the ONE TWO THREE get FOUR FIVE'
myVariable = re.compile(r'(?<=get the) .*')
myVariableSearch = myVariable.search(variable)
mySearchGroup = myVariableSearch.group()
print(mySearchGroup)
#prints ONE TWO THREE get FOUR FIVE
I want my script to exclude the 2nd "get" and everything behind it. My desired result is to be just the "ONE TWO THREE".
How do I exclude this? Any help would be appreciated!

Answer 1

You can use
\bget\s+the\s+(.*?)(?=\s*\bget\b|$)
See the regex demo.
Details
\bget\s+the\s+ - whole word get, 1+ whitespaces, the, 1+ whitespaces
(.*?) - Group 1:
(?=\s*\bget\b|$) - a positive lookahead that requires 0+ whitespaces and then a whole word get, or end of string immediately on the right of the current location.
See the Python demo:
import re
variable = 'get the ONE TWO THREE get FOUR FIVE'
myVariableSearch = re.search(r'\bget\s+the\s+(.*?)(?=\s*\bget\b|$)', variable)
mySearchGroup = ''
if myVariableSearch:
mySearchGroup = myVariableSearch.group(1)
print(mySearchGroup)
# => ONE TWO THREE