Here’s all you need to do:
def doit(text):
import re
matches = re.findall(r'"(.+?)"',text)
# matches is now ['String 1', 'String 2', 'String3']
return ",".join(matches)
doit('Regex should return "String 1" or "String 2" or "String3" ')
result:
'String 1,String 2,String3'
As pointed out by Li-aung Yip:
To elaborate,
.+?
is the “non-greedy” version of.+
. It makes the regular expression match the smallest number of characters it can instead of the most characters it can. The greedy version,.+
, will giveString 1" or "String 2" or "String 3
; the non-greedy version.+?
givesString 1
,String 2
,String 3
.
In addition, if you want to accept empty strings, change .+
to .*
. Star *
means zero or more while plus +
means at least one.