I have following options, but all of them return full string. I need to remove date at the beginning with regex.
d = re.match('^\d{2}.\d{2}.\d{4}(.*?)$', '01.10.2018Any text..')
d = re.match('^[0-9]{2}.[0-9]{2}.[0-9]{4}(.*?)$', '01.10.2018Any text..')
How to do that? Python 3.6
You could use sub to match the date like pattern (Note that that does not validate a date) from the start of the string ^\d{2}\.\d{2}\.\d{4} and replace with an empty string.
And as @UnbearableLightness mentioned, you have to escape the dot \. if you want to match it literally.
import re
result = re.sub(r'^\d{2}\.\d{2}\.\d{4}', '', '01.10.2018Any text..')
print(result) # Any text..
(Note that that does not validate a date) to my answer.
. matching any character works, however for educational purposes it would be good to escape the dot to show OP that metacharacters need escaping.Grab the first group of the match
>>> d = re.match('^\d{2}.\d{2}.\d{4}(.*?)$', '01.10.2018Any text..').group(1)
>>> print (d)
'Any text..'
If you are not sure, if there would be a match, you have to check it first
>>> s = '01.10.2018Any text..'
>>> match = re.match('^\d{2}.\d{2}.\d{4}(.*?)$', s)
>>> d = match.group(1) if match else s
>>> print(d)
'Any text..'
AttributeError: 'NoneType' object has no attribute 'group'
group(1) instead of d[3]? That's weird why I can't use array to print result.
99.10.2018Any text.. which is an invalid date.... matching any character works, however for educational purposes it would be good to escape the dot to show OP that metacharacters need escaping.Use a group to extract the date part:
d = re.match('^(\d{2}.\d{2}.\d{4})(.*?)$', '01.10.2018Any text..')
if d:
print(d.group(1))
print(d.group(2))
Group 0 is the whole string, I added a pair of parentheses in the regex around the date. This is group 1. Group 2 is the text after which is what you're after
99.10.2018Any text.. which is an invalid date...