Regex end with a character or end of line with lookahead

Hanzcerb 0
Viewed 46

I have this string

Book Release Date: 2 June, 2010 [Edition#5]

Book Release Date: 24 October, 1996

I want to use a regex to find the date only like follow:

2 June, 2010

24 October, 1996

I have tried using this pattern that is close to what I want

# this pattern result
# 2 June, 2010 [Edition#5]
# 24 October, 1996
date = re.findall(r"(?<=(Book Release Date: ))(.*?)(?=(\[|\n))", text)

# this pattern result
# 2 June, 2010
# None
date = re.findall(r"(?<=(Book Release Date: ))(.*?)(?=\[)", text)
python-3.x regex python
2 Answers

You don't need any lookaround assertions, just a single capture group that will be returned using re.findall

\bBook Release Date: (\d+ [A-Z][a-z]+, \d{4})\b

Explanation

  • \bBook Release Date:
  • ( Capture group 1
    • \d+ [A-Z][a-z]+ Match 1+ digits, space, uppercase char A-Z, 1+ lowercase chars
    • , \d{4} Match , and 4 digits
  • ) Close group 1
  • \b A word boundary to prevent a partial word match

Regex demo | Python demo

Example

import re
 
pattern = r"\bBook Release Date: (\d+ [A-Z][a-z]+, \d{4})\b"
 
s = ("Book Release Date: 2 June, 2010 [Edition#5]\n"
    "Book Release Date: 24 October, 1996")
 
print(re.findall(pattern, s))

Output

['2 June, 2010', '24 October, 1996']
The fourth bird0

Use this :\s(\d+)\s(\w+,)\s(\d+), code:

import re

s ='''
Book Release Date: 2 June, 2010 [Edition#5]

Book Release Date: 24 October, 1996

'''

print([' '.join(i) for i in re.findall(':\s(\d+)\s(\w+,)\s(\d+)', s)])

Output:

['2 June, 2010', '24 October, 1996']
Shahab Rahnama0
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