This question already has answers here:
How to use regex to find all overlapping matches
(5 answers)
Closed 2 years ago.
I have text as below-
my_text = "My telephone number is 408-555-1234"
on which i am searching the pattern
re.findall(r'\d{3}-\d{1,}',my_text)
My intention was to search for three digit numeric value followed by - and then another set of one or more than one digit numeric value. Hence I was expecting the result to be - ['408-555','555-1234'],
However the result i am getting os only ['408-555'] .
Could anyone suggest me what is wrong in my understaning here. And suggest a pattern that would serve my purpose
you can use:
re.findall(r'(?=(\d{3}-\d+))', my_text)
output:
['408-555', '555-1234']
Related
This question already has answers here:
How to grab number after word in python
(4 answers)
Closed 2 months ago.
I would recieve text like below
CRM NO: 23542536 crmno:# 3542536 crmno:_ 3542536... crm no 43653768754
my desired output will be:
23542536
3542536
3542536
43653768754
I want to write a regex to extract only the number after the string 'CRM NO'.
Also the CRM NO will come in variations like CRM NO or crmno or crm no
I have tried the regex ((?<=CRM NO)\D+\d+) but not compatible with all the entries
You can use a capture group with a case insensitive match and then match the leading part with an optional space
(?i)\bCRM ?NO\D+(\d+)\b
Regex demo
This question already has answers here:
How do I validate a date string format in python?
(5 answers)
Closed 6 months ago.
Im trying to check if a user's input is following the pattern integer/integer/integer(like month/day/year) but i dont know how to use exactly the match function to define that the pattern contains "number",then "/",again "number" and "/"...
Check out https://regex101.com/ for a neat website to check your regex! This is implemented in python using the re library. https://docs.python.org/3/library/re.html
In your case, the pattern would be [0-9]{1,2}\/[0-9]{1,2}\/[0-9]{2,4}
This question already has answers here:
Regex match everything after question mark?
(7 answers)
Closed 12 months ago.
I am trying to compare webpage URLs using regex. I am using the below method.
regex_url = r'https://www.website.com/books/\w{8}$'
is_read = re.match(regex_url, request.url) is not None
if not is_read:
add_to_read(token)
Everything works well for the above regex. But there is a new URL pattern now which I cant seem to get the regex right.
The new URL pattern is
https://www.website.com/books/Ab7us83xI?varient=web
9 characters followed by a question mark and then the word 'varient' and then '=web'. Can anyone help me get the correct regex for this?
Only the first 9 characters change every time. Apologies if this is a stupid question.
Many thanks.
Is this what you need?
https://www.website.com/books/\w{9}\?varient=web$
\w{9} - match 9 characters
\? - match question mark
varient=web - match varient=web
This question already has answers here:
regex for zip-code
(3 answers)
Closed 2 years ago.
I'm trying to write a regex that follows these constraints:
Exactly 5 digits
Sometimes, but not always, followed by a dash with 4 more digits
Zip code needs to be preceded by at least one whitespace
Cannot be at the start of a text
I've arrived at this but it's not giving me the output I want:
r"^[A-Za-z].*\s.*\d{5}(?:[-\s]\d{4})?$"
I would use:
(?<=[ \t])((?:\d{5}(?=[^\d-]|$))|(?:\d{5}-\d{4}(?=[^\d-]|$)))
Demo and explanation
This question already has answers here:
How to use regex to find all overlapping matches
(5 answers)
Closed 3 years ago.
I have a string like:
s = 'ababbabbba'
I'm trying to match all patterns matching any number of b's between a's. This is what I expect the patterns to be for s above:
['aba', 'abba', 'abbba']
This is what I've tried:
import re
re.findall('ab+a', s)
Which gives:
['aba', 'abbba']
I think that happens because any single a can only be part of a single group. Whereas my requirement would make the middle a's be part of two groups. Reading through the re documentation, I can't find any way to do this.
Based on the comment above, the solution is:
re.findall('(?=(ab+a))', s)