Regarding RegEx...

Question

Hi, can you please help. I created a code to check the inputted password if it will pass the given condition which is to contain at least 1 uppercase letter and at least 1 number in any order. If the condition is met, it will print "password created" otherwise "wrong format", the code has a bug which prints "password created" even though I only inputted 1 uppercase letter.
Thanks.
https://code.sololearn.com/cvxv980LXKqi/?ref=app

Answer 1

Jayakrishna🇮🇳 ,

thank you for your great explanation. very well done!

Answer 2

Sandeep
Task need a digit \d and Capital letter [A-B] , for these before and after anything so 2 possibilities: digit first then letter, or letter first then digit.
2 classes we need to define:

1) digit before Letter (.*\d+.*[A-Z]+.*)

( .* before, after, between tells -> there can be any characters (0 Or more) may present)

2) CapsLetter before digit (.*[A-Z].*\d.*)

There 2 classes can be added by | operator, so that can match anyone class.
.* means 0 or more any characters.
\d+ one or more digits, we just need to check single digit so \d is enough..

Combining this and simplified solution is : (.*\d.*[A-Z]+.*)|(.*[A-Z].*\d.*)

Further it can be simplified to @Ani jona mentioned solution:
r"[A-Z].*\d|\d.*[A-Z]"
(It represents : A capital letter next a digit anywhere in string or A digit, next A capital Letter anywhere in string, include of in between .* means there may not need consecutively ).
Hope it helps...

Answer 3

the complete task including counting of specific characters can also be done in a simple way with only very basic regex:

import re

string = "GreenLagune24"
upp = len(re.findall(r"[A-Z]", string))
dig = len(re.findall(r"[0-9]", string))

print(f'upper case: {upp}, digits: {dig}')

Answer 4

Sandeep
\d represents for digit.
\d{2, } represents atleast 2digits sequentially..
(.*\d){2, } represents a class of pattern which matches for atleast 2 digits which may include any number of characters before it..
So it matches : 2asd3, 23, abc23, 2ab3c...,

There may be exist simplest forms for patterns also, am not aware of those now...
As Lothar mentioned, you can use
r"[0-9]" , and findall returns number of matches
Use it as if dig >= 2 : print( "passed" )

Answer 5

\d* means "digit 0 or more", so it accepts 0 digits..
Use \d+ => 1 or more, instead
again you need distinguish to classes for any order..

Answer 6

Try this

pattern = r"[A-Z].*\d|\d.*[A-Z]"

Answer 7

Cris Tradio ,

the current pattern does accept the inputs e.g.:

'passWord' or 'WXayy'

even though the digits are missing.

Answer 8

Lothar Right! and as mentioned by Jayakrishna🇮🇳 changing the * could do the trick

but there is still one problem, If i remember correctly, regex matches are done from left to right, therefore OP regex pattern has to be in order - words , digits, uppercase, words, digits

This means if we input digit first, regex match will fail (like in Samialloh code)

Ex: 13aasJHAM

Above will fail even though it is a right pattern!

Lothar and Jayakrishna, can you tell me how to do this correctly?

Answer 9

thanks! Jayakrishna🇮🇳 :) it looks really simple now! You explained it really well 🙌

..
I can't see Ani Jona answer, looks like some new Sololearn bug(i am getting lots of bugs these days like no-notifs)

Also, Is there a way to add length checking to current regex? (ex- atleast 2 uppercase or 2 digits and they don't have to be side-by-side)

Answer 10

Lothar thanks, I think you misunderstood what I meant,

what I want is something like a password validator, not counting the length of a uppercase or digits!

For ex: "nan2C" string passes the test, but I want the string to include atleast 2 uppercase and 2 digits.

https://code.sololearn.com/cFUWpQ86ikrk/?ref=app

Therefore, "nan2C" input should result in "Not Valid" password and something like "NaN22" results in valid password! (because it contains two [A-B] and two \d)

Answer 11

Jayakrishna🇮🇳 Lothar

it's working! 😁

https://code.sololearn.com/cN05JXMym1Bt/?ref=app

Athough it looks a little too big but I think I can modify this from here. I will also try to add the method mentioned by Lothar to make it more clear!

thanks again! both of you :}

Answer 12

Okay...thank you! Jayakrishna🇮🇳

Answer 13

Thank you.

Answer 14

Hi, thank you all.
Below is my revised coding.
Any suggestions on improving the way I code.? Open for suggestions...

# pw authentication, at least 1 uppercase letter and at least 1 number in any order.
import re
pw = input()

pattern = r"\w*\d*([0-9]+\w*\d*[A-Z]+)|([A-Z]+\w*\d*[0-9]+)\w*\d*"

if re.findall(pattern, pw):
print("Password created")
else:
print("Wrong format")

Answer 15

Cris Tradio
Here,
\w*\d* works same as \w*
Since you are using findall, no need to pattern for entire string.
Pattern part of string is enough so
Can be reduced to... This is enough,

pattern = r"\d.*[A-Z]|[A-Z].*\d"

Answer 16

And what u wanna to know? Ur code working.

Answer 17

Yes it's working, what I want to know is my pattern correct?, because I wanted to meet the condition that it should take at least 1 uppercase letter and at least 1 number, but when I inputted uppercase letter/s only without a number, the output says "password created". It should have been "wrong format", because it needs at least 1 uppercase letter and at least 1 number. I don't know if the condition I wanted to meet is doable. Thanks for your reply.

Answer 18

https://code.sololearn.com/c5HZxhdJbq2n/?ref=app
Cris Tradio