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How do I search for passwords with minimum length of 8 allowing letters, numbers and symbols: ##$% with regex? [closed]

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What is wrong with the below used regular expression? Why does it not match the password?
import re
pattern = re.compile ('^\w##$%{8,}')
password = '12345abcd##$%'
x = pattern.search(password)
print (x)
print (len(password))

You didn't escape the $ which has a special meaning in a regular expression and didn't put the allowed characters in square brackets to allow any of them.
This: ^[\w##\$%]{8,} is the modified version of the regex which matches the password.
Escaping the $ character isn't really necessary within square brackets so ^[\w##$%]{8,} will work as well.
I suggest you check your regular expressions here: https://regex101.com/r/ldvJLf/1 . This site explains in detail the meaning of all single elements of the regular expression, so you can directly see what is wrong if things doesn't work as you expected.

Tip:
check your regexes online https://regexr.com/
I think you want:
pattern = re.compile ('^[\w##$%]{8,}')

Getting two digits from a string in python with regex [closed]

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Closed 4 years ago.
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I have multiple strings like this 90'4 I want to extract the digits from the string and sum them up to get 94.
I tried compiling the pattern.
pattern="\d'\d"
re.compile(pattern)
I tried the methods findall and match, but did not get what I wanted.
I need to use regex I cannot use .split()

Use \d+ with findall to extract numbers and then find their sum:
import re
s = "this is 90'4"
numbers = re.findall(r'\d+', s)
print(sum(map(int, numbers)))
# 94

How to replace a ' from a String in python [closed]

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Closed 5 years ago.
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I am trying to remove a " ' " character from a python string .
Below code gives a syntax error, How to achieve this task
final = string.replace(old_str, '\'', '')

To replace single quotes
final = old_str.replace("'", "")

str = "this is string example....wow!!! this is really string"
print(str.replace("is", "was"))
Here's an example of how to correctly use the replace function. This replaces all the is's with was's. It's hard to see what you're trying to do but use this as a guide.

You can do this
string = "some ' string"
final = string.replace('\'','')
print(final)
output
some string

Using regex to get numbers from a string using Python, incorrectly prints full string [closed]

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Objective
I'm looking to use the regular expression \d+ to extract just the digits from the string, answer_40194.
Problem
I'm targeting a form element with Selenium and I'm printing the formID to the Terminal, but after the line re.findall('\d+', formID) I expect formID to be just the numbers 40194, but instead I'm getting the entire string answer_40194.
script.py
import selenium
import re
form = browser.find_element_by_tag_name('form')
formID = form.get_attribute('id')
re.findall('\d+', formID)
print formIDNumber

You need to assign the result to a variable, e.g.
var1 = re.findall('\d+', formID)
print(var1)
This will generate a list, if you only want one result, use
var1 = re.search('\d+', formID)
print(var1.group(0))
The latter is called a regular expression object, hence the .group(0), see the documentation on python.org for more information.

define a regular expression in python [closed]

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I am trying to use regular expressions in python to say a 4 characters string with 1st character being a digit and 3 other characters being either a digit or a capital letter.
Here's examples of patterns that should match 1CTT, 2IR8, 35TR, 4T1R
I tried many ways, here's the last code I tried :
exp=re.compile("[0-9]{1}([A-Z0-9]{3})")
Thank you for your help !

The expression you've tried last, looks correct and should match the provided test strings. Though you don't have to specify {1} and there is no need for a capturing group (the parenthesis):
>>> import re
>>> text = "text, 1CTT, 2IR8, 35TR, 4T1R, smth else"
>>> pattern = re.compile(r"[0-9][A-Z0-9]{3}")
>>> pattern.findall(text)
['1CTT', '2IR8', '35TR', '4T1R']
You might need to additionally add the word boundary constraint (thanks to #Jon Clements):
>>> text = "text, 1CTT, 2IR8, 35TR, 4T1R, smth else, 35TT35XYZ"
>>> pattern = re.compile(r"\b[0-9][A-Z0-9]{3}\b")
>>> pattern.findall(text)
['1CTT', '2IR8', '35TR', '4T1R']