curriculum/challenges/english/blocks/learn-regular-expressions-by-building-a-password-generator/6565c084627071646f94c4b0.md
Finally, after the for loop, create an if statement to check if count is equal to 4 and break out of the while loop by using the break statement.
You should create an if statement that checks if count is equal to 4 after the for loop.
({ test: () => assert.match(code, /^(\s*)for.+:\s*^\1(\s{4})if\s+constraint\s*<=\s*len\s*\(\s*re\.findall\s*\(\s*pattern\s*,\s*password\s*\)\s*\)\s*:\s*^\1\2\2count\s*\+=\s*1\s*^\1if\s+count\s*==\s*4\s*:/m) })
You should use break inside your new if to break out of the while loop.
({ test: () => {
const commentless_code = __helpers.python.removeComments(code);
const {block_body} = __helpers.python.getBlock(commentless_code, /if\s+count\s*==\s*4\s*/);
assert(block_body.match(/^\s+break\s*$/m));
}
})
import re
import secrets
import string
def generate_password(length, nums, special_chars, uppercase, lowercase):
# Define the possible characters for the password
letters = string.ascii_letters
digits = string.digits
symbols = string.punctuation
# Combine all characters
all_characters = letters + digits + symbols
while True:
password = ''
# Generate password
for _ in range(length):
password += secrets.choice(all_characters)
constraints = [
(nums, r'\d'),
(lowercase, r'[a-z]'),
(uppercase, r'[A-Z]'),
(special_chars, fr'[{symbols}]')
]
--fcc-editable-region--
# Check constraints
count = 0
for constraint, pattern in constraints:
if constraint <= len(re.findall(pattern, password)):
count += 1
--fcc-editable-region--
return password
# new_password = generate_password(8)
# print(new_password)