get LDAP user
0 likes • Nov 18, 2022
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""" Currency Converter----------------------------------------"""import urllib.requestimport jsondef currency_converter(currency_from, currency_to, currency_input):yql_base_url = "https://query.yahooapis.com/v1/public/yql"yql_query = 'select%20*%20from%20yahoo.finance.xchange%20where%20pair' \'%20in%20("'+currency_from+currency_to+'")'yql_query_url = yql_base_url + "?q=" + yql_query + "&format=json&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys"try:yql_response = urllib.request.urlopen(yql_query_url)try:json_string = str(yql_response.read())json_string = json_string[2:json_string = json_string[:-1]print(json_string)yql_json = json.loads(json_string)last_rate = yql_json['query']['results']['rate']['Rate']currency_output = currency_input * float(last_rate)return currency_outputexcept (ValueError, KeyError, TypeError):print(yql_query_url)return "JSON format error"except IOError as e:print(str(e))currency_input = 1#currency codes : http://en.wikipedia.org/wiki/ISO_4217currency_from = "USD"currency_to = "TRY"rate = currency_converter(currency_from, currency_to, currency_input)print(rate)
import itertoolsimport stringimport timedef guess_password(real):chars = string.ascii_lowercase + string.ascii_uppercase + string.digits + string.punctuationattempts = 0for password_length in range(1, 9):for guess in itertools.product(chars, repeat=password_length):startTime = time.time()attempts += 1guess = ''.join(guess)if guess == real:return 'password is {}. found in {} guesses.'.format(guess, attempts)loopTime = (time.time() - startTime);print(guess, attempts, loopTime)print("\nIt will take A REALLY LONG TIME to crack a long password. Try this out with a 3 or 4 letter password and see how this program works.\n")val = input("Enter a password you want to crack that is 9 characters or below: ")print(guess_password(val.lower()))
from math import pidef rads_to_degrees(rad):return (rad * 180.0) / pirads_to_degrees(pi / 2) # 90.0
def clamp_number(num, a, b):return max(min(num, max(a, b)), min(a, b))clamp_number(2, 3, 5) # 3clamp_number(1, -1, -5) # -1
from collections import Counterdef find_parity_outliers(nums):return [x for x in numsif x % 2 != Counter([n % 2 for n in nums]).most_common()[0][0]]find_parity_outliers([1, 2, 3, 4, 6]) # [1, 3]
filename = "data.txt"data = "Hello, World!"with open(filename, "a") as file:file.write(data)