Challenge: Safely Auth
Note
These comments help you find the code that needs to be modified:
========== Modify code below ==========
========== Modify code above ==========
Tâche
Swipe to start coding
You have implemented the login() function, which raises errors. Implement the safely_auth() function that uses the login() function and catches errors.
- Create a code block to track errors (mention the keyword).
- Call the
login()function in the error tracking block. - Catch the
TypeErrorand save its info inside theerrorvariable. - Print the
errorvariable (error info) in the console. - Catch the
ValueErrorand save its info inside theerrorvariable.
Solution
99
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
USER_EMAIL = "top.user@in.the.world"
USER_PASSWORD = "super123secure321password"
def login(email, password):
if not isinstance(email, str):
raise TypeError("The email should be string")
if not isinstance(password, str):
raise TypeError("The password should be string")
if not "@" in email:
raise ValueError("The taken email argument is not email")
if not (email == USER_EMAIL and password == USER_PASSWORD):
raise ValueError("There are no user with this email and password")
else:
print("Authorize is successful")
def safely_auth(email, password):
try:
login(email, password)
except TypeError as error:
print("TypeError:", error)
except ValueError as error:
print("ValueError:", error)
print("Auth with wrong mail")
safely_auth("qw123d", "super123se123321password")
print("-" * 40)
print("Auth with wrong password type")
safely_auth("cat@top.mail", 12512)
Tout était clair ?
Merci pour vos commentaires !
Section 3. Chapitre 6
99
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
USER_EMAIL = "top.user@in.the.world"
USER_PASSWORD = "super123secure321password"
def login(email, password):
if not isinstance(email, str):
raise TypeError("The email should be string")
if not isinstance(password, str):
raise TypeError("The password should be string")
if not "@" in email:
raise ValueError("The taken email argument is not email")
if not (email == USER_EMAIL and password == USER_PASSWORD):
raise ValueError("There are no user with this email and password")
else:
print("Authorize is successful")
# ========== Modify code below ==========
def safely_auth(email, password):
___:
___(email, password)
except ___ as error:
print("TypeError:", ___)
___ ValueError as ___:
print("ValueError:", error)
# ========== Modify code above ==========
print("Auth with wrong mail")
safely_auth("qw123d", "super123se123321password")
print("-" * 40)
print("Auth with wrong password type")
safely_auth("cat@top.mail", 12512)
Demandez à l'IA
Posez n'importe quelle question ou essayez l'une des questions suggérées pour commencer notre discussion
