Trying to solve cash out in intermediate python

Question

Only two test case seem to come out write, and i don't where i'm getting it wrong
This is the code i used:
def withdraw(amount):
print(str(amount) + " withdrawn!")
try:
num=input()
x=withdraw(num)
except (ValueError, TypeError):
print:("Please enter a number")

Answer 1

this code might work ........

def withdraw(amount):
print(str(amount) + " withdrawn!")

try:
a = int(input())
withdraw(a)

except (ValueError, TypeError):
print("Please enter a number")

Answer 2

Jayakrishna🇮🇳 thanks

Answer 3

My other solution

def withdraw(amount):
print(f'{amount} withdrawn!')

while True:
try:
withdraw(int(input()))
except:
print("Please enter a number")
break

Answer 4

You have to typecast num to integer . and remove (:) after print in last line.

Answer 5

Md Sayed 🇧🇩🇧🇩 thanks it worked

Answer 6

Hey Guyz , Here I have Got the perfect solution do check it ... Happy learning !!

def withdraw(amount):
print(str(amount) + " withdrawn!")

#your code goes here
try:
num=int(input())
x=withdraw(num)
except (ValueError, TypeError):
print("Please enter a number")

Answer 7

What is the error is coming out
And also tell me what is the function of that code

Answer 8

Post the problem description pls..
I think you need to take input as a numbers...

Answer 9

calls the corresponding withdrawal method.
In case the input is not a number, the machine should output "Please enter a number".
Use exception handling to take a number as input, call the withdraw() method with the input as its argument, and output "Please enter a number", in case the input is not a number.

Answer 10

An ATM machine takes the amount to be withdrawn as input and calls the corresponding withdrawal method.
In case the input is not a number, the machine should output "Please enter a number".
Use exception handling to take a number as input, call the withdraw() method with the input as its argument, and output "Please enter a number", in case the input is not a number.

Answer 11

try :
int(num)
except :
print('please enter a number')

Take input, and call method with input. Next try this code within the method. If input is not a number then it raise error otherwise don't.. So modify this according to needed in description...

Answer 12

try :
int(num)
except :
print('please enter a number')

Take input, and call method with input. Next try this code within the method. If input is not a number then it raise error otherwise don't.. So modify this according to needed in description...

Answer 13

I tried this and it still did not work
def withdraw(amount):
print(str(amount) + " withdrawn!")
try:
num=input()
int(num)
except :
print:("Please enter a number")

Answer 14

Use it.

def withdraw(amount):
print(str(amount) + " withdrawn!")
try:
num=int(input())
withdraw(num)
except :
print:("Please enter a number")

Answer 15

Just try this code

try :

a = int(input())

except TypeError :

print("please enter a number")

Answer 16

Simisola Osinowo you have : after print in statement
print("Please enter a number").
So its causing problem.
But i said to try this way...

def withdraw(amount):
try:
int(num)
print(str(amount) + " withdrawn!")
except :
print("Please enter a number")

Call this method with input taken....

Answer 17

def withdraw(amount):
print(str(amount) + " withdrawn!")

try:
a = int(input())
withdraw(a)

except (ValueError, TypeError):
print("Please enter a number")

Answer 18

The following answer is my code
def withdraw(amount):
print(str(amount) + " withdrawn!")

#code is entered
try:
withdraw(int(input()))
except (ValueError,TypeError):
print("Please enter a number")
# for simplify
try :
withdraw(int(input()))
except :
print("Please enter a number")