ValueError: invalid literal for int() with base 10: ''

Question

I got this message after my first print:

from random import *
budget = 200
print("ciao hai " + str (budget) + " Euro a disp")

if budget>0:
while budget>0:
num_scelto= int (input("\ninserisci un numero tra 0 e 49 "))
print("\nprocediamo all'estrazione....")

if num_scelto < 0 or num_scelto >50:
print("scegli tra 0 e 49")
#Error in digitation
else:
print("\n hai scelto il numero ... "+ str(num_scelto))

puntata= int(input("\nscegli una somma:"))
if budget- puntata < 0:
puntata = int(input("ripetere la puntata:"))
else:
print("\n ti rimangono" + str(budget - puntata) + "euro..." )

print ("\nhai puntato" + str(puntata) + "euro")
estratto = (randint(0,49))
print( "\nil numero estratto è " + str(estratto))
if estratto != num_scelto:
budget= budget - puntata
print("\nhai perso ora hai " + str(budget))

else:
vincita = puntata *50
budget = budget + vincita
print("hai indovinato ora hai " + str(budget))
else:
print(budget)

print("fine")

Answer 1

What are the inputs and outputs for program?

Answer 2

Paste your code in playground and then give firstly it is small and it has no bold or italic characters it will be not easy to find errors even its not good for reading

Answer 3

Sorry I don't understand what you mean by sum
Can you give an example of inputs ?
like
23
38
45

And like that ,so I can run it and see where error is occurring or can you tell

Exactly at which line you are getting error invalid literal for int()?

Answer 4

The input should be a number from 0 to 49 and specify a sum to bet. You have to guess a number than python will return a random int number. If you guess it you will add points otherwise you will loss

Answer 5

OK error In line 19. Invalid literal for int() with base 10.
I would choose a number between 0 and 49. Then computer extract another one with Randint. If you guess the number you win... Like in casino