'How to exit a program if a blank line is entered?
I just want to exit the program if a blank line is entered. What do I need?
I've tried sys.exit(), but that doesn't quit the program, just returns a blank
while True: # Ask for patient ID
try:
i = int(input("Enter patient ID (or blank to exit): "))
if not i:
sys.exit()
else:
a = len(str(abs(i))) # Checking valid integer
if a != 6: # Checking id is six digits
print("Enter valid patient ID (six digit positive integer)")
else:
break
except ValueError:
print("Enter valid patient ID (six digit positive integer)")
I just expect the program to quit.
Solution 1:[1]
Your mistake was that you were reading and converting the input immediately into an integer! You need to check for blank input in a string format and then convert it to a number.
import sys
while True: # Ask for patient ID
try:
#Get the user's input as a string.
inpu = input("Enter patient ID (or blank to exit): ")
#If it's blank, exit.
if inpu == "":
sys.exit(0)
else:
#Now convert the inputed string into an integer.
i = int(inpu)
a = len(str(abs(i))) # Checking valid integer
if a != 6: # Checking id is six digits
print("Enter valid patient ID (six digit positive integer)")
else:
break
except ValueError:
print("Enter valid patient ID (six digit positive integer)")
Solution 2:[2]
In case of error : sys.exit(1) Else in case of successful run u can use sys.exit() or sys.exit(0)
But your code should be like :
while True: # Ask for patient ID
try:
i = raw_input("Enter patient ID (or blank to exit): ")
if len(i) == 0:
sys.exit()
else:
a = len(str(abs(i))) # Checking valid integer
if a != 6: # Checking id is six digits
print("Enter valid patient ID (six digit positive integer)")
else:
break
except ValueError:
print("Enter valid patient ID (six digit positive integer)")
Maybe replace raw_input by input
Hope that can help you,
Have a good day.
Solution 3:[3]
In python, that's kinda easy..!! :-
import os
val = input("TYPE IN HERE: ")
if val and (not val.isspace()):
print (val)
else:
#print("BLANK")
os._exit(1)
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | babaliaris |
Solution 2 | |
Solution 3 | athrvvv |