How to get input where a newline exists in Python

Question

I want to receive input where there is a newline in Python.

I want to input a number and I have implemented the following code, but I get the following error

Code

string = []

while True:
    input_str = int(input(">"))
    if input_str == '':
        break
    else:
        string.append(input_str)

Error

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

line error

>1
>2
>3
>
Traceback (most recent call last):
  File "2750.py", line 4, in <module>
    input_str = int(input(">"))
ValueError: invalid literal for int() with base 10: ''

I don't see the line that could be causing the error. In which line are you trying to cast a `string` into an `int`? — Mike, Apr 16 '19 at 10:05
Can you show some input you're using and where you're actually calling `int`? It also can't be Python 2.x, as using `input` and just hitting enter will give a `SyntaxError` @roganjosh — Jon Clements, Apr 16 '19 at 10:06
The error occurs at "input_str = int (input ("> "))" when nothing is entered. — prgnewbies, Apr 16 '19 at 10:06
@prgnewbies that's not the code you posted... you've got a check for an empty string there... so if the input's empty - skip it... otherwise, then try to convert it to int? — Jon Clements, Apr 16 '19 at 10:07
Please post the actual code you're using. The code you've posted doesn't have this line `input_str = int(input(">"))` — glhr, Apr 16 '19 at 10:08
@prgnewbies the linked duplicate addresses best practices and the alternate ways you can handle what you're trying to do here. — Jon Clements, Apr 16 '19 at 10:10

score 0 · Answer 1 · answered Apr 16 '19 at 10:10

0

try isdigit function of string, see below example:

string = []

while True:
    input_str = input(">")
    if input_str.isdigit():
        input_str = int(input_str)
    if input_str == '':
        break
    else:
        string.append(input_str)

answered Apr 16 '19 at 10:10

Amit Nanaware

3,203
1
6
19

DirtyBit · Answer 2 · 2019-04-16T10:17:56.153

0

I would suggest a try-except approach:

strList = []      
while True:
    try:
     strList.append(int(input('Enter the number > ')))
    except ValueError:
        print ('Invalid number. Please try again!')
        break
print(strList)

OUTPUT:

Enter the number > 1
Enter the number > 2
Enter the number > 3
Enter the number > 
Invalid number. Please try again!
[1, 2, 3]

edited Apr 16 '19 at 10:17

answered Apr 16 '19 at 10:12

DirtyBit

16,613
4
34
55

I don't think this is how they want it to behave. They want a check for equality to an empty string for `break` to occur, and the rest should be in a `try`/`except` as you have. The test for equality won't throw an exception and still allows incorrect input to not break out of the loop. This will currently break out on any non-integer-like input. – roganjosh Apr 16 '19 at 10:15
@roganjosh my intial thoughts as well. I'd say we see if OP clarifies this and I'll make an edit. – DirtyBit Apr 16 '19 at 10:19

How to get input where a newline exists in Python

2 Answers2