Give hint.

Question

End the loop when the user enters 0.
Output the resulted list after the while loop ends.
(using while loop)

Sample Input
1
2
3
0

Sample Output
[1, 2, 3]

Answer 1

Rohit Wadher , you may try something like this:

items = []
while True:
inp = int(input('enter a number or 0 to quit: '))
if inp == 0:
break
items.append(inp)
print(items)

you are not far away with your code. here some comments:
- your input has to be inside the for loop, otherwise you can not enter multiple times
- an indentation is missing after if... statement
- you don't need items.remove

Answer 2

It can be done like this:

items = []
item = int(input())
while item!=0:
items.append(item)
item = int(input())
print(items)

We first create an empty list `items`

Then create a variable `item` and assign input to it.

We want to stop taking input when input is 0. So use condition `item! =0`. When item is 0 loop will stop. Inside loop keep appending the inputs that you get from user to the list `items`

at the end we print the list.

I'm also not python programmer so anyone please correct if I'm wrong here.

Answer 3

🇮🇳Omkar🕉 , there was nothing wrong in your code 👍, i only mentioned that input() is implemented twice in that code. that is a duplication that should be avoided.

Answer 4

although in code playground python 3.8 has been available for some time, it is rarely used. here is a nice sample solving the current task by using := (walrus operator)
https://code.sololearn.com/ciyLNRet035q/?ref=app

Answer 5

🇮🇳Omkar🕉 , not to blame anyone, but you don't need 2 input() statements. if you switch to a while True:, you only need one input(), but to exit the loop it needs a break .
But there is nothing wrong with it.

Answer 6

Your attempt?

Answer 7

Rohit Wadher , "I tried but" -
Share your code here. We'll try to explain what's wrong and how it can be fixed.

Answer 8

items = []
n = int(input())

while True:
items.append(n)
print(items)
if n = 0:
a = items.remove(n)
print(a)
break

Answer 9

Lothar ok

Answer 10

items = []
while True:
inp = int(input('enter a number or 0 to quit: '))
if inp == 0:
break
items.append(inp)
print(items)

i try my best u can check it help u r not i can say 99.9% it is correct

Answer 11

this worked for me!

items = []

while True:

n = int(input())
items.append(n)

if n == 0:
print(items[:-1])
break

Answer 12

It is a code coach problem?

Answer 13

lst=[]
while True:
input_=int(input("Enter a number :"))
if input_ != 0:
lst.append(input_)
else:
break
print(lst)

you can do like this...
Hope this helps,

Answer 14

this worked for me!

items = []

while True:

n = int(input())
items.append(n)

if n == 0:
print(items[:-1])
break

Answer 15

🇮🇳Omkar🕉 Thank you, i have tried it successfully.

Answer 16

Lothar ,
Thanks for trying to correct. I appreciate that.

The condition when we need to exit loop is included in the while loop itself

while item! =0

Answer 17

🇮🇳Omkar🕉 how can I integrate an external API on WordPress blog?

Answer 18

Emmanuel ,
I have no experience with wordpress.

You can ask it as new question in qa forum. Those who know will answer 🙂.

Answer 19

🇮🇳Omkar🕉
Thanks.

Answer 20

from itertools import *
lst=[]
for i in count(1):
item=int(input("Enter Choice"))
if(item==0):
break
lst.append(item)
print(lst)

#Using for loop