Issue
I have the following code which check if the element is in the list if the element has been found then it would add it to alert thereby in a while loop the element would would show up as exists
once.
test_list = [ 1, 6, 3, 5, 3, 4 ]
alert = []
while True:
i = 1
if i in test_list and (i != alert):
print('exists')
alert.append(i)
But it is endlessly printing exist
. Could you please advise what i need to do so if the element has been found in the list it only prints 1 time as exists
Solution
when you use while True:
the code inside it will run forever until you break
so in your code, I think you need to do this:
test_list = [ 1, 6, 3, 5, 3, 4 ]
alert = []
for i in test_list:
if not i in alert:
print('exists')
alert.append(i)
EDIT: if you want to run it forever:
test_list = [ 1, 6, 3, 5, 3, 4 ]
alert = []
while True:
for i in test_list:
if not i in alert:
print('exists')
alert.append(i)
Answered By - Mhmd Admn
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