Why this code doesn't work....?

Question

p=input()
flag=0
for j in p:
if j=='1':
flag=1
continue
elif j=='0':
if flag==1:
print('ten',end='')
flag=0
else:
print('zero',end='')
elif j=='2':
print('two',end='')
elif j=='3':
print('three',end='')
elif j=='4':
print('four',end='')
elif j=='5':
print ('five',end='')
elif j=='6':
print('six',end='')
elif j=='7':
print('seven',end='')
elif j=='8':
print('eight',end='')
elif j=='9':
print('nine',end='')

elif flag=='1':
print('one',end='')
flag=0
else:
print(j,end='')

Answer 1

What is the expected output u want to help u ?

Answer 2

You need to decide whether your flag is an integer (flag == 1) or a string (flag == '1'). You can't use both in the same code.

Answer 3

It skipping to convert 1 to one. Missing case which have input 1. Check your logic again when j=='1', and it never going true in 'elif flag=='1' : need a change here.

instead of print in every if block, use a string and finally print string . else use replace function..
and also you can simplifies this code lot of ways.. even to 1 liner.
but still, I think ,better to continue first complete code to pass all cases. next try to simplify.. just telling..

edit:

oh. yes ..
change '1' to 1 will works fine . and add a space in

elif flag==1:
print('one',end=' ')
flag=0

Answer 4

Explain your question... also use appropriate tag

Answer 5

I think so it is no numerals code coach