python, using if loop to make sure user input is a number then manipulate number
1
I am very new to programming and tried to make a age checker code, however the loop to make sure user input is a number does not work.
If a string is entered, for example hello, the input passes the .isdigit() condition, and messes up the last if statement.
Any help would or advice on how to construct the number checker loop properly, so that strings are not entered into the last if statement would be greatly appreciated.
Here is my code:
while True:
age = input('Enter age: ')
if age.isdigit():
print('Thank you, number entered is:', age)
else:
print('A number must be entered')
break
print('Please enter age')
used_variable = age
if used_variable > 18:
print('user is older than 18')
python
edited Nov 21 '18 at 15:00
Avishay Cohen
701724
asked Nov 21 '18 at 13:10
Jacob Jacob
91
add a comment |
1
I am very new to programming and tried to make a age checker code, however the loop to make sure user input is a number does not work.
If a string is entered, for example hello, the input passes the .isdigit() condition, and messes up the last if statement.
Any help would or advice on how to construct the number checker loop properly, so that strings are not entered into the last if statement would be greatly appreciated.
Here is my code:
while True:
age = input('Enter age: ')
if age.isdigit():
print('Thank you, number entered is:', age)
else:
print('A number must be entered')
break
print('Please enter age')
used_variable = age
if used_variable > 18:
print('user is older than 18')
python
edited Nov 21 '18 at 15:00
Avishay Cohen
701724
asked Nov 21 '18 at 13:10
Jacob Jacob
91
1
Once you know that it only consists of digits you still have to cast your string into an integer with the function int()
– leoburgy
Nov 21 '18 at 13:19
are you using python3 or python2 ? like @Adelin mentiond below, they have different output depending on the python version soisdigit()might not be a good fit here. also
– Avishay Cohen
Nov 21 '18 at 14:29
add a comment |
1
1
1
I am very new to programming and tried to make a age checker code, however the loop to make sure user input is a number does not work.
If a string is entered, for example hello, the input passes the .isdigit() condition, and messes up the last if statement.
Any help would or advice on how to construct the number checker loop properly, so that strings are not entered into the last if statement would be greatly appreciated.
Here is my code:
while True:
age = input('Enter age: ')
if age.isdigit():
print('Thank you, number entered is:', age)
else:
print('A number must be entered')
break
print('Please enter age')
used_variable = age
if used_variable > 18:
print('user is older than 18')
python
edited Nov 21 '18 at 15:00
Avishay Cohen
701724
asked Nov 21 '18 at 13:10
Jacob Jacob
91
I am very new to programming and tried to make a age checker code, however the loop to make sure user input is a number does not work.
If a string is entered, for example hello, the input passes the .isdigit() condition, and messes up the last if statement.
Any help would or advice on how to construct the number checker loop properly, so that strings are not entered into the last if statement would be greatly appreciated.
Here is my code:
while True:
age = input('Enter age: ')
if age.isdigit():
print('Thank you, number entered is:', age)
else:
print('A number must be entered')
break
print('Please enter age')
used_variable = age
if used_variable > 18:
print('user is older than 18')
python
python
edited Nov 21 '18 at 15:00
Avishay Cohen
701724
asked Nov 21 '18 at 13:10
Jacob Jacob
91
edited Nov 21 '18 at 15:00
Avishay Cohen
701724
asked Nov 21 '18 at 13:10
Jacob Jacob
91
edited Nov 21 '18 at 15:00
Avishay Cohen
701724
edited Nov 21 '18 at 15:00
Avishay Cohen
701724
edited Nov 21 '18 at 15:00
Avishay Cohen
701724
701724
asked Nov 21 '18 at 13:10
Jacob Jacob
91
asked Nov 21 '18 at 13:10
Jacob Jacob
91
asked Nov 21 '18 at 13:10
Jacob Jacob
91
91
1
Once you know that it only consists of digits you still have to cast your string into an integer with the function int()
– leoburgy
Nov 21 '18 at 13:19
are you using python3 or python2 ? like @Adelin mentiond below, they have different output depending on the python version soisdigit()might not be a good fit here. also
– Avishay Cohen
Nov 21 '18 at 14:29
add a comment |
1
Once you know that it only consists of digits you still have to cast your string into an integer with the function int()
– leoburgy
Nov 21 '18 at 13:19
are you using python3 or python2 ? like @Adelin mentiond below, they have different output depending on the python version soisdigit()might not be a good fit here. also
– Avishay Cohen
Nov 21 '18 at 14:29
1
1
Once you know that it only consists of digits you still have to cast your string into an integer with the function int()
– leoburgy
Nov 21 '18 at 13:19
Once you know that it only consists of digits you still have to cast your string into an integer with the function int()
– leoburgy
Nov 21 '18 at 13:19
are you using python3 or python2 ? like @Adelin mentiond below, they have different output depending on the python version so
isdigit() might not be a good fit here. also– Avishay Cohen
Nov 21 '18 at 14:29
are you using python3 or python2 ? like @Adelin mentiond below, they have different output depending on the python version so
isdigit() might not be a good fit here. also– Avishay Cohen
Nov 21 '18 at 14:29
add a comment |
1 Answer
1
active
oldest
votes
-1
You were breaking out of loop when user inserted incorrect value (not digit) instead of letting loop go on to ask him again, you should have break the loop when user typed correct value.
while True:
age = input('Enter age: ')
if age.isdigit():
print('Thank you, number entered is:', age)
break
else:
print('A number must be entered')
#print('Please enter age')
used_variable = int(age)
if used_variable > 18:
print('user is older than 18')
answered Nov 21 '18 at 13:12
Filip MłynarskiFilip Młynarski
1,5881311
1
so, for an 'int' that does a conversion, you copy pasted an answer, with no explanation whatsoever, an answer that was already given countless of times? Yeah, thats a -1
– Adelin
Nov 21 '18 at 13:14
I did not copy pasted any answer, and I think its ok to first post code to let author see and create explanation in meantime.
– Filip Młynarski
Nov 21 '18 at 13:15
If you think this answer is a good answer that was not already posted before, then keep it, but I stick with my downvote
– Adelin
Nov 21 '18 at 13:29
If author's code only problem would be about converting input to integer you would be right, but there's other bugs that I shown and explained.
– Filip Młynarski
Nov 21 '18 at 13:32
Thanks for answer, it worked.
– Jacob
Nov 22 '18 at 12:12
|
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
-1
You were breaking out of loop when user inserted incorrect value (not digit) instead of letting loop go on to ask him again, you should have break the loop when user typed correct value.
while True:
age = input('Enter age: ')
if age.isdigit():
print('Thank you, number entered is:', age)
break
else:
print('A number must be entered')
#print('Please enter age')
used_variable = int(age)
if used_variable > 18:
print('user is older than 18')
answered Nov 21 '18 at 13:12
Filip MłynarskiFilip Młynarski
1,5881311
1
so, for an 'int' that does a conversion, you copy pasted an answer, with no explanation whatsoever, an answer that was already given countless of times? Yeah, thats a -1
– Adelin
Nov 21 '18 at 13:14
I did not copy pasted any answer, and I think its ok to first post code to let author see and create explanation in meantime.
– Filip Młynarski
Nov 21 '18 at 13:15
If you think this answer is a good answer that was not already posted before, then keep it, but I stick with my downvote
– Adelin
Nov 21 '18 at 13:29
If author's code only problem would be about converting input to integer you would be right, but there's other bugs that I shown and explained.
– Filip Młynarski
Nov 21 '18 at 13:32
Thanks for answer, it worked.
– Jacob
Nov 22 '18 at 12:12
|
show 1 more comment
-1
You were breaking out of loop when user inserted incorrect value (not digit) instead of letting loop go on to ask him again, you should have break the loop when user typed correct value.
while True:
age = input('Enter age: ')
if age.isdigit():
print('Thank you, number entered is:', age)
break
else:
print('A number must be entered')
#print('Please enter age')
used_variable = int(age)
if used_variable > 18:
print('user is older than 18')
answered Nov 21 '18 at 13:12
Filip MłynarskiFilip Młynarski
1,5881311
1
so, for an 'int' that does a conversion, you copy pasted an answer, with no explanation whatsoever, an answer that was already given countless of times? Yeah, thats a -1
– Adelin
Nov 21 '18 at 13:14
I did not copy pasted any answer, and I think its ok to first post code to let author see and create explanation in meantime.
– Filip Młynarski
Nov 21 '18 at 13:15
If you think this answer is a good answer that was not already posted before, then keep it, but I stick with my downvote
– Adelin
Nov 21 '18 at 13:29
If author's code only problem would be about converting input to integer you would be right, but there's other bugs that I shown and explained.
– Filip Młynarski
Nov 21 '18 at 13:32
Thanks for answer, it worked.
– Jacob
Nov 22 '18 at 12:12
|
show 1 more comment
-1
-1
-1
You were breaking out of loop when user inserted incorrect value (not digit) instead of letting loop go on to ask him again, you should have break the loop when user typed correct value.
while True:
age = input('Enter age: ')
if age.isdigit():
print('Thank you, number entered is:', age)
break
else:
print('A number must be entered')
#print('Please enter age')
used_variable = int(age)
if used_variable > 18:
print('user is older than 18')
answered Nov 21 '18 at 13:12
Filip MłynarskiFilip Młynarski
1,5881311
You were breaking out of loop when user inserted incorrect value (not digit) instead of letting loop go on to ask him again, you should have break the loop when user typed correct value.
while True:
age = input('Enter age: ')
if age.isdigit():
print('Thank you, number entered is:', age)
break
else:
print('A number must be entered')
#print('Please enter age')
used_variable = int(age)
if used_variable > 18:
print('user is older than 18')
answered Nov 21 '18 at 13:12
Filip MłynarskiFilip Młynarski
1,5881311
edited Nov 21 '18 at 13:14
answered Nov 21 '18 at 13:12
Filip MłynarskiFilip Młynarski
1,5881311
answered Nov 21 '18 at 13:12
Filip MłynarskiFilip Młynarski
1,5881311
answered Nov 21 '18 at 13:12
Filip MłynarskiFilip Młynarski
1,5881311
1,5881311
1
so, for an 'int' that does a conversion, you copy pasted an answer, with no explanation whatsoever, an answer that was already given countless of times? Yeah, thats a -1
– Adelin
Nov 21 '18 at 13:14
I did not copy pasted any answer, and I think its ok to first post code to let author see and create explanation in meantime.
– Filip Młynarski
Nov 21 '18 at 13:15
If you think this answer is a good answer that was not already posted before, then keep it, but I stick with my downvote
– Adelin
Nov 21 '18 at 13:29
If author's code only problem would be about converting input to integer you would be right, but there's other bugs that I shown and explained.
– Filip Młynarski
Nov 21 '18 at 13:32
Thanks for answer, it worked.
– Jacob
Nov 22 '18 at 12:12
|
show 1 more comment
1
so, for an 'int' that does a conversion, you copy pasted an answer, with no explanation whatsoever, an answer that was already given countless of times? Yeah, thats a -1
– Adelin
Nov 21 '18 at 13:14
I did not copy pasted any answer, and I think its ok to first post code to let author see and create explanation in meantime.
– Filip Młynarski
Nov 21 '18 at 13:15
If you think this answer is a good answer that was not already posted before, then keep it, but I stick with my downvote
– Adelin
Nov 21 '18 at 13:29
If author's code only problem would be about converting input to integer you would be right, but there's other bugs that I shown and explained.
– Filip Młynarski
Nov 21 '18 at 13:32
Thanks for answer, it worked.
– Jacob
Nov 22 '18 at 12:12
1
1
so, for an 'int' that does a conversion, you copy pasted an answer, with no explanation whatsoever, an answer that was already given countless of times? Yeah, thats a -1
– Adelin
Nov 21 '18 at 13:14
so, for an 'int' that does a conversion, you copy pasted an answer, with no explanation whatsoever, an answer that was already given countless of times? Yeah, thats a -1
– Adelin
Nov 21 '18 at 13:14
I did not copy pasted any answer, and I think its ok to first post code to let author see and create explanation in meantime.
– Filip Młynarski
Nov 21 '18 at 13:15
I did not copy pasted any answer, and I think its ok to first post code to let author see and create explanation in meantime.
– Filip Młynarski
Nov 21 '18 at 13:15
If you think this answer is a good answer that was not already posted before, then keep it, but I stick with my downvote
– Adelin
Nov 21 '18 at 13:29
If you think this answer is a good answer that was not already posted before, then keep it, but I stick with my downvote
– Adelin
Nov 21 '18 at 13:29
If author's code only problem would be about converting input to integer you would be right, but there's other bugs that I shown and explained.
– Filip Młynarski
Nov 21 '18 at 13:32
If author's code only problem would be about converting input to integer you would be right, but there's other bugs that I shown and explained.
– Filip Młynarski
Nov 21 '18 at 13:32
Thanks for answer, it worked.
– Jacob
Nov 22 '18 at 12:12
Thanks for answer, it worked.
– Jacob
Nov 22 '18 at 12:12
|
show 1 more comment
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1
Once you know that it only consists of digits you still have to cast your string into an integer with the function int()
– leoburgy
Nov 21 '18 at 13:19
are you using python3 or python2 ? like @Adelin mentiond below, they have different output depending on the python version so
isdigit()might not be a good fit here. also– Avishay Cohen
Nov 21 '18 at 14:29