python, using if loop to make sure user input is a number then manipulate number

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')

edited Nov 21 '18 at 15:00

Avishay Cohen

701724

asked Nov 21 '18 at 13:10

Jacob

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

add a comment |

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')

edited Nov 21 '18 at 15:00

Avishay Cohen

701724

asked Nov 21 '18 at 13:10

Jacob

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

add a comment |

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')

edited Nov 21 '18 at 15:00

Avishay Cohen

701724

asked Nov 21 '18 at 13:10

Jacob

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

edited Nov 21 '18 at 15:00

Avishay Cohen

701724

asked Nov 21 '18 at 13:10

Jacob

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

asked Nov 21 '18 at 13:10

Jacob

asked Nov 21 '18 at 13:10

Jacob

asked Nov 21 '18 at 13:10

Jacob

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

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 so isdigit() might not be a good fit here. also

– Avishay Cohen
Nov 21 '18 at 14:29

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

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')

edited Nov 21 '18 at 13:14

answered Nov 21 '18 at 13:12

Filip 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

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53412796%2fpython-using-if-loop-to-make-sure-user-input-is-a-number-then-manipulate-number%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

1 Answer
1

active

oldest

votes

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')

edited Nov 21 '18 at 13:14

answered Nov 21 '18 at 13:12

Filip 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')

edited Nov 21 '18 at 13:14

answered Nov 21 '18 at 13:12

Filip 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')

edited Nov 21 '18 at 13:14

answered Nov 21 '18 at 13:12

Filip 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')

edited Nov 21 '18 at 13:14

answered Nov 21 '18 at 13:12

Filip Młynarski

1,5881311

edited Nov 21 '18 at 13:14

answered Nov 21 '18 at 13:12

Filip Młynarski

1,5881311

answered Nov 21 '18 at 13:12

Filip Młynarski

1,5881311

answered Nov 21 '18 at 13:12

Filip 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

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

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

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Nsryjdtyk