/Python/Valueerror: invalid literal for int() with base 10: 'id'

Valueerror: invalid literal for int() with base 10: 'id'

ValueError: invalid literal for int() with base 10: site:stackoverflow.com

# Just for the record:
print(int('55063.000000'))

#Traceback (most recent call last):
  #File "<stdin>", line 1, in <module>
#ValueError: invalid literal for int() with base 10: '55063.000000'

# use this
print(int(float('55063.000000')))

Valueerror: invalid literal for int() with base 10: 'id'

#another way is to check if the entered number isdigit()
#that way you will ensure unhandled exception incase of erroneus data

number= input("What is your weight:")
if number.isdigit():
    kilos=int(float(number))
    print ("The weight of the person is:" + str(kilos))
else:
    print("Error - Please enter a proper weight")

Source: itsmycode.com

Similar codes

7 votes278 views6 answers
Hello world in python
Python
7 votes273 views1 answer
Pip pickle
Python
21 votes300 views4 answers
Flask quickstart
Python
0 vote297 views1 answer
How to access a file from root folder in python project
Python
-1 vote277 views2 answers
If __name__=='__main__':
Python
3 votes236 views1 answer
Null coalescing assignment operators python
Python
-1 vote292 views2 answers
Turn off grid in matplotlib 3d
Python
0 vote287 views1 answer
How to return only fractional part in python
Python
4 votes300 views2 answers
Use datetime python to get runtime
Python
2 votes284 views1 answer
Tkinter info box
Python