/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

4 votes269 views3 answers
Dice simulator python
Python
4 votes281 views1 answer
Pandas find inf values
Python
12 votes258 views4 answers
Age calculator in python
Python
6 votes253 views2 answers
Unique list values python ordered
Python
9 votes237 views8 answers
Python get first element from list
Python
2 votes281 views2 answers
Does python short circuit
Python
0 vote306 views1 answer
Sqlalchemy one to one foreign key
Python
3 votes278 views1 answer
Json not readable python
Python
16 votes334 views6 answers
Django forms
Python Django
0 vote282 views1 answer
Importing time and sleep. python
Python