/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

1 vote327 views1 answer
Pygame doesnt dedect collision between sprite and image
Python
14 votes311 views3 answers
Exit python terminal
Python
1 vote321 views1 answer
Creating an interface tkinter
Python
0 vote295 views1 answer
Serversocket.accept() python
Python
3 votes311 views1 answer
Python initialize multidimensional list
Python
11 votes313 views6 answers
Pandas distinct
Python
0 vote330 views1 answer
Python add hyphen to string
Python
0 vote310 views1 answer
Example of Python Inline comments
Python
0 vote320 views1 answer
Pandas Timedelta to postgres
Python
0 vote317 views1 answer
Django queryset to form
Python