/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

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