/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

2 votes121 views1 answer
Model pickle file create
Python
1 vote125 views1 answer
Sample based on column pandas
Python
3 votes124 views2 answers
Run python notepad++
Python
0 vote135 views2 answers
Youtube net worth
Python
0 vote120 views2 answers
Is vowel python
Python
0 vote126 views1 answer
Compare input with list python
Python
0 vote118 views1 answer
Who is bayceee roblox id
Python
0 vote80 views1 answer
Unhapppy man with monwy
Python
0 vote122 views1 answer
How to get left click input in pygame
Python
0 vote152 views1 answer
How to make a config txt file on python
Python