/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

0 vote169 views1 answer
Open file in python network url
Python
2 votes174 views1 answer
Remove comma from string python column
Python
11 votes166 views3 answers
How to convert small letters to capital letters in python
Python
89 votes134 views6 answers
Int to string python
Python
1 vote217 views1 answer
Hot reloading flask
Python Flask
13 votes174 views1 answer
Configure funCtion in tkinter
Python
1 vote168 views1 answer
Convert a Video in python to individual Frames
Python
5 votes193 views4 answers
Create python list
Python
1 vote191 views1 answer
Selenium refresh till the element appears python
Python
46 votes167 views6 answers
Dataframe to CSV
Python