/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 vote120 views1 answer
Python get stock prices
Python
0 vote127 views1 answer
Python get words between two words
Python
0 vote159 views1 answer
Scrape phone numbers from website python
Python
3 votes138 views2 answers
Import file from parent directory python
Python
1 vote127 views4 answers
Pandas list comprehension
Python
0 vote94 views1 answer
Headless chrome python
Python
0 vote131 views1 answer
Sklearn import tree
Python
60 votes144 views6 answers
List files in directory python
Python
39 votes116 views6 answers
Python check if int
Python
9 votes129 views2 answers
How to read video in opencv python
Python