Why does my "Try" in Python not work?

Question

x = int(input("Por favor, insira um número: "))
try:
    print(x != int)
except ValueError:
    print("Somente números.")

The mistake that gives is:

ValueError: invalid literal for int() with base 10: 'x'

I don’t know what to do so that Python gives me an error message when the user gives something other than number as information.

Answer 1

You ask for the die and try to do the conversion (the function int() does this), and if the conversion is not possible (because the typed text is unsatisfactory to create an integer) an exception will be thrown (unfortunately the documentation is flawed and says nothing about it). If you want to do something when this occurs then you should capture the exception as you did in except, but capture will only occur in codes within the block try, which was not the case, so when the exception is thrown there is no capture and the application breaks. So do what you want:

try:
    print(int(input("Por favor, insira um número: ")))
except ValueError:
    print("Somente números.")

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

And to be clear, Python is giving exactly one error message when the user types something other than a number. What you want is just the opposite, you want your code to give a message when it happens and not Python.

Answer 2

The try serves exactly for this, to launch an exception of yours without necessarily stopping the program, however, the way you are trying to use this incorrect, the line that throws the exception must be within the try

In your case you start trying to assign a value to x using the input, up ai beauty, but you want to convert it to a whole number using the int In case if typed something it is not possible to convert to a whole number it throws the exception

To fix this problem you can use the try, but remembering, it has to be inside the same, for example

try:
    x = int(input("Por favor, insira um número: "))
except ValueError:
    print("Somente números.")

From what I understand you wanted to put a condition for the error, however, the try does not use condition and yes when an error is released

Bonus

If you want to throw an error using condition use the assert

An example using it with your code, remembering that it is an example with your code, but the best method is the top

x = input("Por favor, insira um número: ")
try:
    assert x.isnumeric() and x.isascii()  # O método "isascii()" funciona somente para as versões mais recentes do Python, porém, o uso dela seria para restringir a quantidade de caracteres para o determinado exemplo, para versões desatualizadas, pode fazer o uso de "ord()" no lugar dos dois.
    # assert all([ord(d) in range(48, 58) for d in x])  # Para versões anteriores há essa opção onde irá restringir os números de 0 a 9
except:
    print("Somente números.")
else:
    x = int(x)

Answer 3

Vincenslau if you want to save the input value in the variable x do as follows:

while True:
    x = input("Por favor, insira um número: ")
    try:
        x = int(x)
        break
    except ValueError:
        print("Somente números.")

and Happy New Year...