Create . txt file in PYTHON if it does not exist

Question

The code:

        arquivo = open(input('Nome do arquivo a ser editado:'), 'r')
        texto = arquivo.readlines()
        texto.append(input('Insira o texto:'))
        arquivo = open(input('Nome do arquivo a ser editado:'), 'w')
        arquivo.writelines(texto)
        arquivo.close()

It is a file . Existing TXT records what is written inside it. And add the text that is inserted, I found this way of doing that requires it to ask 2x the name of the file. Is there any way you don’t need to call 2x the name request ?

        File "C:\Users\Americo\Desktop\TESTE\Trabson.py", line 51, in sys_calls
        arquivo = open(input('Nome do arquivo a ser editado:'), 'r')
        FileNotFoundError: [Errno 2] No such file or directory: 'teste.txt'

And also if the file does not exist, as I do to create and put the text I wish inside warning that the file was created and that it did not exist. Another user had put in another doubt the following code snippet:

      if os.path.isfile(diretorio):
      ...
      ficheiro = open(diretorio, "w") # criamos/abrimos o ficheiro
      #.... .... ... operacoes no ficheiro
      ficheiro.close()

This is applied in this case for the confirmation of the txt file ? The excerpt worked for part of my code served for the whole part of MOVE CREATE DELETE AND RENAME some file or directory.

Answer 1

If you do that, you don’t have to call twice.

arquivo = open(input('Nome do arquivo a ser editado:'), 'r+')
texto = arquivo.readlines()
texto.append(input('insira o valor'))
arquivo.writelines(texto)
arquivo.close()

Now, if the reported file does not exist it can be so.

try:
    nome_arquivo = input('Nome do arquivo a ser editado:')
    arquivo = open(nome_arquivo, 'r+')
except FileNotFoundError:
    arquivo = open(nome_arquivo, 'w+')
    arquivo.writelines(u'Arquivo criado pois nao existia')
#faca o que quiser
arquivo.close()

Answer 2

Maybe you can use the "append" of files to resolve this issue. But, making the least change in your code, Voce could put the input in a variable, IE:

    nome_do_arquivo = input('Nome do arquivo a ser editado:')
    arquivo = open(nome_do_arquivo, 'r')
    texto = arquivo.readlines()
    texto.append(input('Insira o texto:'))
    arquivo = open(nome_do_arquivo, 'w')
    arquivo.writelines(texto)
    arquivo.close()

Or you can make the following code:

    arquivo = open(input('Nome do arquivo a ser editado:'), 'r+')
    print(arquivo.readlines()) #Consegue ler
    arquivo.write(input('Insira o texto:')+"\n") #consegue editar
    arquivo.close()

It is important to remember that the file must exist, otherwise check before and create.