Delete and Query function- Python

The consultarLivro function resolves as follows:

by adding range(len(livro)) .

This resolves why the list index only works with numbers and not with strings that is what you were doing.

Example:

livro[0] , which would return the first element of the list, and then compare it.

What you were doing: livro['Titulo'] , which would be functional if it were a dictionary .

Another change I made was with [0] ahead of livro[i] , which is why you made all livro content lists, see:

[[299], ['t'], ['a'], ['e'], [12], [2017], [1.99]] # Retirado do console

Then it runs without error, but you just asked to reprint the book [i], basically asking to print the ID again (?).

getting:

def consultarLivro():
    id = int(input('Digite o ID:'))
    for i in range(len(livro)):
        if id == livro[i]:
            print(livro[i])

and the output:

>>> [299]

The excluirLivro() function was basically the same thing. I added for and then it deleted, but deleted only the ID, which is wrong, but now it deletes something.

def excluirLivro():
    print('Informe o ID do livro que você deseja excluir')
    idLivro1 = int(input())
    for i in range(len(livro)):
        if idLivro1 == livro[i][0]:
            del(livro[i])
            break # Quando deletar, ele para o For, senão vai causar um erro.

livro before : [[299], ['t'], ['a'], ['e'], [12], [2017], [1.99]]

livro then : [['t'], ['a'], ['e'], [12], [2017], [1.99]] # Deletando apenas o ID

Well, what I did above was just removing the errors from your code, but it still remains wrong in a way, as it does not print the book entirely, well, there it is with you.

If you want a hint, create a dictionary, and in each dictionary value you will have an ID of each book, and within each ID you will have its values (Title, etc ...), example:

livros = {"299":["Titulo","Autor","Editora",2,2017,1.99],
"212":["Titulo2","Autor2","Editora2",2,2017,1.99]
}

I hope I've helped make your code clearer.

Pay attention to the scope of the variables in your program.

I have seen several lists that are created changed but die after the function call because no function of yours has return value.

I'll give you a simple example to understand scope and then you can try to look for the error or errors in your code and tell me.

x = 5
def foo():
    x = 10
foo()
print(x) # valor printado é 5 pois a variável x de dentro da função só existe la dentro, não é possível acessá-la no programa principal.

Another example:

def foo1()
    var1 = 15
foo1()
print(var1) # Erro de sintaxe, variável var1 não existe.

If you want the variable to exist outside of a function, it needs to RETURN one or more Ex values:

def foo2()
    var2 = 13
    return var2
var2 = foo2() # faz a variável var2 receber o valor de retorno da função e existir fora do escopo de foo2. (O nome dessa variável não importa, poderia ser var0 = foo2() que daria na mesma)

To conclude. You can return multiple values in the same function by separating each variable (expression to be more exact) with a comma.

Example:

def insereLivro():
    lista_livro = ['l1', 'l2', 'l3']
    lista_precos = [10, 20, 500]
    lista_editora = ['e1', 'e2', 'e3']
    return lista_livro, lista_precos, lista_editora
lista_livro, lista_precos, lista_editora = insereLivro()  # Novamente, os nomes das variáveis nesse linha podem ser diferentes dos nomes da função retorno, x1, x2, x3 = insereLivro() iria dar na mesma.

EDIT: One more thing when working with lists, which are changeable data structures, if any list is passed by parameter to some function, the passage is done by reference, ie if you pass a list by parameter for a function and within that function make changes to that list, those changes will be reflected in the global scope, even if the list is not returned using the return command.

def appendlista(lista)
    lista.append(10)
lista = [1,2,3]
print(appendlista(lista)) # Resulta em [1,2,3,10]