如果（Integer Console.Readline“”或null）不起作用

Question

I can't figure out after searching and searching (you know how it goes) why this code doesn't work.

I just want it to work like this

if (Nummer == "") {
    Console.WriteLine("0");
}

That's it, and it doesn't work. I have been searching for an hour and a half. Can't understand why there is a simple basic explanation. I only found how to fix it with a string or something, then I tried to convert it and it still didn't work. Can someone help me please?

I appreciate your patience for my limited knowledge. Thanks for your time

static void Main(string[] args)
{
    bool herhaal = true;

    do
    {                
        Console.Write("Geef een getal : ");
        int Nummer = Convert.ToInt16(Console.ReadLine());           

        if (Console.ReadLine() == "" && Console.ReadLine() == null)
        {
            Console.WriteLine("0");
        }

        double kw = Math.Pow(Nummer, 2);

        Console.WriteLine("Kwadraat van {0} is: {1}", Nummer, kw + Environment.NewLine);
    } while (herhaal);
}

Answer 1

static void Main(string[] args)
{
    int Nummer;
    bool herhaal = true;
    do
    {                
        Console.Write("Geef een getal : ");          
        //only read from the Console ONCE per loop iteration, and always read to a string first
        string input = Console.ReadLine(); 

        //TryParse better than Convert for converting strings to integers
        if (!int.TryParse(input, out Nummer))        
        {
            Console.WriteLine("0");
        }
        else  //only do the second part if the conversion worked
        {
            double kw = Math.Pow(Nummer, 2);
            Console.WriteLine("Kwadraat van {0} is: {1}\n", Nummer, kw);
        }

    } while (herhaal);
}

To do this from a WinForms app, as attempted in the comments:

private void button1_Click(object sender, EventArgs e)
{
    double aantalgroep;
    if (!double.TryParse(textBox1.Text, out aantalgroep))        
    {
        textBox1.Text = "0";
    }
    else 
    {
        double kw = Math.Pow(aantalgroep, 2);
        textBox1.Text = String.Format("Kwadraat van {0} is: {1}", aantalgroep, kw);
    }
}

Answer 2

The Console.ReadLine() method 'reads' one line of user input, and stores in whatever the variable you assign it to. Here's more info .

So, your following line of code reads the number you inputs and stores in Nummer .

int Nummer = Convert.ToInt16(Console.ReadLine());           

But then you go on and do two more Console.ReadLine() method calls in your if statement, so what the if statement really does is try to read from the console two more times, and see if the first read is "" and if the second read is null , which is not your desired behavior.

What you want to do is read once, and compare whatever you read. Going by your code it looks like you want to output the square of the number entered by the user, so you probably should check for more than just Nummer == "" , since if the user entered an alphabet character, that would also result in an error. So using int.TryPrase() is a better option for you.

static void Main(string[] args)
{
    bool herhaal = true;
    do
    {
        Console.Write("Geef een getal : ");
        string Nummer = Console.ReadLine();

        if (int.TryParse(Nummer, out int result))
        {
            double kw = Math.Pow(result, 2);
            Console.WriteLine("Kwadraat van {0} is: {1}", Nummer, kw + Environment.NewLine);
            herhaal = false;
        }
        else
        {
            Console.WriteLine("0");
        }

    } while (herhaal);

    Console.ReadLine();
}

Answer 3

According to the C# Documentation , Console.ReadLine does the following

Reads the next line of characters from the standard input stream.

This means that every time you call Console.ReadLine , a new line will be read from the console. From what I can see in your code, this isn't the behavior that you want. To fix your problem, you should store the result of Console.ReadLine in a variable and use that variable instead of the ReadLine method.