简体   繁体   中英

C# While TryParse

char typeClient = ' ';
bool clientValide = false;
while (!clientValide)
{
     Console.WriteLine("\nEntrez le type d'employé (c ou g) : ");
     clientValide = char.TryParse(Console.ReadLine(), out typeClient);
     if (clientValide)
         typeClient = 'c';
}

I'd like to make it so it doesn't exit the while unless the char is 'g' or 'c' help ! :)

string input;
do {
    Console.WriteLine("Entrez le type d'employé (c ou g):");
    input = Console.ReadLine();
} while (input != "c" && input != "g");

char typeClient = input[0];

Is you use Console.ReadLine , the user has to press Enter after pressing c or g . Use ReadKey instead so that the response is instantaneous:

bool valid = false;
while (!valid)
{
    Console.WriteLine("\nEntrez le type d'employé (c ou g) : ");
    var key = Console.ReadKey();
    switch (char.ToLower(key.KeyChar))
    {
        case 'c':
            // your processing
            valid = true;
            break;
        case 'g':
            // your processing
            valid = true;
            break;
        default:
            Console.WriteLine("Invalid. Please try again.");
            break;
    }
}

You're really close, I think something like this will work well for you:

char typeClient = ' ';
while (typeClient != 'c' && typeClient != 'g')
{
    Console.WriteLine("\nEntrez le type d'employé (c ou g) : ");
    var line = Console.ReadLine();
    if (!string.IsNullOrEmpty(line)) { typeClient = line[0]; }
    else { typeClient = ' '; }
}

basically it reads the input into the typeClient variable when the user enters something so the loop will continue until they enter g or c .

You can use ConsoleKeyInfo with Console.ReadKey() :

 ConsoleKeyInfo keyInfo;
 do {
   Console.WriteLine("\nEntrez le type d'employé (c ou g) : ");
   keyInfo = Console.ReadKey();
 } while (keyInfo.Key != ConsoleKey.C && keyInfo.Key != ConsoleKey.G);
   

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM