從 switch 語句返回文件位置

Question

我對編程很陌生，但我試圖讓我的方法返回一個值，該值包含依賴於用戶所需選擇的文件位置。 我已經為此擺弄了一天多，並且堅持如何正確返回值，我一直告訴我並非所有代碼路徑都返回值。 如何解決此問題並將代碼路徑返回到主

public static string fileLocation()
    {
        int fileRequest = 10;
        bool errorCheck = true;
        string filePath;

        while (errorCheck == true)
        {
            Console.Write(">Enter '1' through '9' to choose a hand.");
            Console.Write("Enter '0' for random.");
            fileRequest = Convert.ToInt16(Console.ReadLine());

            switch (fileRequest)
            {
                case 0:
                    Console.WriteLine(">Random selection loading.");
                    Random rnd = new Random();
                    fileRequest = rnd.Next(10);
                    errorCheck = true;
                    return (null);

                case 1:
                    Console.WriteLine(">Loading file one.");
                    filePath = Path.GetFullPath("Flush.txt");
                    errorCheck = false;
                    return (Convert.ToString(filePath));

                case 2:
                    Console.WriteLine(">Loading file two.");
                    filePath = Path.GetFullPath("FourKind.txt");
                    errorCheck = false;
                    return (Convert.ToString(filePath));

                case 3:
                    Console.WriteLine(">Loading file three.");
                    filePath = Path.GetFullPath("FullHouse.txt");
                    errorCheck = false;
                    return (Convert.ToString(filePath));

                case 4:
                    Console.WriteLine(">Loading file four.");
                    filePath = Path.GetFullPath("Pair.txt");
                    errorCheck = false;
                    return (Convert.ToString(filePath));

                case 5:
                    Console.WriteLine(">Loading file five.");
                    filePath = Path.GetFullPath("RoyalFlush.txt");
                    errorCheck = false;
                    return (Convert.ToString(filePath));

                case 6:
                    Console.WriteLine(">Loading file six.");
                    filePath = Path.GetFullPath("Straight.txt");
                    errorCheck = false;
                    return (Convert.ToString(filePath));

                case 7:
                    Console.WriteLine(">Loading file seven.");
                    filePath = Path.GetFullPath("StraightFlush.txt");
                    errorCheck = false;
                    return (Convert.ToString(filePath));

                case 8:
                    Console.WriteLine(">Loading file eight.");
                    filePath = Path.GetFullPath("ThreeKind.txt");
                    errorCheck = false;
                    return (Convert.ToString(filePath));

                case 9:
                    Console.WriteLine(">Loading file nine.");
                    filePath = Path.GetFullPath("TwoPair.txt");
                    errorCheck = false;
                    return (Convert.ToString(filePath));

                default:
                    Console.WriteLine(">Invalid request.");
                    filePath = "Invalid";
                    errorCheck = true;
                    return (null);
            }
        }

Answer 1

我假設您要做的是將 0 到 9 之間的整數作為輸入。 如果是0，你想把它隨機當作1到9。如果是別的，你想再次要求輸入。 這應該這樣做（未經測試）：

public static string FileLocation()
{
    while (true)
    {
        Console.Write(">Enter '1' through '9' to choose a hand.");
        Console.Write("Enter '0' for random.");
        int fileRequest = Convert.ToInt16(Console.ReadLine());
        if (fileRequest == 0)
            fileRequest = (new Random()).Next(1, 10);

        switch (fileRequest)
        {
            case 1:
                Console.WriteLine(">Loading file one.");
                return Path.GetFullPath("Flush.txt");

            case 2:
                Console.WriteLine(">Loading file two.");
                return Path.GetFullPath("FourKind.txt");

            case 3:
                Console.WriteLine(">Loading file three.");
                return Path.GetFullPath("FullHouse.txt");

            case 4:
                Console.WriteLine(">Loading file four.");
                return Path.GetFullPath("Pair.txt");

            case 5:
                Console.WriteLine(">Loading file five.");
                return Path.GetFullPath("RoyalFlush.txt");

            case 6:
                Console.WriteLine(">Loading file six.");
                return Path.GetFullPath("Straight.txt");

            case 7:
                Console.WriteLine(">Loading file seven.");
                return Path.GetFullPath("StraightFlush.txt");

            case 8:
                Console.WriteLine(">Loading file eight.");
                return Path.GetFullPath("ThreeKind.txt");

            case 9:
                Console.WriteLine(">Loading file nine.");
                return Path.GetFullPath("TwoPair.txt");

            default:
                Console.WriteLine("Invalid request.");
                break;
        }
    }
}

Answer 2

好吧，您遇到了編譯器無法理解的情況。

您正在 while 循環中檢查 errorCheck 並且內部有一個案例。 這種情況總是會返回，但是當編譯器看到 errorCheck 在沒有返回的情況下為真的可能性時，它會抱怨執行路徑不存在返回的可能性。

首先，您在所有情況下都在執行 return，因此可以安全地刪除 while，因為它什么都不做。 否則，如果您計划始終返回並始終循環，直到用戶選擇正確的選項忽略ERRORCHECK，只需執行一段時間（true）{...}，因為在選擇正確的選項時，您的交換機將返回。

Answer 3

一個用 VB 編寫的示例，但它應該顯示邏輯。 我的變量來自在 OnLoad 事件中填充的數據庫變量，但除此之外，它幾乎就是您要執行的操作。

Private Sub ShowPDF()

    Dim mySelection As Integer = _myDb.MyHuntingArea
    'the path where the pdf files are located
    Dim filePath As String = MyMachine.AssemblyDirectory & "\Regulations\"
    'variable for the appropriate file name to get
    Dim fileName As String = ""
    Select Case mySelection

        Case 1 
            fileName = filePath & "Area1.pdf"
        Case 2 
            fileName = filePath & "Area2.pdf"
        Case 3  
            fileName = filePath & "Area3.pdf"
        Case Else
            MessageBox.Show("We cannot determine what area you are requesting regulations for.  Make sure it is set under Setup.", Application.ProductName, MessageBoxButtons.OK, MessageBoxIcon.Error)
    End Select

    If Not fileName = "" Then
        Try
            System.Diagnostics.Process.Start(fileName)
        Catch ex As Exception
            MsgBox(ex.Message)
        End Try
    End If
End Sub