即使条件为真,也会跳过wpf

问题描述 投票:0回答:3

一个简单的IF函数编写用于测试,条件显然是正确的,但它每次都跳到ELSE。有人能告诉我,我做错了什么吗?

namespace SANK
{
/// <summary>
/// Interaction logic for SignIN.xaml
/// </summary>
public partial class SignIn : Window
{
    string username;
    string password;
    string user = "user";
    string pass = "pass";
    public SignIn()
    {
        InitializeComponent();
    }

    private void LogInButton_Click(object sender, RoutedEventArgs e)
    {
        password = PassBox.ToString();
        username = UserBox.ToString();
        bool jednako = String.Equals(password, pass, StringComparison.OrdinalIgnoreCase);
        bool jednako2 = String.Equals(username, user, StringComparison.OrdinalIgnoreCase);
        if (jednako&&jednako2)
        {
            SignedIn.Visibility=Visibility.Visible;
            Application.Current.MainWindow.Show();
            this.Hide();
        }
        else
        {
            Wrong.Visibility = Visibility.Visible;
        }

    }
}
c# wpf
3个回答
5
投票

你在用TextBoxes吗?如果是这样,请将其更改为使用.Text属性:

password = PassBox.Text;
username = UserBox.Text;

1
投票

你不应该在类的密码和用户名的输出上调用toString方法

password = PassBox.Text;
username = UserBox.Text;

是“System.Windows.Controls.TextBox”这没有任何意义!要访问textbox的值,您需要调用TextBox.Text属性。

快乐的编码


0
投票

您仍然可以使用PaswordBox,但要从Passwordbox获取文本,您必须使用此:

password = PassBox.Password;

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