从你写的code，似乎问题是，你比较字符串重的presentation 员工及其雇员属性 Employee.EmployeeID.ToString（）== Employee.ToLower（）。这条线将随时返回假除非覆盖的ToString（） 员工的方法类返回的财产雇员（我presume你没有）。 试试这个，而不是（假设参数用户包含了用户的姓名）：

 使用（var数据=新ExampleDataSet（））
{
    VAR的loggedIn = dataSet.Employee.Any（E =＆GT; e.UserName ==用户放大器;＆安培; e.Password ==通行证）;
    VAR消息=的loggedIn？ 您登录：您还没有登录;
    的MessageBox.show（消息）;
    返回的loggedIn;
}
 

I wrote the following code to log in to my application using C# and LINQ. It connected to a SQL service oriented database that I have created in Visual Studio. The problem that I am having is one that I do not understand and am hoping that someone can help me about here. I have created two message boxed to try to see the output of my code buy I am not getting anything from it.

If anyone could help that would be great!

public bool UserLogin(string User, string Pass)
{
    var Database = new ExampleDataSet();

    var query = from Employee in Database.Employee
        where (Employee.EmployeeID.ToString() == Employee.ToLower() && Employee.Password == Pass)
        select Employee;

    if (query.Count() != 0)
    {
        return true;
        MessageBox.Show("You are logged in");
    }

    return false;
    MessageBox.Show("You are not logged in");
}

private void cmdLogin_Click(object sender, EventArgs e)
{
    string User = (txtUser.Text);
    string Pass = (txtPass.Text);
    UserLogin(User, Pass);
}

From the code you wrote, it seems that the problem is that you compare the string representation of an Employee with its EmployeeId property Employee.EmployeeID.ToString() == Employee.ToLower(). This line will always return false unless you override ToString() method of Employee class to return the property EmployeeId (which I presume you didn't). Try this instead (assuming that parameter User contains the name of the user):

using(var dataSet = new ExampleDataSet())
{
    var loggedIn = dataSet.Employee.Any(e=>e.UserName == User && e.Password == Pass);
    var message = loggedIn ? "You are logged in" : "You are not logged in";
    MessageBox.Show(message);
    return loggedIn;
}