VB.net和Mysql使用两张表登录问题

Question

How can i login using doctor level in his table.Because everytime i log the receptionist form is always popping even when i log using the user and pass of the doctor maybe my syntax is wrong please help me PS: Doctors Forms and Receptionists Forms are different.

This is my code:

Try
    mycon.Open()
    Dim query1 As String
    Dim query2 As String
    Dim name As String
    query1 = "SELECT * FROM receptionisttbl"
    command = New MySqlCommand(query1, mycon)
    reader = command.ExecuteReader

    With reader
        .Read()

        If .Item("level") = "1" Then
            With reader
                .Read()
                name = .Item("name")
            End With
            MsgBox("Welcome " + name)
            Form1.Show()

            command.Dispose()
            reader.Close()
            mycon.Close()
        Else
            mycon.Open()
            query2 = "SELECT * FROM doctortbl"
            command = New MySqlCommand(query2, mycon)
            reader = command.ExecuteReader

            With reader
                .Read()

                If .Item("level") = "2" Then
                    With reader
                        .Read()
                        name = .Item("name")
                    End With
                    MsgBox("Welcome " + name)
                    Form2.Show()

                    command.Dispose()
                    reader.Close()
                    mycon.Close()

                End If
            End With

        End If
    End With

Catch ex As Exception
    MsgBox(ex.ToString)
Finally
    mycon.Dispose()
End Try

Answer 1

First, let's consider your requirements.

1. Provide a login form to validate users.

2. If the user is valid (userName and matching password are found in the database) then display a Form based on the level of the user.

Considering the requirements look at the design of the table in the database. Let's call it MDUsers. Now to the columns.

UserName as an VarChar size 100 and Primary key of the table
UserPassword as VarChar size 100
Level an Integer

Now you have fields for all the data you need for your requirements.

Add some sample data to your table.

Next think about what you need from the user to accomplish your requirements. Design your user interface accordingly. Your login form will have 2 text boxes (userName and password) and a button (Login).

Then you need to think about the code for the Login button. You need to connect to your database and check if there is a match to userName and password. Once you find a match you only need to retrieve a single piece of data to complete your requirements, the level.

Now how can we ask the database to do all that?

Select level From MDUsers Where userName = @userName And UserPassword = @pword;

Since userName is the Primary Key of the table we know we will find no more than one record with that userName. If the password matches too then we have a valid user and we will return the single piece of data, level for that record.

Using...End Using blocks ensure that your database objects are closed and disposed even if there is an error. You can pass your connection string directly to the constructor of the connection and pass the query string and connection to the constructor of the command.

Always use parameters.

Since we only need a single piece of data we can use .ExecuteScalar

Private Sub OpCode()
    Dim retVal As Object
    Using cn As New MySqlConnection("Your connection string"),
            cmd As New MySqlCommand("Select level From MDUsers Where userName = @userName And UserPassword = @pword;", cn)
        cmd.Parameters.Add("@userName", MySqlDbType.VarChar, 100).Value = txtName.Text
        cmd.Parameters.Add("@pword", MySqlDbType.VarChar, 100).Value = txtPassword.Text
        cn.Open()
        retVal = cmd.ExecuteScalar
    End Using
    If retVal Is Nothing Then
        MessageBox.Show("Login failed")
        Return
    End If
    If CInt(retVal) = 1 Then
        Form1.Show()
    ElseIf CInt(retVal) = 2 Then
        Form2.Show()
    Else
        MessageBox.Show("Level not recognized")
    End If
End Sub

You will have to check MySql rules for column names and table names being aware of reserved words.