Visual Basic

Christian Graus wrote:

"I am working on a project that will convert a FORTRAN code to corresponding C++ code.I am not aware of FORTRAN syntax" ( spotted in the C++/CLI forum )

Hakmeh Mohannad wrote:

and right now i dont care about SQL injection all what i want is run this form (Step by Step)

Hakmeh Mohannad wrote:

Hello i am new with vb 2005 and the programming world

Christian Graus wrote:

Hakmeh Mohannad wrote:
and right now i dont care about SQL injection all what i want is run this form (Step by Step)

So no-one is ever going to use this form ?

Hakmeh Mohannad wrote:

right now i dont care about SQL injection all what i want is run this form (Step by Step)

Public Class SqlHelpers
    Public Shared Function GetSqlConnection() As SqlConnection
        Dim connStr As String = GetSqlConnectionString()   ' You have to define this method!
        Dim newConn As New SqlConnection(connStr)
        Return newConn
    End Function
 
    Public Shared Function GetSqlConnectionString() As String
        ' You would normally get the connection string from the app.config file, or some other
        ' source.  With a few changes, this could also support multiple connection strings for
        ' multiple databases and database types quite easily.
        '
        ' In this example, we'll just return the string you have.
        Return "Data Source=HAKMEH;Initial Catalog=HMS;Integrated Security=True;"
    End Function

SELECT COUNT(UserId) FROM tblEmployees WHERE EmployeeCode=@EmployeeCode AND EmployeePassword=@Password

Public Function Login(ByVal EmployeeCode As String, ByVal   EmployeePassword As String) As Boolean
    Using conn As SqlConnection = GetSqlConnection()
        Dim SelectQuery As String = "SELECT COUNT(UserId) FROM tblEmployees WHERE EmployeeCode=@EmployeeCode AND EmployeePassword=@Password"
        Dim comm As New SqlCommand(SelectQuery, conn)
        comm.Parameters.Add(New SqlParameter("@EmployeeCode", SqlDbType.NVarChar, 25)).Value = EmployeeCode
        comm.Parameters.Add(New SqlParameter("@EmployeePassword", SqlDbType.NVarChar, 25)).Value = EmployeePassword
        Dim result As Integer
        conn.Open()
        result = comm.ExecuteScalar
        If result > 1 Then
            ' Something weird happened.  We had more than one user with the same ID in the database!
            ' Yikes!  Someone is going to get shot over this one!  :(
            ' Handle this error however you want.
        End If
        
        ' If the result came back 1, the login was successfull, otherwise it should be 0.
        ' So, If result = 1, return True to the caller so it knows the login was successfull.
        Return (result = 1)
    End Using
End Function