VB.Net not able to connect to SQL - sql

Question

Private Sub Button5_Click_1(sender As Object, e As EventArgs) Handles Button5.Click
Dim connetionString As String
Dim cnn As SqlConnection
connetionString = "Server=localhost;Initial Catalog=acernis;User ID=root;Password=password"
cnn = New SqlConnection(connetionString)
Try
cnn.Open()
MsgBox("Connection Open ! ")
cnn.Close()
Catch ex As Exception
MsgBox("Can not open connection ! ")
End Try
End Sub
Currently in my code I have the code above the SQL username and pass are correct as is the DB. I have WAMP running and am using Navicat to make sure I can connect. What is wrong?

Answer 1

if you are using MS SQL server then your issue is with the connection string. try this one :
connetionString = "Server=localhost;Database=acernis;User ID=root;Password=password"
however if you mean MySQL then you need to change more than just the connectio string
try this:
Private Sub Button5_Click_1(sender As Object, e As EventArgs) Handles Button5.Click
Dim connetionString As String
connetionString = "Server=localhost;Database=acernis;Uid=root;Pwd=password;"
Dim conn As New MySqlConnection(connetionString)
Try
conn.Open()
MsgBox("Connection Open ! ")
Catch ex As MySqlException
MsgBox("Can not open connection Error: " & ex.ToString())
Finally
conn.Close()
End Try
End Sub
to find out what is the connection string that best suites you refer to the website
http://www.connectionstrings.com/

Answer 2

Private Sub Button5_Click_1(sender As Object, e As EventArgs) Handles Button5.Click
Dim connetionString As String
Dim cnn As SqlConnection
connetionString = "Server=localhost\<InstanceName>;Initial Catalog=acernis;User ID=root;Password=password"
cnn = New SqlConnection(connetionString)
Try
if not cnn.State=ConnectionState.Open then cnn.Open()
MsgBox("Connection Open ! ")
cnn.Close()
Catch ex As Exception
MsgBox("Can not open connection ! ")
End Try
End Sub
As per observation, if you are using SQL Server, you need to specify the <Instance Name> in Server property.
Plus, do not use class Exception in catch section, Instead, use specific exceptoin class; in this case, use SQLException