使用 VB.Net 和 SQL Server 上传图片

Question

 Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
    Dim command As New SqlCommand("insert into rent(Image,Status)values(@Image,@Status)", connection)
    Dim ms As New MemoryStream
    PictureBox1.Image.Save("ms", PictureBox1.Image.RawFormat)

    command.Parameters.Add("@Image", SqlDbType.VarChar).Value = ms.ToArray
    command.Parameters.Add("@Status", SqlDbType.VarChar).Value = TextBox5.Text

    connection.Open()

    If command.ExecuteNonQuery = 1 Then
        MessageBox.Show("Successfully uploaded")
    Else
        MessageBox.Show("Not uploaded")

    End If
    connection.Close()

End Sub

I'm trying to upload an image into my SQL Server using Visual Studio; everything is working except when I click the upload button, I keep getting the following error:

I tried every possible solution and no luck, I tried enabling the tcp and changing the ip even in SQL Server.

Answer 1

The error you get means that you can't connect to SQL Server. Make sure your connection string is correct, and you don't have a firewall blocking the connection between the computer that runs the code and the computer that hosts SQL Server.

However, once you sort the connection error, you still have a few problems with your code.

• change PictureBox1.Image.Save("ms", PictureBox1.Image.RawFormat) to PictureBox1.Image.Save(ms, PictureBox1.Image.RawFormat) to save the image into the memory stream.

• Change command.Parameters.Add("@Image", SqlDbType.VarChar).Value = ms.ToArray to command.Parameters.Add("@Image", SqlDbType.VarBinary).Value = ms.ToArray because memoryStream.ToArray returns a byte array, not a string.

• make sure the Image column in your table is, in fact, VarBinary.

• SqlCommand , SqlConnection and MemoryStream all implements the IDisposable interface, therefor you should use all of them as local variable inside the using statement. Your code suggest you are using a class level SqlConnecion instance. That should be changed.

• All communication with the database should be inside a try...catch block, since too many things you can't control can go wrong (network disconnected, for instance).

Your code should look more like this:

Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
    Dim RowsEffected as int = 0
    Using Dim connection As NewSqlConnection(ConnectionString
        Using Dim command As New SqlCommand("insert into rent(Image,Status)values(@Image,@Status)", connection)
            Using Dim ms As New MemoryStream
                PictureBox1.Image.Save(ms, PictureBox1.Image.RawFormat)

                command.Parameters.Add("@Image", SqlDbType.VarBinary).Value = ms.ToArray
                command.Parameters.Add("@Status", SqlDbType.VarChar).Value = TextBox5.Text
                Try
                connection.Open()
                RowsEffected = command.ExecuteNonQuery()
                End Try
                Catch Exception ex
                    MessageBox.Show("Failed to upload image:"& VbCrLf & ex.Message)
                End Catch           
            End Using 
        End Using                                         
    End Using

    If RowsEffected = 1 Then
        MessageBox.Show("Successfully uploaded")
    Else
        MessageBox.Show("Not uploaded")
    End If

End Sub