上一节我们看到在程序运行的时候出现了错误，为什么读取二进制数据会出现错误呢？@H_403_1@

我重新改写了一下代码，测试读取的二进制数据：@H_403_1@

    Private Sub Button4_Click(sender As Object,e As EventArgs) Handles Button4.Click
        Dim odc As New OleDbConnection()
        odc.ConnectionString = "Provider = Microsoft.ACE.OLEDB.12.0;data source=D:\save\博客教程\08 数据库\Northwind.mdb"

        Dim sql As String = "select 图片 from 类别 where 类别ID=3"
        Dim imageCmd As New OleDbCommand(sql,odc)


        Dim datareader As OleDbDataReader

        odc.Open()

        Try
            datareader = imageCmd.ExecuteReader(CommandBehavior.SequentialAccess And CommandBehavior.SingleResult)
            Dim readerData() As Byte

            datareader.Read()
            readerData = datareader(0)

            ''读取图片二进制数据
            Dim sw As New StreamWriter("d:\ddd.txt",False)
            Dim hexData As String

            For i As Integer = 0 To readerData.Length - 1
                If (i + 1) Mod 8 = 0 Then
                    hexData = Hex(readerData(i))
                    sw.Write(IIf(hexData.Length = 1,"0" & hexData,hexData))
                    sw.Write(ControlChars.CrLf)
                Else
                    hexData = Hex(readerData(i))
                    sw.Write(IIf(hexData.Length = 1,hexData) & " ")
                End If

            Next
            sw.Close()

        Catch ex As Exception
            MessageBox.Show(ex.Message)
        Finally
            datareader.Close()
            odc.Close()
        End Try
    End Sub

根据BMP文件的格式可知，BMP前两个字节是BMP文件头，为"BM"，对应的十六进制值为 42 4D@H_403_1@

可是我们获得的数据前两个字节并不是424D，但是经过仔细查看会发现数据中间有42 4D 出现在第79字节处：@H_403_1@

@H_403_1@

由于.net平台下C#和vb.NET很相似，本文也可以为C#爱好者提供参考。@H_403_1@

学习更多vb.net知识，请参看vb.net 教程 目录@H_403_1@ 原文链接：https://www.f2er.com/vb/256245.html