从打开文件对话框路径中选择 [英] select from open file dialog path

问题描述

我有这个select语句来选择要插入数据库的文件。

i have this select statement to select the file to insert into database.

Dim query As String = "INSERT INTO  test (School, Campus, AdminNo, ModuleCode, ModuleGrp) " & _
       "SELECT F1 as School, F2 as Campus, F3 as AdminNo, F4 as ModuleCode, F5 as ModuleGrp FROM [Text;HDR=NO;DATABASE=J:\FYP\;].[SEG1.txt]"

i我可以将select从文件更改为openfiledialog路径/文件吗？



i尝试过：

i can i change the select from file to the openfiledialog path/file?

i tried:

Dim query As String = "INSERT INTO  test (School, Campus, AdminNo, ModuleCode, ModuleGrp) " & _
        "SELECT F1 as School, F2 as Campus, F3 as AdminNo, F4 as ModuleCode, F5 as ModuleGrp FROM [Text;DATABASE=" & System.IO.Path.GetDirectoryName(Me.OpenFileDialog1.FileName) & ";HDR=NO].[" & Me.OpenFileDialog1.SafeFileName & "]"

提前致谢！

Thanks in advance!

推荐答案

是，你可以 OpenFileDialog [ ^ ]，但以不同的方式，使用字符串变量而不是OpenFileDialog属性。



请创建新的WindowsApplication项目并在表单上放置一个按钮，然后双击它。复制并粘贴以下代码：

Yes, you can ust it OpenFileDialog[^], but in different way, using string variable instead OpenFileDialog properties.

Please, create new WindowsApplication project and put one button on the form, then double-click on it. Copy and paste below code:

Public Class Form1

    Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
        Dim ofd As OpenFileDialog = Nothing
        Dim sFullFileName As String = String.Empty
        Try
            ofd = New OpenFileDialog()
            With ofd
                .InitialDirectory = Environment.SpecialFolder.MyDocuments
                .Multiselect = False
                .DefaultExt = "txt"
                .Filter = "txt files (*.txt)|*.txt"
                If .ShowDialog = Windows.Forms.DialogResult.Cancel Then Exit Try
                sFullFileName = .FileName
            End With

            MsgBox("Directory name: '" & System.IO.Path.GetDirectoryName(sFullFileName) & "'" & vbCr & _
                "Short file name: '" & System.IO.Path.GetFileName(sFullFileName) & "'", MsgBoxStyle.Information, "Information...")

        Catch ex As Exception
            MsgBox(ex.Message, MsgBoxStyle.Exclamation, "Error...")
        Finally
            ofd = Nothing
        End Try
    End Sub
End Class

尝试这个...

try to this...

if (openFileDialog1.ShowDialog(this) == DialogResult.OK)
{
    strfilename = openFileDialog1.InitialDirectory + openFileDialog1.FileName;
}

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