我有一个打开文件的打开文件按钮。
现在我想使用文件的路径,这样我就可以单击一个保存按钮并保存文件。
这是打开按钮的代码:
Private Sub OpenINIButton_Click(sender As Object, e As EventArgs) Handles OpenINIButton.Click
Dim OpenDLG As New OpenFileDialog
OpenDLG.Filter = "Configuration File (*.ini)|*.ini"
OpenDLG.Title = "Open INI File"
OpenDLG.InitialDirectory = "C:\"
OpenDLG.RestoreDirectory = True
DialogResult = OpenDLG.ShowDialog
If DialogResult = Windows.Forms.DialogResult.OK Then
Dim OpenFile = OpenDLG.FileName.ToString()
wValue.Text = ReadIni(OpenFile, Isolation, Value, "")
ElseIf DialogResult = Windows.Forms.DialogResult.Cancel Then
End If
End Sub
我想要保存按钮中的OpenFile变量,我想要使用的代码是:
Private Sub Button2_Click(sender As System.Object, e As System.EventArgs) Handles saveINI.Click
System.IO.File.WriteAllText(OpenFile, "")
writeIni(OpenFile, BuildOptions, Isolation, w.Value.Text)
End Sub
但是OpenFile变量不可用。
是否有可能设置OpenFile变量Global? 我无法将其移动到SUB之外,因为“打开文件”按钮不再起作用。
谢谢!
答案 0 :(得分:0)
解决方案非常简单。只需在子
之外声明OpenFile
即可
Private OpenFile as String
Private Sub OpenINIButton_Click( ...
删除Dim
语句Dim OpenFile =
替换为OpenFile =
可选择测试变量是否设置在Button2_Click
If OpenFile is Nothing then Exit Sub