vb.net 使用 SaveFileDialog 将数据保存到文本图块？
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Save Data to text tile using SaveFileDialog?
提问by Matt
I have already viewed the MSDN Examplebut I am still having problems.
我已经查看了MSDN 示例，但我仍然遇到问题。
I created a super-simple program to multiply two numbers, and display the output in the textbox. Now I need to be able to read that text box value and put the value in a text file, bringing up the save to file dialog when the "Save To File" button is clicked.
我创建了一个超级简单的程序来将两个数字相乘，并在文本框中显示输出。现在我需要能够读取该文本框值并将该值放入文本文件中，当单击“保存到文件”按钮时打开保存到文件对话框。
Private Sub MutiplyBtn_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MutiplyBtn.Click
Dim FirstNum As Double = Num1.Text
Dim SecondNum As Double = Num2.Text
Dim Answer2 As Double = FirstNum * SecondNum
Answerbox.Text = Answer2
End Sub
Private Sub SaveResultToFile_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles SaveResultToFile.Click
Dim myStream As Stream
Dim saveFileDialog1 As New SaveFileDialog()
saveFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*"
saveFileDialog1.FilterIndex = 2
saveFileDialog1.RestoreDirectory = True
If saveFileDialog1.ShowDialog() = DialogResult.OK Then
myStream = saveFileDialog1.OpenFile()
If (myStream IsNot Nothing) Then
System.IO.File.WriteAllText(Answerbox.Text)
myStream.Close()
End If
End If
End Sub
Currently, Visual Studio is giving me an error: Overload resolution failed because no accessible 'WriteAllText' accepts this number of arguments.
目前，Visual Studio 给我一个错误： Overload resolution failed because no accessible 'WriteAllText' accepts this number of arguments.
回答by Steve
WriteAllTextstatic method requires the name of the file where the data should be written to.
You could use directly the name selected in the saveFileDialog1
WriteAllText静态方法需要数据应写入的文件的名称。
您可以直接使用在 saveFileDialog1 中选择的名称
If saveFileDialog1.ShowDialog() = DialogResult.OK Then
System.IO.File.WriteAllText(saveFiledialog1.FileName, Answerbox.Text)
End If
instead if you really want to use the stream opened by OpenFile() method your code should be
相反，如果你真的想使用 OpenFile() 方法打开的流，你的代码应该是
If saveFileDialog1.ShowDialog() = DialogResult.OK Then
Dim sw As StreamWriter = new StreamWriter(saveFileDialog1.OpenFile())
If (sw IsNot Nothing) Then
sw.WriteLine(Answerbox.Text)
sw.Close()
End If
End If
The code is an example, you need to add a bit of error handling
代码是一个例子，需要加一点错误处理
回答by DEV RAJ
Hi I tried above method but I succeed in this way....
嗨，我尝试了上述方法，但我以这种方式成功了....
Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
SaveFileDialog1.Filter = "TXT Files (*.txt*)|*.txt"
If SaveFileDialog1.ShowDialog = Windows.Forms.DialogResult.OK _
Then
My.Computer.FileSystem.WriteAllText _
(SaveFileDialog1.FileName, RichTextBox1.Text, True)
End If
End Sub
