从保存的字节流打开docx文件时,我们始终收到文件损坏的错误消息,evry其他文件类型正常工作
以下是重新说明问题的示例表单中的代码
Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
'Objective is to be able to copy a file to a bytestream then create a new document from that stream and then opne it.
'This replicates the behaviour of our primary application where it stores and retrieves the stream from a database
'With docx files we consistently experience some sort of corruption in the write of the original file
'When selecting .doc or other format files we do not encounter the same problem
'use selected file
Dim _o1 As String = TextBox1.Text
'get its bytestream
Dim fs As New FileStream(_o1, FileMode.Open, FileAccess.Read)
Dim byteStream(Convert.ToInt32(fs.Length)) As Byte
fs.Read(byteStream, 0, Convert.ToInt32(fs.Length))
'create a new file and use the bytestream to create it and save to disk
Dim _o As String = "C:\" & Now.Ticks & getNewFileName()
Dim fs1 As New FileStream(_o, FileMode.OpenOrCreate, FileAccess.Write)
Using bw As New BinaryWriter(fs1)
bw.Write(byteStream)
bw.Flush()
End Using
'open the new document
System.Diagnostics.Process.Start(_o)
Application.DoEvents()
End Sub
Private Function getNewFileName() As String
Dim fi As New FileInfo(TextBox1.Text)
Return Now.Ticks.ToString & fi.Name
End Function
Private Sub Button2_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button2.Click
OpenFileDialog1.InitialDirectory = "c:\"
OpenFileDialog1.FilterIndex = 2
OpenFileDialog1.RestoreDirectory = True
OpenFileDialog1.Filter = "docx files |*.docx"
If OpenFileDialog1.ShowDialog() = System.Windows.Forms.DialogResult.OK Then
TextBox1.Text = OpenFileDialog1.FileName
End If
End Sub
答案 0 :(得分:1)
原谅我,但这是一些混乱的代码。
Dim _o As String = "C:\" & Now.Ticks & getNewFileName()
将成为......
Dim _o As String = "C:\" & Now.Ticks & Now.Ticks.ToString & fi.Name
示例结果“C:\”“634015010433498951”“634015010433498951”“FileName.txt”可能不是您所期望的,除非您打算减去两个滴答计数以确定填充FileInfo所需的时间。
您的FileStream损坏可能是一个编码问题,一个文件长度问题关闭,甚至深度路径中的长文件名可能会出现问题。而不是使用FileStream,这段代码应该可以正常工作:
Dim sourceFile As String = TextBox1.text
Dim fi As New System.IO.FileInfo(sourceFile)
Dim destFile = "C:\" & Now.Ticks & fi.Name
fi.CopyTo(destFile)
'open the new document
System.Diagnostics.Process.Start(destFile)