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从保存的字节流中打开 docx 文件时，我们始终收到文件损坏错误消息，其他文件类型都可以正常工作

以下是复制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
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1 回答 1

1

请原谅我，但那是一些混乱的代码。

Dim _o As String = "C:\" & Now.Ticks & getNewFileName()

会变成...

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，甚至深层路径中的长文件名可能是一个问题。这段代码应该可以正常工作，而不是使用 FileStream：

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)
于 2010-02-11T21:31:01.807 回答