Dropbox API上传等待异步函数
Dropbox API上传等待异步函数
提问于 2020-07-11 04:17:21
伙计们
我在上传Dropbox API时遇到了困难。我有一个文件列表，在for的每一次传递中，我都会调用上传的函数：
Public Function Upload(ByVal ArquivoOrigem As String, ByVal ArquivoDestino As String)
Try
Dim fileStream As FileStream = New FileStream(ArquivoOrigem, FileMode.Open)
Me.CtrlTask = Task.Run(Function() Upload_Executar(fileStream, ArquivoDestino))
Me.CtrlTask.Wait()
Catch ex As Exception
End Try
End Function
Private Async Function Upload_Executar(ByVal fileStream As FileStream, ByVal ArquivoDestino As String) As Task
Me.ArquivoUploadIs = False
Try
Dim numChunks As Integer = CInt(Math.Ceiling(CDbl(fileStream.Length) / Me.chunkSize))
Dim buffer As Byte() = New Byte(Me.chunkSize - 1) {}
Dim sessionId As String = Nothing
Console.WriteLine("Chunk upload file...")
Console.WriteLine("fileStream.Length: " + fileStream.Length.ToString())
Console.WriteLine("chunkSize: " + Me.chunkSize.ToString())
Console.WriteLine("numChunks: " + numChunks.ToString())
For idx = 0 To numChunks - 1
Dim Porc1 As Integer = 0
Dim Porc2 As Integer = 0
Porc1 = CInt((idx / numChunks) * 100)
Porc2 = idx * Me.chunkSize
Console.WriteLine("Posicao: " + idx.ToString() + " / Total: " + numChunks.ToString() + " / Porc1: " + Porc1.ToString() + " / Total Transferido: " + FormatBytes(Porc2) + " / Tamanho Total: " + FormatBytes(fileStream.Length))
Dim byteRead = fileStream.Read(buffer, 0, Me.chunkSize)
Using memStream As MemoryStream = New MemoryStream(buffer, 0, byteRead)
If idx = 0 Then
Console.WriteLine("memStream.Length: " + memStream.Length.ToString())
Console.WriteLine("UploadSessionStartAsync")
Dim result = Await Dbx.Files.UploadSessionStartAsync(False, memStream)
Console.WriteLine(result)
sessionId = result.SessionId
Console.WriteLine("sessionId: " + sessionId)
Else
Dim cursor As UploadSessionCursor = New UploadSessionCursor(sessionId, CULng((Me.chunkSize * idx)))
If idx = numChunks - 1 Then
Console.WriteLine("UploadSessionFinishAsync")
Dim CtrlUp = Await Dbx.Files.UploadSessionFinishAsync(cursor, New CommitInfo(ArquivoDestino), memStream)
If CtrlUp.Id <> "" Then
Me.ArquivoUploadIs = True
End If
Else
Console.WriteLine("UploadSessionAppendV2Async")
Await Dbx.Files.UploadSessionAppendV2Async(cursor, body:=memStream)
End If
End If
End Using
Next
Catch ex As Exception
ShowMsgError(ex)
End Try
End Function我通过任务调用她，使用" wait“等待提交完成，然后转到下一个文件：
Upload("C:\Arq1.pdf", "/Arq1.pdf");
Upload("C:\Arq2.pdf", "/Arq2.pdf");
Upload("C:\Arq3.pdf", "/Arq3.pdf");
Upload("C:\Arq4.pdf", "/Arq4.pdf");然而，当上传完成时，应用程序被卡住了。然而，为了测试put线程，这会导致列表中的所有文件同时发送，我想发送一个，等待它完成，然后发送下一个文件。
有人有什么建议吗？
Stack Overflow用户
发布于 2020-07-12 19:56:00
您应该避免使用.Wait，因为它可能会导致死锁。
Public Async Function Upload(ByVal ArquivoOrigem As String, ByVal ArquivoDestino As String) As Task
Try
Dim fileStream As FileStream = New FileStream(ArquivoOrigem, FileMode.Open)
Await Upload_Executar(fileStream, ArquivoDestino))
Catch ex As Exception
End Try
End FunctionAwait Upload("C:\Arq1.pdf", "/Arq1.pdf");
Await Upload("C:\Arq2.pdf", "/Arq2.pdf");
Await Upload("C:\Arq3.pdf", "/Arq3.pdf");
Await Upload("C:\Arq4.pdf", "/Arq4.pdf");页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/62841719
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