Question

我使用this example，我的代码是

private void dialogBtn_Click(object sender, EventArgs e)
{
    string file;
    DialogResult result = openFileDialog1.ShowDialog(); // Show the dialog.
    if (result == DialogResult.OK) // Test result.
    {
        file = openFileDialog1.FileName;

        System.IO.FileInfo fileInfo = new System.IO.FileInfo(file);
        TransferServiceClient clientUpload = new TransferServiceClient();
        RemoteFileInfo uploadRequestInfo = new RemoteFileInfo();

        using (System.IO.FileStream stream = new System.IO.FileStream(file, System.IO.FileMode.Open, System.IO.FileAccess.Read))
        {
            uploadRequestInfo.FileName = file;
            uploadRequestInfo.Length = fileInfo.Length;
            uploadRequestInfo.FileByteStream = stream;
            //clientUpload.UploadFile(uploadRequestInfo);
            clientUpload.UploadFile(fileInfo.Name, fileInfo.Length, stream);                      
        }                
    }           
}

在clientUpload.UploadFile(fileInfo.Name, fileInfo.Length, stream);我收到错误

期间

mscorlib.dll中发生了'System.ServiceModel.FaultException`1'类型的未处理异常附加信息：找不到路径'C：\ upload \ Book1.xlsx'的一部分。

但是，Book1.xlsx不在上传文件夹中。它在桌面上。

Answer 1

嗯，示例中UploadFile方法的服务实现如下：

public void UploadFile(RemoteFileInfo request)
{
    FileStream targetStream = null;
    Stream sourceStream =  request.FileByteStream;

    string uploadFolder = @"C:\upload\";

    string filePath = Path.Combine(uploadFolder, request.FileName);

    using (targetStream = new FileStream(filePath, FileMode.Create, 
                          FileAccess.Write, FileShare.None))
    {
        //read from the input stream in 65000 byte chunks

        const int bufferLen = 65000;
        byte[] buffer = new byte[bufferLen];
        int count = 0;
        while ((count = sourceStream.Read(buffer, 0, bufferLen)) > 0)
        {
            // save to output stream
            targetStream.Write(buffer, 0, count);
        }
        targetStream.Close();
        sourceStream.Close();
    }
}

注意这一行

string uploadFolder = @"C:\upload\";

您需要将其更改为您希望上传文件存储的现有文件夹。

wcf传输文件之前的异常

1 个答案: