0

当我使用以下代码时,文件在 中正确上传uploadPath,但该GetLocalFileName方法返回错误的BodyPart名称。

文件名中的 guid(例如BodyPart_f4a1e725-0194-4bda-a666-b6ffea7b0618:)将是其他一些不是刚刚上传的文件的 guid。

我正在尝试获取上传文件的文件名。

private string lastUploadFileName

public async Task AsyncReceive(HttpRequestMessage incomingRequest)
{
    request = incomingRequest;
    streamProvider = new MultipartFormDataStreamProvider(uploadPath);
    var uploadTask = request.Content.ReadAsMultipartAsync(streamProvider);

    await uploadTask.ContinueWith(t =>
    {
        if (t.IsFaulted || t.IsCanceled)
        {
            throw new Exception("Error uploading file.");

        }
    });

    lastUploadFileName = streamProvider.GetLocalFileName(
        request.Content.Headers);
}
4

0 回答 0