c# - 在 ASP.NET 中重命名并发上传的文件

Question

我使用以下代码来接收上传的文件并将它们存储在服务器上的uploadedFiles文件夹中。

public class CustomMultipartFormDataStreamProvider : MultipartFormDataStreamProvider
{
    public CustomMultipartFormDataStreamProvider(string path) : base(path) { }

    public override string GetLocalFileName(HttpContentHeaders headers)
    {
        return headers.ContentDisposition.FileName.Replace("\"", string.Empty);
    }
}

[HttpPost]
public async Task<HttpResponseMessage> ReceiveFileupload()
{
    if (!Request.Content.IsMimeMultipartContent("form-data"))
        throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);

    // Prepare CustomMultipartFormDataStreamProver in which our multipart form data will be loaded
    string fileSaveLocation = HttpContext.Current.Server.MapPath("~/UploadedFiles");
    CustomMultipartFormDataStreamProvider provider = new CustomMultipartFormDataStreamProvider(fileSaveLocation);
    List<string> files = new List<string>();
    Duck duck = null;

    try
    {
        // Read all contents of multipart message into CustomMultipartFormDataStreamProvider
        await Request.Content.ReadAsMultipartAsync(provider);

        // Parse an ID which is passed in the form data
        long id;
        if (!long.TryParse(provider.FormData["id"], out id))
        {
            return Request.CreateResponse(HttpStatusCode.ExpectationFailed, "Couldn't determine id.");
        }

        duck = db.Ducks.Find(id);
        if (null == duck)
            return Request.CreateResponse(HttpStatusCode.NotAcceptable, "Duck with ID " + id + " could not be found.");

        // Loop through uploaded files
        foreach (MultipartFileData file in provider.FileData)
        {
            // File ending needs to be xml
            DoSomething();

            // On success, add uploaded file to list which is returned to the client
            files.Add(file.LocalFileName);
        }

        // Send OK Response along with saved file names to the client.
        return Request.CreateResponse(HttpStatusCode.OK, files);
    }

    catch (System.Exception e) {
        return Request.CreateResponse(HttpStatusCode.InternalServerError, e);
    }
}

现在我想重命名上传的文件。据我了解，存储和重命名必须在关键部分完成，因为当另一个用户同时上传同名文件时，第一个文件将被覆盖。

这就是我想象的解决方案，但是 await 不允许出现在 lock 语句中。

lock(typeof(UploadController)) {
    await Request.Content.ReadAsMultipartAsync(provider);  //That's the line after which the uploaded files are stored in the folder.
    foreach (MultipartFileData file in provider.FileData)
    {
        RenameFile(file); // Rename according to file's contents
    }
}

我如何保证文件将自动存储和重命名？

Answer 1

与其使用原始名称然后重命名，不如在上传代码中即时生成一个名称？

例如如果文件显示为“foo.xml”，您将其作为“BodyPart_be42560e-863d-41ad-8117-e1b634e928aa”写入磁盘？

另请注意，如果您将文件上传到 "~/UploadedFiles"，任何人都可以使用 www.example.com/UploadedFiles/name.xml 这样的 URL 下载它 - 您应该将文件存储在 ~/App_Data/ 中以防止出现这种情况。

更新:

为什么不删除您对 GetLocalFileName 方法的重写，并简单地使用 MultipartFileStreamProvider 中的原始方法？基类？

public virtual string GetLocalFileName(HttpContentHeaders headers)
{
    if (headers == null)
    {
        throw Error.ArgumentNull("headers");
    }

    return String.Format(CultureInfo.InvariantCulture, "BodyPart_{0}", Guid.NewGuid());
}

这应该将每个文件保存为唯一名称，而无需稍后重命名。