Question

content.Add的以下行是否会使对象无法正确处理？如果是这样，处理这个问题的正确方法是什么。

 public string UploadGameToWebsite(string filename, string url, string method = null)
        {
            var client = new HttpClient();
            var content = new MultipartFormDataContent();
            content.Add(new StreamContent(File.Open(filename, FileMode.Open, FileAccess.Read)), "Game", "Game.txt");
            var task = client.PutAsync(url, content);
            var result = task.Result.ToString();
            return result;
        }

Answer 1

如果您要在方法中调用异步操作，请使您的方法异步。
处置您的文件流以及客户端。在下面的示例中，通过部署StreamContent，它还会处理基础FileStream。
我更喜欢使用finally块来处理多个一次性对象，它也非常适合嵌套using语句。
不确定为什么要返回ToString上的HttpResponseMessage，也许状态代码会更有用，或者看看StatusCode = 200并返回布尔值（true / false）？

代码：

public async Task<string> UploadGameToWebsiteAsync(string filename, string url, string method = null)
{
    HttpClient client = null;
    StreamContent fileStream = null;
    try
    {
        client = new HttpClient();
        fileStream = new StreamContent(System.IO.File.Open(filename, FileMode.Open, FileAccess.Read))
        var content = new MultipartFormDataContent();
        content.Add(fileStream, "Game", "Game.txt");
        HttpResponseMessage result = await client.PutAsync(url, content);
        return result.ToString();
    }
    finally 
    {
        // c# 6 syntax
        client?.Dispose();
        fileStream?.Dispose(); // StreamContent also disposes the underlying file stream
    }
}

使用using块的代码版本＃2。

public async Task<string> UploadGameToWebsiteAsync(string filename, string url, string method = null)
{
    using (var client = new HttpClient())
    {
        using (var fileStream = new StreamContent(System.IO.File.Open(filename, FileMode.Open, FileAccess.Read)))
        {
            var content = new MultipartFormDataContent();
            content.Add(fileStream, "Game", "Game.txt");
            HttpResponseMessage result = await client.PutAsync(url, content);
            return result.ToString();
        }
    }
}

Answer 2

是的，它会。 File.Open返回应该处理的流。最简单的方法是使用＆＃34;使用＆＃34;块，像这样：

 public string UploadGameToWebsite(string filename, string url, string method = null)
    {
        var client = new HttpClient();
        var content = new MultipartFormDataContent();
        using (var fileStream = File.Open(filename, FileMode.Open, FileAccess.Read))
        {
            content.Add(new StreamContent(fileStream), "Game", "Game.txt");
            var task = client.PutAsync(url, content);
            var result = task.Result.ToString();            
            return result;
        }
    }

处理File.Open或StreamContent

2 个答案: