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如何正确编写从webapi接收的文件

asp.net-core-webapi c#

Miamy Dan Barrett · 技术社区 · 7 年前

我尝试制作一个简单的webapi来从服务器接收文件。

以下是服务器部分:

var result = new HttpResponseMessage(HttpStatusCode.OK);
var stream = new FileStream(itemPath, FileMode.Open, FileAccess.Read);
result.Content = new StreamContent(stream);
result.Content.Headers.ContentType = 
      new MediaTypeHeaderValue("application/octet-stream");
result.Content.Headers.ContentDisposition = 
      new ContentDispositionHeaderValue("attachment") { FileName = itemPath };
return result;

客户方:

var client = new HttpClient();
var responce = Task.Run(() => client.GetAsync(@"https://localhost:...")).Result;
var bytes = Task.Run(() => responce.Content.ReadAsByteArrayAsync()).Result;
var writer = 
  new BinaryWriter(new FileStream(fileName, FileMode.CreateNew, FileAccess.Write));
writer.Write(bytes);
writer.Close();

以及书面文件内容:

{"version":{"major":1,"minor":1,"build":-1,"revision":-1,"majorRevision":-1,
"minorRevision":-1},"content":{"headers":[{"key":"Content-Type","value":["application/octet-stream"]},{"key":"Content-Disposition",
"value":["attachment; filename=\"=?utf-8?B?ZDpcTGlicmFyeVBhdGhc0J/QsNGH0LXQutC+XEV4YW1wbGUgR
DVcQ2QxN1xDTElQQlJEXE1haW5Gcm0ucGFz?=\""]}]},"statusCode":200,
"reasonPhrase":"OK","headers":[],"requestMessage":null,"isSuccessStatusCode":true}

我希望这个文件将包含一个来自服务器本身的文件,但不包含描述它的JSON。

我错过了什么?

1 回复 | 直到 7 年前

Nkosi 7 年前

响应的内容不应该是 HttpResponseMessage .

Asp.net核心Web API正在读取 Http响应消息

重构服务器端代码以在控制器中使用建议的语法

public IActionResult MyAction(...) {
    //...
    var stream = new FileStream(itemPath, FileMode.Open, FileAccess.Read);
    return File(stream, "application/octet-stream", "desired file name here");
}

返回一个 IActionResult 派生 FileStreamResult

.Result . 还不清楚为什么异步调用被包装在 Task.Run 当他们已经可以等待的时候。

始终保持代码异步

var client = new HttpClient();
var response = await client.GetAsync(@"https://localhost:..."); ;
var bytes = await response.Content.ReadAsByteArrayAsync();
var fileName = response.Content.Headers.ContentDisposition.FileName;
using (var stream = new FileStream(fileName, FileMode.CreateNew, FileAccess.Write)) {
    await stream.WriteAsync(bytes, 0, bytes.Length);
}

var client = new HttpClient();
var response = await client.GetAsync(@"https://localhost:..."); ;
var stream = await response.Content.ReadAsStreamAsync();
var fileName = response.Content.Headers.ContentDisposition.FileName;
using (var fileStream = new FileStream(fileName, FileMode.CreateNew, FileAccess.Write)) {
    await stream.CopyToAsync(fileStream);
}

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