Question

我有以下MVC帖子。它将文件内容发布到API。

[HttpPost]
public ActionResult FileUpload_Post()
{
    if (Request.Files.Count > 0)
    {
        var file = Request.Files[0];

        using (HttpClient client = new HttpClient())
        {
            using (var content = new MultipartFormDataContent())
            {
                byte[] fileBytes = new byte[file.InputStream.Length + 1];                     file.InputStream.Read(fileBytes, 0, fileBytes.Length);
                var fileContent = new ByteArrayContent(fileBytes);
                fileContent.Headers.ContentDisposition = new System.Net.Http.Headers.ContentDispositionHeaderValue("attachment") { FileName = file.FileName };
                content.Add(fileContent);
                var result = client.PostAsync(requestUri, content).Result;
                if (result.StatusCode == System.Net.HttpStatusCode.Created)
                {
                    ViewBag.Message= "Created";
                }
                else
                {
                    ViewBag.Message= "Failed";
                }
            }
        }
    }
    return View();
}

如果我想传递其他自定义对象（最好是json格式）以及文件内容怎么办？

CustomObject obj = new CustomObject;
obj.FirstName = "A";
object.LastName = "B";

注意：以下是接收上述请求的Api方法。

[HttpPost]
public HttpResponseMessage Upload()
{
    if(!Request.Content.IsMimeMultipartContent())
    {
        throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
    }
    if (System.Web.HttpContext.Current.Request.Files.Count > 0)
    {
        var file = System.Web.HttpContext.Current.Request.Files[0];
        ....
        // save the file
        ....
        return new HttpResponseMessage(HttpStatusCode.Created);
    }
    else
    {
        return new HttpResponseMessage(HttpStatusCode.BadRequest);
    }
}

Answer 1

首先，您需要将CustomObject序列化为json。例如使用Json.NET

var jsonString = Newtonsoft.Json.JsonConvert.SerializeObject(obj);

然后您可以将jsonString添加到MultipartFormDataContent，例如：

var jsonContent = new StringContent(jsonString);
content.Add(jsonContent, "CustomObject");

在Upload API方法中，按

获取发布的json内容

var jsonString = System.Web.HttpContext.Current.Request.Form["CustomObject"];

如果API项目引用了类CustomObject，您可以使用以下代码反序列化jsonString：

var obj = Newtonsoft.Json.JsonConvert.DeserializeObject<CustomObject>(jsonString);

如果没有，您也可以将其反序列化为dynamic对象：

var obj = Newtonsoft.Json.Linq.JObject.Parse(jsonString);

如何将其他对象发送到client.PostAsync（以及文件内容）

1 个答案: