C#：从 body 发布模型并文件为 MultipartFormDataContent

Question

我正在寻找一种在一个请求中发送模型和图像的方法。我试图从 body 发送我的模型，但我不知道如何发送文件。除了在不同的文件中发送图像和模型之外，还有其他方法吗？

这是我的 API 中的 POST 方法：

[HttpPost]
[Route("UploadNewEvent")]
public async Task<IActionResult> CreateNewEventAsync([FromBody] EventModel model)
{
    var file = this.Request.Form.Files.LastOrDefault();

    if (file != null)
    {
        var uploads = Path.Combine(_environment.WebRootPath, "uploads");

        using (var fileStream = new FileStream(Path.Combine(uploads, "test.jpg"), FileMode.Create))
        {
            await file.CopyToAsync(fileStream);
        }
     }

     // do sth with model later    
     return Ok();
}

这是来自我的应用程序的代码：

public async Task SendNewEvent(EventModel model, MediaFile photo)
{
    var uri = $"{baseUri}api/User/Event/CreateNewEvent";

    if (photo != null)
    {
        var multipartContent = new MultipartFormDataContent();
        multipartContent.Add(new StreamContent(photo.GetStream()), "\"file\"", $"\"{photo.Path}\"");

        var httpClient = new HttpClient();               
        var jsonObject = JsonConvert.SerializeObject(model);
        var stringContent = new StringContent(jsonObject, Encoding.UTF8, "application/json");
        var httpResponseMessage = await httpClient.PostAsync(uri, stringContent);
    }
}

Answer 1

要通过 File 参数传递模型，您需要 post 数据作为表单数据。

按照以下步骤操作：

将FromBody改为FromForm

    [HttpPost]
[Route("UploadNewEvent")]
public async Task<IActionResult> CreateNewEventAsync([FromForm] EventModel model)
{
    // do sth with model later    
    return Ok();
}

更改客户端代码以发送表单数据而不是 json 字符串

    var uri = $"https://localhost:44339/UploadNewEvent";
    FileStream fileStream = new FileStream(@"filepath\T1.PNG", FileMode.Open);
    var multipartContent = new MultipartFormDataContent();
    multipartContent.Add(new StreamContent(fileStream), "\"file\"", @"filepath\T1.PNG");


    // EventModel other fields
    multipartContent.Add(new StringContent("2"), "Id");
    multipartContent.Add(new StringContent("Tom"), "Name");
    var httpClient = new HttpClient();
    var httpResponseMessage = httpClient.PostAsync(uri, multipartContent).Result;

事件模型

public class EventModel
{
public int Id { get; set; }
public string Name { get; set; }
public IFormFile File { get; set; }
}

Answer 2

我已经通过以下代码解决了此类问题。如果您发现任何困难，请告诉我。在我的示例中，我使用了文件、模型对象和简单的字符串，这将在将来有所帮助。

 using Newtonsoft.Json;
using System;
using System.IO;
using System.Net.Http;
using System.Net.Http.Headers;
using System.Threading.Tasks;

namespace TestingProject
{
    class Program1
    {
        static void Main(string[] args)
        {
            UploadFile();
        }

        public static async Task UploadFile()
        {


            HttpClient httpClient = new HttpClient
            {
                BaseAddress = new Uri("https://google.com/")
            };


            var filepath = @"C:\Sample Documents.pdf";
            var filename = "Sample Documents.pdf";
            using (MultipartFormDataContent content = new MultipartFormDataContent())
            {
                ByteArrayContent fileContent = new ByteArrayContent(System.IO.File.ReadAllBytes(filepath));
                fileContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment") { FileName = filename };
                fileContent.Headers.ContentType = new MediaTypeHeaderValue("application/pdf");

                Request request = new Request()
                {
                    UserId = "1",
                    UserName = "test",
                    address = new AddressModel()
                    {

                        City = "test",
                        Country = "India"
                    },
                    FileDetails = fileContent





                };
                var addressSerialize = JsonConvert.SerializeObject(request.address);

                FileStream fileStream = new FileStream(filepath, FileMode.Open);
                var multipartContent = new MultipartFormDataContent();
                multipartContent.Add(new StreamContent(fileStream), "\"file\"", filepath);
                multipartContent.Add(new StringContent(request.UserId), "UserId");
                multipartContent.Add(new StringContent(request.UserName), "UserName");
                multipartContent.Add(new StringContent(addressSerialize), "address");
         



                HttpResponseMessage httpResponseMessage = await httpClient.PostAsync("upload", multipartContent).ConfigureAwait(false);
                var personResponse = await httpResponseMessage.Content.ReadAsStringAsync().ConfigureAwait(false);

            }

        }


        public class Request
        {
            /// <summary>
            /// userId 
            /// </summary>
            [JsonProperty(PropertyName = "userId")]
            public string UserId { get; set; }


            /// <summary>
            /// UserName 
            /// </summary>
            [JsonProperty(PropertyName = "UserName")]
            public string UserName { get; set; }


            /// <summary>
            /// docType 
            /// </summary>
            [JsonProperty(PropertyName = "FileDetails")]
            public ByteArrayContent FileDetails { get; set; }

            /// <summary>
            /// address 
            /// </summary>
            [JsonProperty(PropertyName = "address")]
            public AddressModel address { get; set; }


        }

        public class AddressModel
        {
            /// <summary>
            /// City 
            /// </summary>
            [JsonProperty(PropertyName = "City")]
            public string City { get; set; }


            /// <summary>
            /// Country 
            /// </summary>
            [JsonProperty(PropertyName = "Country")]
            public string Country { get; set; }
        }

    }
}

C#：从 body 发布模型并文件为 MultipartFormDataContent

C#: Posting Model from body and file as MultipartFormDataContent

c#

multipartform-data

asp.net-core