Question

我正在寻找一种在一个请求中发送模型和图像的方法。我试图从身体发送模型，但是我不知道如何发送文件。除了在其他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();
}

这是我的应用程序中的代码：

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Model，您需要将数据作为表单数据发布。

请按照以下步骤操作：

将FromBody更改为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;

EventModel

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

C＃：从正文和文件中将模型发布为MultipartFormDataContent

1 个答案: