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如何在ASP中接收字节数组和标头内容。NET Core Web API控制器

webclient asp.net-core asp.net-web-api visual-studio c#

Lakmal · 技术社区 · 8 年前

我需要在c#ASP中接收内容和字节数组。NET核心Web API应用程序。

**--HttpClient (console Application)**  

                Dictionary<int, byte[]> fileparts = new Dictionary<int, byte[]>();
                int bufferSize = 1000 * 1024;

                byte[] buffer;
                string filePath = @"D:\Extra\Songs\test.mp3";
                using (FileStream fileData = File.OpenRead(filePath))
                {
                    int index = 0;
                    int i = 1;
                    while (fileData.Position < fileData.Length)
                    {

                        if (index + bufferSize > fileData.Length)
                        {
                            buffer = new byte[(int)fileData.Length - index];
                            fileData.Read(buffer, 0, ((int)fileData.Length - index));
                        }
                        else
                        {
                            buffer = new byte[bufferSize];
                            fileData.Read(buffer, 0, bufferSize);
                        }

                        fileparts.Add(i, buffer);
                        index = (int)fileData.Position;
                        i++;
                    }
                }
                
 while (fileparts.Count() != 0)
 {               
      var data = fileparts.First();          
  var fileContent = new ByteArrayContent(data);//byte content
    fileContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment") { FileName = "test" };
                        fileContent.Headers.Add("id", id);
                        fileContent.Headers.Add("name", name);
                        fileContent.Headers.Add("length", length);
                        if (fileparts.Count() == 1)
                          fileContent.Headers.Add("IsCompleted", "true");
                        else
                          fileContent.Headers.Add("IsCompleted", "false");
                          
                    using (var content = new MultipartFormDataContent())
                    {
                       content.Add(fileContent);
                       // Make a call to Web API 
                        var result = client.PostAsync("/api/file", fileContent).Result;
                        if (result.StatusCode == System.Net.HttpStatusCode.OK)
                            fileparts.Remove(data.Key);
                    }

**--WebApi Core ApiController**

    public class FileController : Controller
    {
        [HttpPost]
        public async Task<IActionResult> Post()
        {
    /*i need to get the content here , for some reason existing .NET libraries does not work here
    like 
        MultipartMemoryStreamProvider provider = new MultipartMemoryStreamProvider();
        FilePart part = null;
       
        await Request.Content.ReadAsMultipartAsync(provider); -- request.content not available 
        using (Stream fileStream = await provider.Contents[0].ReadAsStreamAsync())
                {
                    part = provider.Contents[0].Headers.GetData(); //public static FilePart GetData name,length,id 
                    part.Data = fileStream.ReadFully();
    */    
    }

2 回复 | 直到 8 年前

Niladri 8 年前

您可以修改控制器操作方法,如下所示,以接受 ByteArrayContent 或 MultipartFormDataContent 根据需要。我用过 [FromBody] POST中模型绑定的属性。

[来自正文]

https://docs.microsoft.com/en-us/aspnet/core/mvc/models/model-binding

public class FileController : Controller
    {
        [HttpPost]
        public async Task<IActionResult> Post([FromBody] MultipartFormDataContent content)
        {
         MultipartMemoryStreamProvider provider = new MultipartMemoryStreamProvider();
        FilePart part = null;
         // access the content here 
         await content.ReadAsMultipartAsync(provider);
        // rest of the code
       }
   }

如前所述,您应该使用 content 如下所示发布到此API

var result = client.PostAsync("/api/file", content).Result;

alec 8 年前

[HttpPost]
public async Task<IActionResult> Post(IFormFile formFile)
{
    using (Stream stream = formFile.OpenReadStream())
    {
         //Perform necessary actions with file stream
    }
}

我认为您还需要更改客户端代码以匹配参数名称:

fileContent.Headers.Add("name", "formFile");