2

我正在按照本教程从 android 将文件上传到服务器，但我似乎无法在服务器端正确获取代码。有人可以帮我编写适用于该 android java 上传器的 Web Api post 方法吗？我当前的 web api 控制器类如下所示：

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Net;
using System.Net.Http;
using System.Threading.Tasks;
using System.Web;
using System.Web.Http;

namespace WSISWebService.Controllers
{
    public class FilesController : ApiController
    {
        // GET api/files
        public IEnumerable<string> Get()
        {
            return new string[] { "value1", "value2" };
        }

        // GET api/files/5
        public string Get(int id)
        {
            return "value";
        }

        // POST api/files
        public string Post([FromBody]string value)
        {
            var task = this.Request.Content.ReadAsStreamAsync();
            task.Wait();
            Stream requestStream = task.Result;

            try
            {
                Stream fileStream = File.Create(HttpContext.Current.Server.MapPath("~/" + value));
                requestStream.CopyTo(fileStream);
                fileStream.Close();
                requestStream.Close();
            }
            catch (IOException)
            {
                //  throw new HttpResponseException("A generic error occured. Please try again later.", HttpStatusCode.InternalServerError);
            }

            HttpResponseMessage response = new HttpResponseMessage();
            response.StatusCode = HttpStatusCode.Created;
            return response.ToString();
        }          

        // PUT api/files/5
        public void Put(int id, [FromBody]string value)
        {
        }

        // DELETE api/files/5
        public void Delete(int id)
        {
        }
    }
}

由于截止日期是星期二，我非常渴望得到这个工作。如果有人可以提供帮助，将不胜感激。

4

1 回答 1

5

您可以将文件发布为 multipart/form-data

    // POST api/files
    public async Task<HttpResponseMessage> Post()
    {
        // Check if the request contains multipart/form-data.
        if (!Request.Content.IsMimeMultipartContent())
        {
            throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
        }

        string root = HttpContext.Current.Server.MapPath("~/App_Data");
        var provider = new MultipartFormDataStreamProvider(root);

        string value;

        try
        {
            // Read the form data and return an async data.
            var result = await Request.Content.ReadAsMultipartAsync(provider);

            // This illustrates how to get the form data.
            foreach (var key in provider.FormData.AllKeys)
            {
                foreach (var val in provider.FormData.GetValues(key))
                {
                    // return multiple value from FormData
                    if (key == "value")
                        value = val;
                }
            }                       

            if (result.FileData.Any())
            {                    
                // This illustrates how to get the file names for uploaded files.
                foreach (var file in result.FileData)
                {
                    FileInfo fileInfo = new FileInfo(file.LocalFileName);
                    if (fileInfo.Exists)
                    {
                       //do somthing with file
                    }
                }
            }


            HttpResponseMessage response = Request.CreateResponse(HttpStatusCode.Created, value);
            response.Headers.Location = new Uri(Url.Link("DefaultApi", new { id = files.Id }));
            return response;
        }
        catch (System.Exception e)
        {
            return Request.CreateErrorResponse(HttpStatusCode.InternalServerError, e);
        }
    }
于 2013-11-13T10:20:48.420 回答