C# 将文件发布到ApiController
我不熟悉ApicController,也不熟悉如何将文件从客户端发送到api并从这里保存。下面的代码确实有效,文件从客户端“复制”到服务器,但是响应消息是“204无内容”-我应该注意这里的任何事情吗 ApiController代码如下所示:C# 将文件发布到ApiController,c#,asp.net,asp.net-web-api,model-view-controller,C#,Asp.net,Asp.net Web Api,Model View Controller,我不熟悉ApicController,也不熟悉如何将文件从客户端发送到api并从这里保存。下面的代码确实有效,文件从客户端“复制”到服务器,但是响应消息是“204无内容”-我应该注意这里的任何事情吗 ApiController代码如下所示: [HttpPost] [Route("api/UploadFile")] public async Task UploadFile() { string fileName = "myfil
[HttpPost]
[Route("api/UploadFile")]
public async Task UploadFile()
{
string fileName = "myfilename.txt";
Stream requestStream = await Request.Content.ReadAsStreamAsync();
using (FileStream fileStream = File.Create(@"C:\myDropFolder\" + fileName))
{
await requestStream.CopyToAsync(fileStream);
}
}
对客户端的调用如下所示,从控制台应用程序进行测试:
class Program
{
static void Main(string[] args)
{
try
{
HttpWebRequest request = (HttpWebRequest)HttpWebRequest.Create("http://localhost:1741/api/UploadFile/");
request.Method = WebRequestMethods.Http.Post;
byte[] fileToSend = File.ReadAllBytes(@"C:\myDropFolder\tester.txt");
request.ContentLength = fileToSend.Length;
using (Stream requestStream = request.GetRequestStream())
{
// Send the file as body request.
requestStream.Write(fileToSend, 0, fileToSend.Length);
requestStream.Close();
}
using (HttpWebResponse response = (HttpWebResponse)request.GetResponse())
Console.WriteLine("HTTP/{0} {1} {2}", response.ProtocolVersion, (int)response.StatusCode, response.StatusDescription);
Console.ReadLine();
}
catch (Exception)
{
throw;
}
}
}
POST的Web API默认响应状态代码为204,GET的默认响应状态代码为200()。除非返回
IHttpActionResult
,然后显式返回200:
[HttpPost]
[Route("api/UploadFile")]
public async Task<IHttpActionResult> UploadFile()
{
string fileName = "myfilename.txt";
Stream requestStream = await Request.Content.ReadAsStreamAsync();
using (FileStream fileStream = File.Create(@"C:\myDropFolder\" + fileName))
{
await requestStream.CopyToAsync(fileStream);
}
return Ok();
}
[HttpPost]
[路由(“api/上传文件”)]
公共异步任务上载文件()
{
字符串fileName=“myfilename.txt”;
Stream requestStream=wait Request.Content.ReadAsStreamAsync();
使用(FileStream FileStream=File.Create(@“C:\myDropFolder\”+fileName))
{
等待requestStream.CopyToAsync(文件流);
}
返回Ok();
}
Web API的POST默认响应状态代码为204,GET默认响应状态代码为200()。除非返回IHttpActionResult
,然后显式返回200:
[HttpPost]
[Route("api/UploadFile")]
public async Task<IHttpActionResult> UploadFile()
{
string fileName = "myfilename.txt";
Stream requestStream = await Request.Content.ReadAsStreamAsync();
using (FileStream fileStream = File.Create(@"C:\myDropFolder\" + fileName))
{
await requestStream.CopyToAsync(fileStream);
}
return Ok();
}
[HttpPost]
[路由(“api/上传文件”)]
公共异步任务上载文件()
{
字符串fileName=“myfilename.txt”;
Stream requestStream=wait Request.Content.ReadAsStreamAsync();
使用(FileStream FileStream=File.Create(@“C:\myDropFolder\”+fileName))
{
等待requestStream.CopyToAsync(文件流);
}
返回Ok();
}
非常感谢您对Arin的解释!非常感谢你的解释,阿林!