我正在使用asp.net mvc 4 webapi beta来构建一个休息服务。我需要能够接受张贴的图像/文件从客户端应用程序。这可能使用webapi吗?下面是如何行动，我目前正在使用。有人知道一个例子吗?
[HttpPost]
public string ProfileImagePost(HttpPostedFile profileImage)
{
string[] extensions = { ".jpg", ".jpeg", ".gif", ".bmp", ".png" };
if (!extensions.Any(x => x.Equals(Path.GetExtension(profileImage.FileName.ToLower()), StringComparison.OrdinalIgnoreCase)))
{
throw new HttpResponseException("Invalid file type.", HttpStatusCode.BadRequest);
}
// Other code goes here
return "/path/to/image.png";
}
下面是一个快速而简单的解决方案，它从HTTP正文中获取上传的文件内容并将其写入文件。我为文件上传包含了一个“骨架”HTML/JS片段。
Web API方法:
[Route("api/myfileupload")]
[HttpPost]
public string MyFileUpload()
{
var request = HttpContext.Current.Request;
var filePath = "C:\\temp\\" + request.Headers["filename"];
using (var fs = new System.IO.FileStream(filePath, System.IO.FileMode.Create))
{
request.InputStream.CopyTo(fs);
}
return "uploaded";
}
HTML文件上传:
<form>
<input type="file" id="myfile"/>
<input type="button" onclick="uploadFile();" value="Upload" />
</form>
<script type="text/javascript">
function uploadFile() {
var xhr = new XMLHttpRequest();
var file = document.getElementById('myfile').files[0];
xhr.open("POST", "api/myfileupload");
xhr.setRequestHeader("filename", file.name);
xhr.send(file);
}
</script>
[HttpPost]
public JsonResult PostImage(HttpPostedFileBase file)
{
try
{
if (file != null && file.ContentLength > 0 && file.ContentLength<=10485760)
{
var fileName = Path.GetFileName(file.FileName);
var path = Path.Combine(Server.MapPath("~/") + "HisloImages" + "\\", fileName);
file.SaveAs(path);
#region MyRegion
////save imag in Db
//using (MemoryStream ms = new MemoryStream())
//{
// file.InputStream.CopyTo(ms);
// byte[] array = ms.GetBuffer();
//}
#endregion
return Json(JsonResponseFactory.SuccessResponse("Status:0 ,Message: OK"), JsonRequestBehavior.AllowGet);
}
else
{
return Json(JsonResponseFactory.ErrorResponse("Status:1 , Message: Upload Again and File Size Should be Less Than 10MB"), JsonRequestBehavior.AllowGet);
}
}
catch (Exception ex)
{
return Json(JsonResponseFactory.ErrorResponse(ex.Message), JsonRequestBehavior.AllowGet);
}
}
下面是一个快速而简单的解决方案，它从HTTP正文中获取上传的文件内容并将其写入文件。我为文件上传包含了一个“骨架”HTML/JS片段。
Web API方法:
[Route("api/myfileupload")]
[HttpPost]
public string MyFileUpload()
{
var request = HttpContext.Current.Request;
var filePath = "C:\\temp\\" + request.Headers["filename"];
using (var fs = new System.IO.FileStream(filePath, System.IO.FileMode.Create))
{
request.InputStream.CopyTo(fs);
}
return "uploaded";
}
HTML文件上传:
<form>
<input type="file" id="myfile"/>
<input type="button" onclick="uploadFile();" value="Upload" />
</form>
<script type="text/javascript">
function uploadFile() {
var xhr = new XMLHttpRequest();
var file = document.getElementById('myfile').files[0];
xhr.open("POST", "api/myfileupload");
xhr.setRequestHeader("filename", file.name);
xhr.send(file);
}
</script>
朝着同样的方向，我发布了一个客户端和服务器片段，使用WebApi, c# 4发送Excel文件:
public static void SetFile(String serviceUrl, byte[] fileArray, String fileName)
{
try
{
using (var client = new HttpClient())
{
client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));
using (var content = new MultipartFormDataContent())
{
var fileContent = new ByteArrayContent(fileArray);//(System.IO.File.ReadAllBytes(fileName));
fileContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment")
{
FileName = fileName
};
content.Add(fileContent);
var result = client.PostAsync(serviceUrl, content).Result;
}
}
}
catch (Exception e)
{
//Log the exception
}
}
服务器的webapi控制器:
public Task<IEnumerable<string>> Post()
{
if (Request.Content.IsMimeMultipartContent())
{
string fullPath = HttpContext.Current.Server.MapPath("~/uploads");
MyMultipartFormDataStreamProvider streamProvider = new MyMultipartFormDataStreamProvider(fullPath);
var task = Request.Content.ReadAsMultipartAsync(streamProvider).ContinueWith(t =>
{
if (t.IsFaulted || t.IsCanceled)
throw new HttpResponseException(HttpStatusCode.InternalServerError);
var fileInfo = streamProvider.FileData.Select(i =>
{
var info = new FileInfo(i.LocalFileName);
return "File uploaded as " + info.FullName + " (" + info.Length + ")";
});
return fileInfo;
});
return task;
}
else
{
throw new HttpResponseException(Request.CreateResponse(HttpStatusCode.NotAcceptable, "Invalid Request!"));
}
}
和自定义MyMultipartFormDataStreamProvider，需要自定义文件名:
PS:我从另一个帖子http://www.codeguru.com/csharp/.net/uploading-files-asynchronously-using-asp.net-web-api.htm取了这个代码
public class MyMultipartFormDataStreamProvider : MultipartFormDataStreamProvider
{
public MyMultipartFormDataStreamProvider(string path)
: base(path)
{
}
public override string GetLocalFileName(System.Net.Http.Headers.HttpContentHeaders headers)
{
string fileName;
if (!string.IsNullOrWhiteSpace(headers.ContentDisposition.FileName))
{
fileName = headers.ContentDisposition.FileName;
}
else
{
fileName = Guid.NewGuid().ToString() + ".data";
}
return fileName.Replace("\"", string.Empty);
}
}
