时间：2019-03-17 标签：c#webrequestpostimagetowebapi

我在将图像上传到我正在运行的 Web API 时遇到问题。使用 GET 请求时，我可以从 Web API 检索数据，但在处理 POST 请求时遇到问题。我需要将 BMP 图像上传到 Web API，然后发回 json 字符串。

[HttpPost]
public IHttpActionResult TestByte()
{
    Log("TestByte function entered");
    //test to see if i get anything, not sure how to do this
    byte[] data = Request.Content.ReadAsByteArrayAsync().Result;
    byte[] test = Convert.FromBase64String(payload);

    if(test == null || test.Length <= 0)
    {
        Log("No Payload");
        return NotFound();
    }

    if (data == null || data.Length <= 0)
    {
        Log("No payload");
        return NotFound();
    }

    Log("Payload received");
    return Ok();

}

发送图像的 MVC 端如下所示：

// Create a request using a URL that can receive a post. 
WebRequest request = WebRequest.Create(url);
// Set the Method property of the request to POST.
request.Method = "POST";

// Create POST data and convert it to a byte array.
byte[] byteArray = GetImageData(Path.Combine(AppDomain.CurrentDomain.BaseDirectory, content, barcodeUri));
string base64String = Convert.ToBase64String(byteArray);
byte[] dataArray = Encoding.Default.GetBytes(base64String);

// Set the ContentType property of the WebRequest.
request.ContentType = "multipart/form-data";
// Set the ContentLength property of the WebRequest.
request.ContentLength = dataArray.Length;

// Get the request stream.
Stream dataStream = request.GetRequestStream();
// Write the data to the request stream.
dataStream.Write(dataArray, 0, dataArray.Length);
// Close the Stream object.
dataStream.Close();

// Get the response.
WebResponse response = request.GetResponse();
// Get the stream containing content returned by the server.
dataStream = response.GetResponseStream();
// Open the stream using a StreamReader for easy access.
StreamReader reader = new StreamReader(dataStream);
// Read the content.
string responseFromServer = reader.ReadToEnd();
// Clean up the streams.
reader.Close();
dataStream.Close();
response.Close();

由于某种原因我总是得到 404WebException on

WebResponse response = request.GetResponse();

我已经检查过 URL 应该是正确的。是我格式化帖子 URL 的方式还是我犯了其他错误？

编辑，添加 webconfig 路由：

public static void Register(HttpConfiguration config)
{
    // Web API configuration and services

    // Web API routes
    config.MapHttpAttributeRoutes();

    config.Routes.MapHttpRoute(
        name: "DefaultApi",
        routeTemplate: "api/{controller}/{action}/{id}",
        defaults: new { id = RouteParameter.Optional }
    );
}

你可以使用multipart/form-data来传输文件。以下是如何在 Web API 操作中读取上传文件内容的示例：

[HttpPost]
[Route("api/upload")]
public async Task<IHttpActionResult> Upload()
{
    if (!Request.Content.IsMimeMultipartContent())
    {
        return this.StatusCode(HttpStatusCode.UnsupportedMediaType);
    }

    var filesProvider = await Request.Content.ReadAsMultipartAsync();
    var fileContents = filesProvider.Contents.FirstOrDefault();
    if (fileContents == null)
    {
        return this.BadRequest("Missing file");
    }

    byte[] payload = await fileContents.ReadAsByteArrayAsync();
    // TODO: do something with the payload.
    // note that this method is reading the uploaded file in memory
    // which might not be optimal for large files. If you just want to
    // save the file to disk or stream it to another system over HTTP
    // you should work directly with the fileContents.ReadAsStreamAsync() stream

    return this.Ok(new
    {
        Result = "file uploaded successfully",
    });
}

现在使用以下命令编写客户端是一项简单的任务HttpClient:

class Program
{
    private static readonly HttpClient client = new HttpClient();

    static void Main()
    {
        string responsePayload = Upload().GetAwaiter().GetResult();
        Console.WriteLine(responsePayload);
    }

    private static async Task<string> Upload()
    {
        var request = new HttpRequestMessage(HttpMethod.Post, "http://localhost:8180/api/upload");
        var content = new MultipartFormDataContent();

        byte[] byteArray = ... get your image payload from somewhere
        content.Add(new ByteArrayContent(byteArray), "file", "file.jpg");
        request.Content = content;

        var response = await client.SendAsync(request);
        response.EnsureSuccessStatusCode();

        return await response.Content.ReadAsStringAsync();
    }
}