使用 ember-upload 上传文件，如何用 servicestack 的附加数据填充请求？

Question

作为介绍，我在通过 REST 用 ember.js 编写的服务堆栈和应用程序之间的通信存在问题，我正在使用 ember-uploader 组件将文件上传到服务堆栈。

查看hbs：

<table class="table table-bordered table-hover">
{{file-upload}}
</table>

咖啡脚本中的组件

  ABC.FileUploadComponent = Ember.FileField.extend(
  url: "/api/upload"
  filesDidChange: (->
    uploadUrl = @get("url")
    console.log uploadUrl
    files = @get("files")
    test = { fileName: "test" }
    uploader = Ember.Uploader.create(
        url: uploadUrl
    )

    uploader.upload(files[0],test) unless Ember.isEmpty(files)
    console.log files
    return
  ).observes("files")
)

javascript 中的组件

ABC.FileUploadComponent = Ember.FileField.extend({
  url: "/api/upload",
  filesDidChange: (function() {
    var files, test, uploadUrl, uploader;
    uploadUrl = this.get("url");
    console.log(uploadUrl);
    files = this.get("files");
    test = {
      fileName: "test"
    };
    uploader = Ember.Uploader.create({
      url: uploadUrl,
      data: test
    });
    if (!Ember.isEmpty(files)) {
      uploader.upload(files[0], test);
    }
    console.log(files);
  }).observes("files")
});

我的服务模式：

namespace ABC.Service.ServiceModel
{
    public class Upload
    {

        [Route("/upload")]
        public class UploadRequest : IRequiresRequestStream
        {
            public System.IO.Stream RequestStream { set; get; }

            public object FileName { set; get; }

        }

        public class UploadResponse
        {
            public int Successed { set; get; }

        }
    }
}

我的服务方式

namespace ABC.Service.Service
{
    public class UploadService : ServiceBase // Service base inherites from ServiceStack.Service
    {
        public Upload.UploadResponse Post(Upload.UploadRequest request)
        {
            var req = base.Request;
            var reqThatIwant = request.FileName;

            return new Upload.UploadResponse() { Successed = 1 };
        }

    }
}

这是手表的屏幕：

所以我的问题是，我必须如何更改代码以将标记为“2”的数据转换为标记为“1”的请求对象（在屏幕上标记）？

Answer 1

处理原始请求流

当您使用IRequiresRequestStream 时，您是在说您想要take over deserializing the Request 并以流的形式访问原始输入HTTP 请求正文。因此，ServiceStack 不会尝试从请求正文中读取，而是注入 HTTP 请求流 - 在这种情况下，它能够填充的唯一请求 DTO 参数是 /pathinfo 或 ?QueryString 上的参数，例如：

[Route("/upload/{FileName}")]
public class Upload : IRequiresRequestStream
{
    public Stream RequestStream { set; get; }

    public string FileName { set; get; }
}

访问 FormData HTTP POST

但是，如果 JavaScript 组件向您发送 HTTP POST FormData（即application/x-www-form-urlencoded 或 multipart/form-data），那么您不太可能希望将其视为原始请求流，而是访问 Request.FormData 或 Request.Files已发布。

处理文件上传示例

根据您的屏幕截图，HTTP 请求 Content-Type 是 multipart/form-data，在这种情况下您很可能能够使用 access any uploaded files using Request.Files。

Live Demos 中提供了一些访问 HTTP 上传文件的示例：

Imgur - Save uploaded files to a MemoryStream

public object Post(Upload request)
{
    foreach (var uploadedFile in Request.Files
       .Where(uploadedFile => uploadedFile.ContentLength > 0))
    {
        using (var ms = new MemoryStream())
        {
            uploadedFile.WriteTo(ms);
            WriteImage(ms);
        }
    }
    return HttpResult.Redirect("/");
}

Rest Files - Save to FileSystem

public void Post(Files request)
{
    var targetDir = GetPath(request);

    var isExistingFile = targetDir.Exists
        && (targetDir.Attributes & FileAttributes.Directory) != FileAttributes.Directory;

    if (isExistingFile)
        throw new NotSupportedException(
        "POST only supports uploading new files. Use PUT to replace contents of an existing file");

    if (!Directory.Exists(targetDir.FullName))
        Directory.CreateDirectory(targetDir.FullName);

    foreach (var uploadedFile in base.Request.Files)
    {
        var newFilePath = Path.Combine(targetDir.FullName, uploadedFile.FileName);
        uploadedFile.SaveTo(newFilePath);
    }
}

HTTP Benchmarks - Handle multiple and .zip uploaded files

public object Post(UploadTestResults request)
{
    //...
    foreach (var httpFile in base.Request.Files)
    {
        if (httpFile.FileName.ToLower().EndsWith(".zip"))
        {
            using (var zip = ZipFile.Read(httpFile.InputStream))
            {
                var zipResults = new List<TestResult>();
                foreach (var zipEntry in zip)
                {
                    using (var ms = new MemoryStream())
                    {
                        zipEntry.Extract(ms);
                        var bytes = ms.ToArray();

                        var result = new MemoryStream(bytes).ToTestResult();
                        zipResults.Add(result);
                    }
                }
                newResults.AddRange(zipResults);
            }
        }
        else
        {
            var result = httpFile.InputStream.ToTestResult();
            newResults.Add(result);
        }
    }
}