获取内容配置参数

Question

如何使用WebAPI从WebClient控制器返回内容配置参数？

WebApi控制器

    [Route("api/mycontroller/GetFile/{fileId}")]
    public HttpResponseMessage GetFile(int fileId)
    {
        try
        {
                var file = GetSomeFile(fileId)

                HttpResponseMessage response = new HttpResponseMessage(HttpStatusCode.OK);
                response.Content = new StreamContent(new MemoryStream(file));
                response.Content.Headers.ContentDisposition = new System.Net.Http.Headers.ContentDispositionHeaderValue("attachment");
                response.Content.Headers.ContentDisposition.FileName = file.FileOriginalName;

                /********* Parameter *************/
                response.Content.Headers.ContentDisposition.Parameters.Add(new NameValueHeaderValue("MyParameter", "MyValue"));

                return response;

        }
        catch(Exception ex)
        {
            return Request.CreateErrorResponse(HttpStatusCode.InternalServerError, ex);
        }

    }

客户端

    void DownloadFile()
    {
        WebClient wc = new WebClient();
        wc.DownloadDataCompleted += wc_DownloadDataCompleted;
        wc.DownloadDataAsync(new Uri("api/mycontroller/GetFile/18"));
    }

    void wc_DownloadDataCompleted(object sender, DownloadDataCompletedEventArgs e)
    {
        WebClient wc=sender as WebClient;

        // Try to extract the filename from the Content-Disposition header
        if (!String.IsNullOrEmpty(wc.ResponseHeaders["Content-Disposition"]))
        {
           string fileName = wc.ResponseHeaders["Content-Disposition"].Substring(wc.ResponseHeaders["Content-Disposition"].IndexOf("filename=") + 10).Replace("\"", ""); //FileName ok

        /******   How do I get "MyParameter"?   **********/

        }
        var data = e.Result; //File OK
    }

我从WebApi控制器返回一个文件，我在响应内容头中附加文件名，但我也想返回一个属性值。

在客户机中，我可以获得文件名，但是如何获得参数？

Answer 1

如果您正在使用.NET 4.5或更高版本，请考虑使用System.Net.Mime.ContentDisposition类：

string cpString = wc.ResponseHeaders["Content-Disposition"];
ContentDisposition contentDisposition = new ContentDisposition(cpString);
string filename = contentDisposition.FileName;
StringDictionary parameters = contentDisposition.Parameters;
// You have got parameters now

编辑：

否则，您需要根据它的规格说明解析内容配置标头。

下面是一个执行解析的简单类，接近规范：

class ContentDisposition {
    private static readonly Regex regex = new Regex(
        "^([^;]+);(?:\\s*([^=]+)=((?<q>\"?)[^\"]*\\k<q>);?)*$",
        RegexOptions.Compiled
    );

    private readonly string fileName;
    private readonly StringDictionary parameters;
    private readonly string type;

    public ContentDisposition(string s) {
        if (string.IsNullOrEmpty(s)) {
            throw new ArgumentNullException("s");
        }
        Match match = regex.Match(s);
        if (!match.Success) {
            throw new FormatException("input is not a valid content-disposition string.");
        }
        var typeGroup = match.Groups[1];
        var nameGroup = match.Groups[2];
        var valueGroup = match.Groups[3];

        int groupCount = match.Groups.Count;
        int paramCount = nameGroup.Captures.Count;

        this.type = typeGroup.Value;
        this.parameters = new StringDictionary();

        for (int i = 0; i < paramCount; i++ ) {
            string name = nameGroup.Captures[i].Value;
            string value = valueGroup.Captures[i].Value;

            if (name.Equals("filename", StringComparison.InvariantCultureIgnoreCase)) {
                this.fileName = value;
            }
            else {
                this.parameters.Add(name, value);
            }
        }
    }
    public string FileName {
        get {
            return this.fileName;
        }
    }
    public StringDictionary Parameters {
        get {
            return this.parameters;
        }
    }
    public string Type {
        get {
            return this.type;
        }
    }
} 

然后你就可以这样使用它了：

static void Main() {        
    string text = "attachment; filename=\"fname.ext\"; param1=\"A\"; param2=\"A\";";

    var cp = new ContentDisposition(text);       
    Console.WriteLine("FileName:" + cp.FileName);        
    foreach (DictionaryEntry param in cp.Parameters) {
        Console.WriteLine("{0} = {1}", param.Key, param.Value);
    }        
}
// Output:
// FileName:"fname.ext" 
// param1 = "A" 
// param2 = "A"  

在使用这个类时，唯一需要考虑的是它不处理没有双引号的参数(或文件名)。

编辑2:

它现在不需要引用就可以处理文件名。

Answer 2

可以使用以下框架代码解析内容配置：

var content = "attachment; filename=myfile.csv";
var disposition = ContentDispositionHeaderValue.Parse(content);

然后把处理实例中的碎片取下来。

disposition.FileName 
disposition.DispositionType

Answer 3

对于.NET Core3.1和更多版本，最简单的解决方案是：

using var response = await Client.SendAsync(request);
response.Content.Headers.ContentDisposition.FileName