![]() ![]() ![]() Result.Content = new StringContent(BuidCSV) ZipStream.Write(fileBytes, 0, fileBytes.Length) Var fileBytes = (item) įileEntry = new ZipEntry("DailyReport" ".csv") įileEntry = new ZipEntry("Data2" ".csv") Using (var zipStream = new ZipOutputStream(tempFileStream)) / I do not want to create C:\Data2.zip on server.I want to send file stream directly to HTTP ResponseįileStream tempFileStream = new FileStream( FileMode.Open, VisitsRecordDaily.Add(new ChartRecordDaily(int.Parse(row),Ĭ(visitsRecordDaily.OrderByDescending(q =>įoreach (var item in Charttemp.RecordDaily)īuidCSV = "\"" ssion "\"" "," "\"" item.moy "\"" Request.Dimensions = "ga:month,ga:day,ga:year" īuidCSV = "Session" "," "Date" "\r\n" HttpResponseMessage result = new HttpResponseMessage(HttpStatusCode.OK) ĭ request = .Get( public HttpResponseMessage GetExportData(DateTime st, DateTime ed) How I can Dynamically create zip file (in memory) and return as part of HttpResponseMessage as a zip file. ![]()
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