コード例 #1
0
 public DataContextReport()
 {
     try
     {
         var command = new CommandSnuOneAuto();
         ModelApi = ConfigFile.ResultGetTemplate <ModelServiceDataBase>(ConfigFile.ServiceModelInventory);
         ModelApi.ForEach(service => service.ApiService = string.Format(service.ApiService, ConfigFile.HostNameService));
         Report = new ReportXlsxMethod(ConfigFile.ExcelReportFile);
         ReportJournalAndFile  = new ReportJournalMethod(ConfigFile.PathJurnal, ConfigFile.PathInn, ModelApi);
         LabelModel            = new LabelModel();
         DeleteJournal         = new DelegateCommand(() => { ReportJournalAndFile.DeleteXmlReportJournal(); });
         DeleteReport          = new DelegateCommand(() => { Report.DeleteReportFile(); });
         OpenReport            = new DelegateCommand(() => { Report.OpenReport(); });
         OpenFile              = new DelegateCommand(() => { command.ConvertXslToXmlAndOpen(Report, ReportJournalAndFile, ConfigFile.ExcelReportFile); });
         FileToServerApiReport = new DelegateCommand(() => command.FileToServerApiReport(LabelModel, ModelApi, ReportJournalAndFile));
         Update = new DelegateCommand(() =>
         {
             ReportJournalAndFile.AddFileXml(ConfigFile.PathInn);
             ReportJournalAndFile.AddJournal(ConfigFile.PathJurnal);
         });
     }
     catch (Exception e)
     {
         Console.WriteLine(e);
         throw;
     }
 }
コード例 #2
0
        /// <summary>
        /// Команда конвертации xml в Excel и открытия файла Excel
        /// </summary>
        /// <param name="reportExcel">отч</param>
        /// <param name="reportJournal"></param>
        /// <param name="pathReport"></param>
        public void ConvertXslToXmlAndOpen(ReportXlsxMethod reportExcel, ReportJournalMethod reportJournal,
                                           string pathReport)
        {
            var fileFullPath =
                LibaryXMLAuto.Converts.ConvertXmlToXslx.ConvertXmltoXlsx.ConvertXmlToXls(reportJournal.XmlFile.Path,
                                                                                         pathReport);

            reportExcel.UpdateColectFile(fileFullPath.DirectoryName);
            reportJournal.OpenFile(fileFullPath.FullName);
        }
コード例 #3
0
        /// <summary>
        /// Отправка файла на сервер для анализа или отчета
        /// </summary>
        /// <param name="model">Возвращаемая модель</param>
        /// <param name="serverApi">Моделька с api</param>
        /// <param name="reportJournal">Выбор файла в журнале</param>
        public void FileToServerApiReport(LabelModel model, List <ModelServiceDataBase> serverApi, ReportJournalMethod reportJournal)
        {
            var modelFileApi = serverApi.FirstOrDefault(api => api.ModelNameFileXml == reportJournal.XmlFile.Name);

            if (modelFileApi != null)
            {
                XmlConvert xmlConverter = new XmlConvert();
                var        type         = Type.GetType($"{modelFileApi.TypeFileNameSpaceClass},{modelFileApi.FileNameDll}");
                var        reports      = xmlConverter.DeserializationXmlToClass(reportJournal.XmlFile.Path, type);
                var        report       = (ModelPathReport)ResultPost(modelFileApi.ApiService, reports);
                model.MessageReport = report.Note;
                model.Url           = report.Url;
                model.Color         = Brushes.Green;
            }
            else
            {
                model.MessageReport = $"Для данной команды выбран не тот файл!!!";
                model.Color         = Brushes.Red;
            }
        }
コード例 #4
0
 /// <summary>
 /// Команда Update вынесена в отдельный класс чтобы не загромождать логику MVVM
 /// </summary>
 /// <param name="xmlUseMethod">XmlUseMethod Метод файла xml</param>
 /// <param name="reportJournalMethod">ReportJournalMethod метод журнала файла xml</param>
 /// <param name="pathFileInn">Путь к файлу FullName</param>
 /// <param name="pathJournal">Просто путь к журналу</param>
 /// <param name="pathInn">Просто путь к ИНН</param>
 public void UpdateModel(XmlUseMethod xmlUseMethod, ReportJournalMethod reportJournalMethod, string pathFileInn, string pathJournal, string pathInn)
 {
     xmlUseMethod.UpdateFileXml(pathFileInn);
     reportJournalMethod.AddFileXml(pathInn);
     reportJournalMethod.AddJournal(pathJournal);
 }