Question

我可以访问Schematron xsl文件和Schematron sch文件。如何使用C＃将其转换为XSLT样式表？

xsl + sch - ＆gt; [??? XSLT处理器??? ] - ＆gt; xslt stylesheet

Answer 1

回答我自己的问题...... 遗憾的是，Schematron文件只支持使用XslCompiledTransform的非常简单的语法。到SAXON看它有效:(

        string xmlFile = @"sch\patient.sch";
        string xslFile = @"xsl\conformance1-5.xsl";
        XslCompiledTransform xsltransform = new XslCompiledTransform();
        xsltransform.Load(xslFile);

        XmlDocument doc = new XmlDocument();
        doc.Load(xmlFile);
        XPathNavigator nav = doc.CreateNavigator();

        System.IO.MemoryStream st = new System.IO.MemoryStream();
        xsltransform.Transform(nav, null, st);
        st.Position = 0;
        System.IO.StreamReader rd = new System.IO.StreamReader(st);
        string xslt = rd.ReadToEnd();

        System.Diagnostics.Debug.WriteLine(xslt);

        XmlReader reader = XmlReader.Create(new System.IO.StringReader(xslt));
        xsltransform.Load(reader);

        var patient = PatientFactory.GeneratePatientBySOAPClasses();
        patient.identifier[0].period.end.value = DateTime.Now.ToString("yyyy-MM-dd");
        patient.identifier[0].period.start.value = DateTime.Now.AddYears(15).ToString("yyyy-MM-dd");
        patient.identifier[0].period.start = null;

        string xml = Serialization.SerializeXML(patient, "http://hl7.org/fhir");

        xml = xml.Replace("xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\" xmlns=\"http://hl7.org/fhir\"", "");
        doc.LoadXml(xml);
        nav = doc.CreateNavigator();

        st = new System.IO.MemoryStream();
        xsltransform.Transform(nav, null, st);
        st.Position = 0;
        rd = new System.IO.StreamReader(st);
        string scematronresult = rd.ReadToEnd();

如何创建XSLT样式表

1 个答案: