Json zu Xml mit xslt

Question

Ich bin in der Lage, JSON zu XML mit XSLT 3.0 zu transformieren. Bei der Transformation in XML-Felder fehlt den Daten in Subjekten die XML-Struktur.

{ 
    "student" : "john", 
    "class" : "Bachelors", 
    "subjects" : "<college><subject><subjects>maths</subjects><term>spring</term></subject></college>" 
} 

XSLT:

<xsl:stylesheet version="3.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> 
    <xsl:output indent="yes"/> 
    <xsl:strip-space elements="*"/> 
    <xsl:param name="jsonText"/> 

    <xsl:template match="@*|node()"> 
    <xsl:copy> 
     <xsl:apply-templates select="@*|node()"/> 
    </xsl:copy> 
    </xsl:template> 

    <xsl:template name="init"> 
    <xsl:apply-templates select="json-to-xml($jsonText)"/> 
    </xsl:template> 
</xsl:stylesheet>` 

Java-Code:

public static void main(String[] args){ 
    final String XSLT_PATH = "src/so/test1/json2xml.xsl"; 
    final String JSON = ""{\n" +" \"student\": \"john\",\n" + 
    " \"class\": \"Bachelors\"\n" + 
    " \"subjects\": \"<college><subject><subjects>maths</subjects><term>spring</term></subject></college>"\n" 
       "}"; 

    OutputStream outputStream = System.out;  
    Processor processor = new Processor(false);  
    Serializer serializer = processor.newSerializer(); 
    serializer.setOutputStream(outputStream);  
    XsltCompiler compiler = processor.newXsltCompiler(); 
    XsltExecutable executable = compiler.compile(new StreamSource(new File(XSLT_PATH)));  
    XsltTransformer transformer = executable.load(); 
    transformer.setInitialTemplate(new QName("init")); 
    transformer.setParameter(new QName("jsonText"), new 
XdmAtomicValue(JSON)); 
    transformer.setDestination(serializer); 
    transformer.transform(); 
} 

Fehler:

Error at char 12 in xsl:apply-templates/@select on line 8 column 58 of json2xml.xsl: 
    FOJS0001: Invalid JSON input: Unescaped control character (xd) 
Exception in thread "main" net.sf.saxon.s9api.SaxonApiException: Invalid JSON input: Unescaped control character (xd) 
     at net.sf.saxon.s9api.XsltTransformer.transform(XsltTransformer.java:599) 
     at com.xmltojson.sampleclass.SimpleJaxp.main(SimpleJaxp.java:49) 
Caused by: net.sf.saxon.trans.XPathException: Invalid JSON input: Unescaped control character (xd) 

Answer 1

ein T hinzufügen emplate

<xsl:template match="string[@key = 'subjects']" xpath-default-namespace="http://www.w3.org/2005/xpath-functions"> 
    <xsl:copy> 
    <xsl:copy-of select="@*"/> 
    <xsl:sequence select="parse-xml(.)/node()"/> 
    </xsl:copy> 
</xsl:template> 

, um sicherzustellen, dass die Zeichenfolgedaten in XML analysiert werden.

Beachten Sie, dass Sie mit XSLT 3.0 <xsl:mode on-no-match="shallow-copy"/> verwenden können, anstatt die Identitätstransformation buchstabieren zu müssen.

Ein kurzes Beispiel ist

<xsl:stylesheet version="3.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> 
    <xsl:output indent="yes"/> 
    <xsl:strip-space elements="*"/> 
    <xsl:param name="jsonText"><![CDATA[{ 
    "student" : "john", 
    "class" : "Bachelors", 
    "subjects" : "<college><subject><subjects>maths</subjects><term>spring</term></subject></college>" 
    }]]></xsl:param> 
    <xsl:mode on-no-match="shallow-copy"/> 
    <xsl:template name="init"> 
     <xsl:apply-templates select="json-to-xml($jsonText)"/> 
    </xsl:template> 

    <xsl:template match="string[@key = 'subjects']" xpath-default-namespace="http://www.w3.org/2005/xpath-functions"> 
    <xsl:copy> 
    <xsl:copy-of select="@*"/> 
    <xsl:sequence select="parse-xml(.)/node()"/> 
    </xsl:copy> 
</xsl:template> 

</xsl:stylesheet> 

, die (wenn sie mit Saxon 9.8.0.3 HE mit it:init laufen) gibt

<?xml version="1.0" encoding="UTF-8"?> 
<map xmlns="http://www.w3.org/2005/xpath-functions"> 
    <string key="student">john</string> 
    <string key="class">Bachelors</string> 
    <string key="subjects"> 
     <college xmlns=""> 
     <subject> 
      <subjects>maths</subjects> 
      <term>spring</term> 
     </subject> 
     </college> 
    </string> 
</map> 

ist hier ein Programm vollständig Java basiert auf der Code-Schnipsel in verschiedenen geschrieben Bearbeitungen:

import java.io.File; 
import java.io.OutputStream; 
import javax.xml.transform.stream.StreamSource; 
import net.sf.saxon.s9api.Processor; 
import net.sf.saxon.s9api.QName; 
import net.sf.saxon.s9api.SaxonApiException; 
import net.sf.saxon.s9api.Serializer; 
import net.sf.saxon.s9api.XdmAtomicValue; 
import net.sf.saxon.s9api.XsltCompiler; 
import net.sf.saxon.s9api.XsltExecutable; 
import net.sf.saxon.s9api.XsltTransformer; 

public class Saxon98HETest1 { 

    public static void main(String[] args) throws SaxonApiException { 
     final String XSLT_PATH = "sheet1.xsl"; 
     String JSON = "{\n" + 
" \"student\" : \"john\",\n" + 
" \"class\" : \"Bachelors\",\n" + 
" \"subjects\" : \"<college><subject><subjects>maths</subjects><term>spring</term></subject></college>\"\n" + 
"}"; 
     testJsonToXml(XSLT_PATH, JSON); 
     System.out.println(); 

     JSON = "{\n" + 
          " \"color\": \"red\",\n" + 
          " \"value\": \"#f00\"\n" + 
          "}"; 
     testJsonToXml(XSLT_PATH, JSON); 
     System.out.println(); 
    } 

    static void testJsonToXml(String xsl, String json) throws SaxonApiException { 
     OutputStream outputStream = System.out; 
     Processor processor = new Processor(false); 
     Serializer serializer = processor.newSerializer(); 
     serializer.setOutputStream(outputStream); 
     XsltCompiler compiler = processor.newXsltCompiler(); 
     XsltExecutable executable = compiler.compile(new StreamSource(new File(xsl))); 
     XsltTransformer transformer = executable.load(); 
     transformer.setInitialTemplate(new QName("init")); 
     transformer.setParameter(new QName("jsonText"), new XdmAtomicValue(json)); 
     transformer.setDestination(serializer); 
     transformer.transform();  
    } 

} 

Wenn kompiliert und ausgeführt gegen Saxon 9.8.0.3 HE I Holen Sie sich die Ausgabe

<?xml version="1.0" encoding="UTF-8"?> 
<map xmlns="http://www.w3.org/2005/xpath-functions"> 
    <string key="student">john</string> 
    <string key="class">Bachelors</string> 
    <string key="subjects"> 
     <college xmlns=""> 
     <subject> 
      <subjects>maths</subjects> 
      <term>spring</term> 
     </subject> 
     </college> 
    </string> 
</map> 

<?xml version="1.0" encoding="UTF-8"?> 
<map xmlns="http://www.w3.org/2005/xpath-functions"> 
    <string key="color">red</string> 
    <string key="value">#f00</string> 
</map> 

und keine Fehler. Das Stylesheet ist, wie oben gezeigt, ohne den voreingestellten param Inhalt:

<?xml version="1.0" encoding="UTF-8"?> 
<xsl:stylesheet version="3.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> 
    <xsl:output indent="yes"/> 
    <xsl:strip-space elements="*"/> 
    <xsl:param name="jsonText"></xsl:param> 
    <xsl:mode on-no-match="shallow-copy"/> 
    <xsl:template name="init"> 
     <xsl:apply-templates select="json-to-xml($jsonText)"/> 
    </xsl:template> 

    <xsl:template match="string[@key = 'subjects']" xpath-default-namespace="http://www.w3.org/2005/xpath-functions"> 
    <xsl:copy> 
    <xsl:copy-of select="@*"/> 
    <xsl:sequence select="parse-xml(.)/node()"/> 
    </xsl:copy> 
</xsl:template> 

</xsl:stylesheet>