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  1. 05-24-2009, 09:02 AM #1
    khdani
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    Default xml sort using xslt

    hello,
    i try to sort xml using this xslt:
    Java Code: 
    <?xml version="1.0" ?>
<xsl:stylesheet xmlns:xsl="link to w3c" Version="1.0">
<xsl:output method="text"/>
<xsl:template match="Messages">
<xsl:apply-templates>
<xsl:sort select="EMAIL" order="descending"/>
</xsl:apply-templates>
</xsl:template>
<xsl:template match = "Message">
ID: <xsl:apply-templates select="ID"/>EMAIL: <xsl:apply-templates select="EMAIL"/>
SUBJECT: <xsl:apply-templates select="SUBJECT"/>
DATE: <xsl:apply-templates select ="DATE"/>
BODY: <xsl-apply-templates select = "BODY"/>
<xsl:text></xsl:tetx>
</xsl:template></xsl:stylesheet>
    and this code in java:
    Java Code: 
    		Source xsl = new StreamSource(xslBuilder.toString());
		Source xml = new StreamSource(this.path);
			DOMResult result = new DOMResult(this.doc);
			TransformerFactory transFactory = TransformerFactory.newInstance();
			javax.xml.transform.Transformer transformer = transFactory.newTransformer(xsl);
			transformer.transform(xml, result);
    (the xsl is in string)
    but i get an error: no protocol
    and fatal error: could not compile stylesheet

    please help...
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  2. 05-24-2009, 04:33 PM #2
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    OrangeDog
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    Is your xslBuilder correctly initialised? Is this.path correct? Does both your xml and your xsl follow the w3c standards exactly?

    (possibly useful link: XSL Transformations (XSLT) Version 2.0)
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  3. 05-24-2009, 04:57 PM #3
    khdani
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    xslBuilder is StringBuilder, the whole xsl is created in StringBuilder
    'this.path' is also correct
    both xml and xsl follow w3c:
    xml
    Java Code: 
    <?xml version="1.0" encoding="utf-8" ?>
<Mail>
	<Message>
		<ID>1</ID>
		<EMAIL>mail@mail.com</EMAIL>
		<SUBJECT>Subject1</SUBJECT>
		<DATE>10/05/2009</DATE>
		<BODY><![CDATA[Hello :)]]></BODY>
	</Message>
</Mail>
    xsl
    Java Code: 
    		StringBuilder xslBuilder = new StringBuilder();
		xslBuilder.append("<?xml version=\"1.0\" encoding =\"UTF-8\" ?>");
		xslBuilder.append("<xsl:stylesheet xmlns:xsl=\"link wo w3c" Version=\"1.0\">");
		xslBuilder.append("<xsl:output method=\"text\"/>");
		xslBuilder.append("<xsl:template match=\"Messages\">");
		xslBuilder.append("<xsl:apply-templates>");
		xslBuilder.append("<xsl:sort select=\"" + sortArg+"\" order=\"" + order + "\"/>");
		xslBuilder.append("</xsl:apply-templates>");
		xslBuilder.append("</xsl:template>");
		
		xslBuilder.append("<xsl:template match = \"Message\">");
		xslBuilder.append("ID: <xsl:apply-templates select=\"ID\"/>");
		xslBuilder.append("EMAIL: <xsl:apply-templates select=\"EMAIL\"/>");
		xslBuilder.append("SUBJECT: <xsl:apply-templates select=\"SUBJECT\"/>");
		xslBuilder.append("DATE: <xsl:apply-templates select =\"DATE\"/>");
		xslBuilder.append("BODY: <xsl-apply-templates select = \"BODY\"/>");
		xslBuilder.append("<xsl:text>");
		xslBuilder.append("</xsl:tetx>");
		
		xslBuilder.append("</xsl:template>");
		
		xslBuilder.append("</xsl:stylesheet>");
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  4. 05-24-2009, 10:36 PM #4
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    OrangeDog
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    Does this operation work when not executed in Java? That is, is the output as expected if you simply apply the stylesheet to the xml and view in a browser?
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  5. 05-25-2009, 01:19 PM #5
    khdani
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    well actually i did had some errors in xsl, but i fixed them and verified with XMLSpy. XMLSpy transforms the xml correctly but the java code gives same error
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  6. 05-25-2009, 01:21 PM #6
    RamyaSivakanth's Avatar
    RamyaSivakanth
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    Could u paste the complete code with xsl?
    Ramya:cool:
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  7. 05-25-2009, 01:53 PM #7
    khdani
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    i think i fixed the problem,
    i should set systemID to some xslt file.
    but i can't parse the result dom.
    Java Code: 
    			DOMSource source = new DOMSource(this.doc);
			DOMResult result = new DOMResult();			
			TransformerFactory transFactory = TransformerFactory.newInstance();
			javax.xml.transform.Transformer transformer = transFactory.newTransformer(xsl);
			transformer.transform(source, result);	
			this.doc = (Document)result.getNode();
    if i call:
    [code]
    this.doc.getDocumentElement().getElementsByTagName ("tagname");
    i get an empty list...
    but if i check in debugger the result node indeed has the xml dom tree transformed...
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  8. 05-25-2009, 02:08 PM #8
    khdani
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    fixed that too
    thank you
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  9. 06-20-2010, 09:04 AM #9
    DSS
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    Sort XML by Date using XSL | Easy Tips and Tricks

    Follow this link for the solution. Works great for me. Cheers!
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