RE: [xsl] root siblings contain xmlns=""

[christoph.klocker@xxxxxx]

Hi Mike,

tried to follow your advice and declared the namespace in the stylesheet
element.
unfortunately I still get the wrong output.
below my testfiles.

test.xml:
<test >
    <element1>test</element1>
    <element2>test</element2>
</test>

test.xsl:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="2.0"
    xmlns:xs="http://www.w3.org/2001/XMLSchema";  xmlns="a-namespace-uri"
exclude-result-prefixes="xs">
    <xsl:template match="node()|@*">
        <xsl:copy>
            <xsl:apply-templates/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="test">
        <Fxml Version="2_0_2">
            <xsl:apply-templates/>
        </Fxml>
    </xsl:template>
</xsl:stylesheet>

output:
<?xml version="1.0" encoding="UTF-8"?><Fxml xmlns="a-namespace-uri"
Version="2_0_2">
    <element1 xmlns="">test</element1>
    <element2 xmlns="">test</element2>
</Fxml>

my desired output:
<?xml version="1.0" encoding="UTF-8"?><Fxml xmlns="a-namespace-uri"
Version="2_0_2">
    <element1>test</element1>
    <element2 >test</element2>
</Fxml>

thx Christoph




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