Subject: [xsl] how to remove xmls="" From: "Joga Singh Rawat jrawat@xxxxxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Date: Wed, 4 Sep 2019 06:03:20 -0000 |
Dear Expert, I am getting <front xmlns=""> and <body xmlns=""> as output from below combination of input xml and xslt. Please let us know how to remove xmlns=bb. INPUT <article xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns:oasis="http://www.niso.org/standards/z39-96/ns/oasis-exchange/table" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:ali="http://www.niso.org/schemas/ali/1.0/" article-type="research-article" dtd-version="1.1" xml:lang="en"> <front> ... </front> <body> ...</body> </article> XSLT <xsl:template match="article"> <article xmlns="http://specifications.silverchair.com/xsd/1/18/SCJATS-journalpublishin g.xsd"> <xsl:if test="@article-type"> <xsl:attribute name="article-type" select="@article-type"/> </xsl:if> <xsl:if test="@xml:lang"> <xsl:attribute name="xml:lang" select="@xml:lang"/> </xsl:if> <xsl:attribute name="xsi:schemaLocation">http://specifications.silverchair.com/xsd/1/19/SCJA TS-journalpublishing.xsd http://specifications.silverchair.com/xsd/1/19/SCJATS-journalpublishing.xsd</ xsl:attribute> <xsl:apply-templates/> </article> </xsl:template> <xsl:template match="node() | @*"> <xsl:copy copy-namespaces="no" inherit-namespaces="no"> <xsl:apply-templates select="node() | @*[not(name()='xmlns')]"/> </xsl:copy> </xsl:template> OUTPUT <article xmlns="http://specifications.silverchair.com/xsd/1/18/SCJATS-journalpublishin g.xsd" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:mml="http://www.w3.org/1998/Math/MathML" article-type="research-article" xml:lang="en" xsi:schemaLocation="http://specifications.silverchair.com/xsd/1/19/S CJATS-journalpublishing.xsd http://specifications.silverchair.com/xsd/1/19/SCJATS-journalpublishing.xsd"> <front xmlns=""> ... </front> <body xmlns=""> ... </body> </article> thanks in advance ...JSR
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