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POPreserving whitespaces from the XSL stylesheet
 

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  1. POPreserving whitespaces from the XSL stylesheet
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    <p>I'm trying to convert this XML :- </p> <pre><code>&lt;list&gt; &lt;unit&gt; &lt;data1&gt;a&lt;/data1&gt; &lt;data2&gt;b&lt;/data2&gt; &lt;data3&gt;c&lt;/data3&gt; &lt;/unit&gt; &lt;/list&gt; </code></pre> <p>to this :- </p> <pre><code>&lt;list&gt; &lt;unit&gt; &lt;category1&gt; &lt;data1&gt;a&lt;/data1&gt; &lt;data2&gt;b&lt;/data2&gt; &lt;/category1&gt; &lt;category2&gt; &lt;data3&gt;c&lt;/data3&gt; &lt;/category2&gt; &lt;/unit&gt; &lt;/list&gt; </code></pre> <p>using XSL. I'm using the following XSL:- </p> <pre><code>&lt;xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:s="some_namespace"&gt; &lt;xsl:template match="@*|node()"&gt; &lt;xsl:copy&gt; &lt;xsl:apply-templates select="@*|node()" /&gt; &lt;/xsl:copy&gt; &lt;/xsl:template&gt; &lt;xsl:template match="//s:unit" xml:space="preserve"&gt; &lt;xsl:copy&gt; &lt;category1&gt; &lt;xsl:apply-templates select="./s:data1"/&gt; &lt;xsl:apply-templates select="./s:data2"/&gt; &lt;/category1&gt; &lt;category2&gt; &lt;xsl:apply-templates select="./s:data3"/&gt; &lt;/category2&gt; &lt;/xsl:copy&gt; &lt;/xsl:template&gt; &lt;/xsl:stylesheet&gt; </code></pre> <p>Now, this preserves the indentation within but completely messes it up w.r.t. list. This is what I get :- </p> <pre><code> &lt;list&gt; &lt;unit&gt; &lt;category1&gt; &lt;data1&gt;a&lt;/data1&gt; &lt;data2&gt;b&lt;/data2&gt; &lt;/category1&gt; &lt;category2&gt; &lt;data3&gt;c&lt;/data3&gt; &lt;/category2&gt; &lt;/unit&gt; &lt;/list&gt; </code></pre> <p>What am I missing here? </p>
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    1. USDimitre Novatchev
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      1. USSean B. Durkin
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    1. COWhat are you searching for? Indentation?
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      1. USEmiliano Poggi
    2. COyes. indentation in the output xml.
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    3. COThis is implementation - dependent. Most processors produce good indentation when `<xsl:output indent="yes"/>` is used -- together or without `<xsl:strip-space select="*"/>` . Which XSLT processor are you using?
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      1. USDimitre Novatchev
 

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