XPath Select All Attribute Name Value Pairs from Entire XML Document

0
votes

I am new to XPath and XML. I was wondering if there is an XPath expression to select all the attribute names and their respective values from an entire XML document.

As a simpler example to what I have:

<bookstore>
    <book category="COOKING">
        <title lang="en">Everyday Italian</title>
        <author>Giada De Laurentiis</author>
        <year>2005</year>
        <price>30.00</price>
            <money currency="Dollars"></money>
    </book>
    <book category="CHILDREN">
        <title lang="en">Harry Potter</title>
        <author>J K. Rowling</author>
        <year>2005</year>
        <price>29.99</price>
            <money currency="Dollars"></money>
    </book>
    <book category="WEB">
        <title lang="en">XQuery Kick Start</title>
        <author>James McGovern</author>
        <author>Per Bothner</author>
        <author>Kurt Cagle</author>
        <author>James Linn</author>
        <author>Vaidyanathan Nagarajan</author>
        <year>2003</year>
        <price>49.99</price>
            <money currency="Euros"></money>
    </book>
        <book category="WEB">
        <title lang="en">Learning XML</title>
        <author>Erik T. Ray</author>
        <year>2003</year>
        <price>39.95</price>
            <money currency="Dollars"></money>
    </book>
</bookstore>

I want this:

category : COOKING
lang : en
currency : Dollars

category : CHILDREN
lang : en
currency : Dollars

category : WEB
lang : en
currency : Euros

category : WEB
lang : en
currency : Dollars

The thing is I want this expression to work for any XML document. Any other suggestions as to how I can do this are welcome as well. Thank you.

1
xmlxsltxpath
It is possible.. Please provide an example input and the expected output. – Lingamurthy CS
Strictly speaking, the value of an attribute is not a node and cannot be selected by an XPath expression. But it is certainly possible to output a list of name/value pairs using XSLT. Show us (at least) the expected output. – michael.hor257k
I provided a simpler example. My document is to large to share but the goal is the same. Thank you. – pseudorandom
"I want this expression to work for any XML document" In your example, there is a clear grouping by book. That cannot work for any document - XML is far more extensible than that. – michael.hor257k
@michael.hor257k: +1 for your helpful explanations. However, just because something is not a node doesn't mean it can't be "selected" (in some sense) by an XPath expression. E.g. string(/*/book[1]/@category) I'm sure you know this, but the way you phrased it might give the unaware a mistaken impression. – LarsH

1 Answers

1
votes

This will work for your input (after you fix the mismatching tags <money> and </currency>!) and other reasonably similar XML documents - but it cannot be guaranteed to work with any XML document:

XSLT 1.0

<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text" encoding="UTF-8"/>
<xsl:strip-space elements="*"/>

<xsl:template match="/*/*">
    <xsl:for-each select=".//@*">
        <xsl:value-of select="concat(name(), ': ', ., '&#10;')"/>
    </xsl:for-each>
    <xsl:text>&#10;</xsl:text>
</xsl:template>

</xsl:stylesheet>

OTOH, this will work with any XML - but the output will be just a single list of all attribute name/value pairs in the entire document, with no separation:

XSLT 1.0

<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text" encoding="UTF-8"/>

<xsl:template match="/">
    <xsl:for-each select="//@*">
        <xsl:value-of select="concat(name(), ': ', ., '&#10;')"/>
    </xsl:for-each>
</xsl:template>

</xsl:stylesheet>

Result

category: COOKING
lang: en
currency: Dollars
category: CHILDREN
lang: en
currency: Dollars
category: WEB
lang: en
currency: Euros
category: WEB
lang: en
currency: Dollars