I want to get all ancestors of current node:
XML:
<root>
<item title="a">
<item title="b">
<item title="c"></item> <!--CURRENT-->
<item title="d"></item>
</item>
<item title="x">
<item title="y"></item>
<item title="z"></item>
</item>
</item>
</root>
Result:
<item title="a">...</item>
<item title="b">...</item>
Edit: Answers with axes ancestor are fine. My problem was elsewhere, in XSLT
XSLT:
<xsl:variable name="curr" select="//item[@title = 'c']"></xsl:variable>
<xsl:variable name="test" select="$curr/ancestor::item"></xsl:variable>
<xsl:for-each select="$test/item">
<xsl:value-of select="@title"></xsl:value-of>
</xsl:for-each>
Returns:
bcdx
Edit2: for dimitre and for all who have a similar problem
All the answers to my question were good.
Just XSLT (up) returns to me a strange result and @Mads Hansen corrected me.
FINAL WORKING EXAMPLE:
XML:
<?xml version="1.0" encoding="utf-8"?>
<root>
<item title="a">
<item title="b">
<item title="c"></item>
<item title="d"></item>
</item>
<item title="x">
<item title="y"></item>
<item title="z"></item>
</item>
</item>
</root>
XSLT:
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<xsl:variable name="curr" select="//item[@title = 'c']"></xsl:variable>
<xsl:variable name="test" select="$curr/ancestor::item"></xsl:variable>
<xsl:for-each select="$test">
<xsl:value-of select="@title"></xsl:value-of>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
Returns:
ab
Congradulations to Adam for a very quick first answer.
Your listed expected result does not match your words. the root element also an ancestor node and the document is also an ancestor node.
ancestor::node()
... will return a sequence in this order:
item[@title='b']
item[@title='a']
root
element (a.k.a. the document element)/
To get the specific result you listed, you need:
ancestor::item/.
The effect of the /. is to change the ordering back to forward document order. The native order of ancestor:: is reverse document order.
This style-sheet (with OP's input)...
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text"/>
<xsl:template match="/">
<xsl:for-each select="//item[@title='c']">
<xsl:value-of select="ancestor::item[1]/@title" />
<xsl:value-of select="ancestor::item[2]/@title" />
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
... will output 'ba' illustrating the point that ancestor:: is indeed a reverse axis. And yet this style-sheet ...
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text"/>
<xsl:template match="/">
<xsl:for-each select="//item[@title='c']">
<xsl:value-of select="(ancestor::item/@title)[1]" />
<xsl:value-of select="(ancestor::item/@title)[2]" />
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
... has the opposite result 'ab' . This is instructive because it shows that in XSLT 1.0 (not so in XSLT 2.0), the brackets remove the reverse nature, and it becomes a document ordered node-set.
The OP has asked about a transform something like....
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text"/>
<xsl:template match="/">
<xsl:for-each select="//item[@title='c']">
<xsl:for-each select="ancestor::item">
<xsl:value-of select="@title" />
</xsl:for-each>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
This one returns 'ab' (in XSLT 2.0 it would return 'ba'). Why? Because in XSLT 1.0, the xsl:for-each instruction ignores the reverse-ness of the axis and processes in document order (unless an xsl:sort instruction says otherwise).