You might select the nearest preceding sibling like this,
<xsl:for-each select="/root/orange/preceding-sibling::*">
<xsl:if test="position() = last()">
<xsl:value-of select="name(.)"/>
...
</xsl:if>
</xsl:for-each>
Hope this is what you want.
Lisa
-----Original Message-----
From: Duane Gran [mailto:duane(_dot_)gran(_at_)gmail(_dot_)com]
Sent: Tuesday, January 03, 2006 1:20 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] trouble with preceding axis
Lisa,
Thank you very much. If you don't mind, I've run into another issue
today with referencing the preceding element. Here is an example:
<root>
<apple/>
<pear/>
<orange/>
</root>
Given an xpath to orange (/root/orange) I want to reference the
previous element which is a peer. I tried the following with no
success:
/root/preceding::orange
I'm sure this is dreadfully simple, but I'm drawing a blank and I'm
not seeing the answer in the tutorials I'm reading.
Duane
On Jan 3, 2006, at 2:52 PM, Xia Li wrote:
If you want to select the preceding <item> element of the <item>
element
whose child <label> element satisfies the predicate [label =
'subTitle'], you might try the following XPath expression,
//item[label = 'subTitle']/preceding::item
Lisa
-----Original Message-----
From: Duane Gran [mailto:duane(_dot_)gran(_at_)gmail(_dot_)com]
Sent: Tuesday, January 03, 2006 10:35 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] trouble with preceding axis
I'm having some trouble using the preceding axis in XPath and would
be much obliged if someone could steer me in the right direction. I
have the following document:
<root>
<item order="3">
<label>Title</label>
</item>
<item order="4">
<label>subTitle</label>
</item>
</root>
I have the following XPath to select the second item, based on the
value of its label child:
//item[label = 'subTitle']
Now I want to modify this so that I refer to the preceding item
element, the first one in the list. I tried the following without
success (it returns the second item element):
//preceding::item[label = 'subTitle']
Suggestions are most welcome on this.
Duane
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