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xsltxpathxslt-1.0msxsl

Using xslt get node value at X position

How can I get using xslt, node value at X position, without using foreach

<items>
<item1>x</item1>
<item2>x</item2>
<item3>x</item3>
</items>

This is explained in programming sense:

<xsl:value-of select="Items/Item[2]"/>

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Just to little expand question, in the following xml:

<items>
    <about>xyz</about>
    <item1>
       <title>t1</title>
       <body>b1</body>
    </item1>
    <item2>
       <title>t2</title>
       <body>b2</body>
    </item2>
    <item3>
       <title>3</title>
       <body>3</body>
   </item3>
</items>

How can I select second's item title.

Solution

  • Answer to expanded question. You can use the positional value if you select a node-set of the wanted elements:

    <xsl:value-of select="(items//title)[2]"/>

    or:

    <xsl:value-of select="(items/*/title)[2]"/>

    Note the usage of the parenthesis required to return wanted node-set before selecting by position.

    You can use what you called "in programming sense". However you need * due to the unknown name of the children elements:

    <xsl:value-of select="items/*[2]"/>

    Note that nodes-sets in XSLT are not zero-based. In the way above you are selecting the second item, not the third one.

    You really need position() when you want compare the current position with a number as in:

    <xsl:value-of select="items/*[position()>2]"/>

    to select all item with position grater than 2. Other case where position() is indespensible is when position value is a variable of type string:

    <xsl:variable name="pos" select="'2'"/>
<xsl:value-of select="items/*[position()=$pos]"/>