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.
Answer
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]"/>