Thread: xsl:variable question
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Old July 2nd, 2009, 05:02 AM
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First get rid of the inner quotes.

A variable holds a value (in this case, a set of nodes), it doesn't hold a path expression. When you write <xsl:variable name="x" select="foo/bar"/>, it evaluates foo/bar in the current context to return a set of nodes. You seem to be under the incorrect impression that the value of the variable will be a path expression that is evaluated in the context where it is used.

In XSLT 2.0, you can write f/b[foo/bar = ('C1', 'C2')]. In XSLT 1.0, you can write f/b[foo/bar[. = 'C1' or . = 'C2']].
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