Re: Re: Defining a Variable that depends on an unknown

> I don't think you can do this join in a single XPath 1.0 expression. It
> either needs XPath 2.0, or some kind of xsl:for-each logic if you really
> must do it in XSLT 1.0.

If you can find a solution in XPath 2.0 or XSLT 1.0, I will take it.

I don't know how you can define a variable using xsl:for-each: you
cannot redefine variables and there is no breaking from a loop.

-- 
Andrew Ferk