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Re: [xsl] how to achieve this ?

2012-01-23 04:44:43
Sounds like HTML/javascript problem, is another list.

On 1/23/12, Roelof Wobben <rwobben(_at_)hotmail(_dot_)com> wrote:

Thanks,



It solved the id problem but the menu still don't work right.

It still don't work as as accordian menu. See http://test.tamarawobben.nl



Roelof









---------------------------------------- > From: bbosgoed(_at_)bkwi(_dot_)nl 
Date:
Mon, 23 Jan 2012 08:35:36 +0100 > To: 
xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com >
Subject: Re: [xsl] how to achieve this ? > > > Op 23 jan 2012, om 08:26
heeft Roelof Wobben het volgende geschreven: > > Not knowing how the input
looks like, i would say that: > <ul class="menu" id={concat('menu',
@value)}> > > should do it for you, its a short notation for xsl:value-of
select > > > > For a menu I have to take care that every <menu class> get a
unique id. > > > > I thought this would work : > > > > <xsl:template
match="year"> > > <ul class="menu" id= "concat ('menu', @value)" > > > <li>
<a href="#"> <xsl:value-of select="@value"/> </a> > > <ul
class="acitem"> > > <xsl:apply-templates select="month" /> > > </ul> > >
</li> > > </ul> > > </xsl:template> > > > > But it don't work. > > > > Do I
have to use here <xsl:value-of> or is there a better way ? > > >
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