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Re: [xsl] how to achieve this ?
Subject: Re: [xsl] how to achieve this ? From: Michel Hendriksen <michel.hendriksen@xxxxx> Date: Mon, 23 Jan 2012 11:44:26 +0100 |
Sounds like HTML/javascript problem, is another list. On 1/23/12, Roelof Wobben <rwobben@xxxxxxxxxxx> 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@xxxxxxx > Date: > Mon, 23 Jan 2012 08:35:36 +0100 > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > > 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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