In order to set a variable based on a condition, you have to do something
like this:-
<xsl:variable name="stadt">
<xsl:choose>
<xsl:when test="$lang = '1' ">Stadt</xsl:when>
<xsl:otherwise>city</xsl:otherwise>
</xsl:choose>
</xsl:variable>
Ritu Kama
Sarvega Inc
745, McLintock Dr
Burr Ridge
IL - 60527
----- Original Message -----
From: "Hubert Holtz" <Turnhose_alt(_at_)gmx(_dot_)net>
To: <XSL-List(_at_)lists(_dot_)mulberrytech(_dot_)com>
Sent: Tuesday, January 28, 2003 1:05 PM
Subject: [xsl] value of variable inside a condition doesn't work?
Hy,
first of all I know there is this <i18n> thing to make multilanguage
sites, but that's not the topic.
I have enabled xslt-with-parameter in my sitemap, in my xsl file i have a
global parameter 'lang' this is the parameter which should
contain the value of the equal url-parameter, so far so good.
Now I want to output text fragments in 2 languages, depending on this
parameter, so I thought of sth. like this:
-- code --
<xsl:if test="($lang)='1' ">
<xsl:variable name="stadt" select="Stadt"/>
<xsl:variable name="Texteingabe" select="Hier Text eingeben"/>
<xsl:variable name="berichtstatus" select="aktuell"/>
</xsl:if>
<xsl:if test="($lang)='2' ">
<xsl:variable name="stadt" select="city"/>
<xsl:variable name="Texteingabe" select="Please enter text"/>
<xsl:variable name="berichtstatus" select="current"/>
</xsl:if>
..
<td><xsl:value-of select="$stadt"/></td>
-- code --
But I get an error message, that there is no 'stadt' variable, if I delete
the <xsl.if> part then there is no error message, but then I can't change
the
value of the variable depending on the 'lang' paramter, of course.
So could it be that variables can't be set in an if statement and if
that's true what would be the solution?
thanks
Homer30
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list