Hi;
Thanks Jenni and Kenny for your replies. Both of your fixes worked well and
I learned a lot.
Now I have another problem but I will post that in another thread.
thansk again.
Myee
Jeni Tennison
<jeni(_at_)jenitenniso To:
Myee(_dot_)Riri(_at_)hic(_dot_)gov(_dot_)au
n.com> cc:
xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] One Strange
Problem
07/05/2004 09:01 Classification: |-------------------|
PM | ( ) In-Confidence |
Please respond to | ( ) Unclassified |
Jeni Tennison |-------------------|
Hi Myee,
I have a xsl:param called sortby, which I change to the value of
SortVal using javascript in the HTML page.
By default, the param is set to the string 'DateVal' -- in other
words, it holds the name of the element by which you want to sort.
[Actually, you're setting it to a result tree fragment whose string
value is 'DateVal'. If you want a variable or parameter to hold a
string, you should set it with the select attribute, as in:
<xsl:param name="sortby" select="'DateVal'" />
<xsl:param name="datatype" select="'number'" />
This is better because it means the processor doesn't have to build
any nodes to set these parameters.]
So I want to sort each viewentry by $sortval, however I don't really
have any idea what I am doing.
The normal way to do this is to select the child element of the
elements that you're sorting whose name is equal to the parameter
$sortby, as in:
<xsl:for-each select="viewentry">
<xsl:sort select="*[name() = $sortby]"
data-type="{$datatype}" />
...
</xsl:for-each>
Note that the things that you want to sort are the <viewentry>
elements. In your current code, you're saying that each <viewentry>
element should create a <div> element and then selecting its <DateVal>
element to be sorted -- since each <viewentry> element has only one
<DateVal> element, there's no point sorting it.
Also, I've assumed that the $datatype parameter is supplying the
datatype of the thing that you want to sort by.
Additionally I need to sort by date and I am aware there is no way
to sort by date, so I made somthing up.
It's a shame that DateVal doesn't use a format for dates that can be
easily sorted (e.g. the ISO format YYYY-MM-DD) as this does complicate
things a little.
You can do a variation on what you're doing at the moment, using a
<xsl:choose> to test whether $sortby is 'DateVal' and doing something
special if it is. Note that if your options are exclusive then it's
better to use an <xsl:choose> than a series of <xsl:if>s, only one of
which can be true:
<xsl:template match="viewentries">
<div class="child" id="Sheet Entry">
<table class="standard" cellspacing="0" cellpadding="0">
<xsl:choose>
<xsl:when test="$sortby = 'DateVal'">
<xsl:apply-templates select="viewentry">
<xsl:sort select="substring(DateVal, 7, 4)" />
<xsl:sort select="substring(DateVal, 4, 2)" />
<xsl:sort select="substring(DateVal, 1, 2)" />
</xsl:apply-templates>
</xsl:when>
<xsl:otherwise>
<xsl:apply-templates select="viewentry">
<xsl:sort select="*[name() = $sortby]"
data-type="{$datatype}" />
</xsl:apply-templates>
</xsl:otherwise>
</xsl:choose>
...
</table>
</div>
</xsl:template>
Note that the first <xsl:sort> when sorting by DateVal needs to select
the entire four digits of the year, not just the first two (you had
"substring(., 7, 2)").
In the above, I've used <xsl:apply-templates> to apply templates to
the <viewentry> elements. This enables you to have the same code
used to transform the <viewentry> elements. Just have a template
that matches <viewentry> elements and does the appropriate thing with
them:
<xsl:template match="viewentry">
<tr>
...
</tr>
</xsl:template>
Cheers,
Jeni
---
Jeni Tennison
http://www.jenitennison.com/
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