[xsl] Re: Generating a 2 column table from only _some_ of the nodes.

["Dimitre Novatchev" <dnovatchev@xxxxxxxxx>]

See:

   http://www.topxml.com/code/default.asp?p=3&id=v20020514091249

This snippet shows how to create from the nodes of an arbitrary node-set
(may not be siblings) a table withN columns, and alternating row colour.


=====
Cheers,

Dimitre Novatchev.
http://fxsl.sourceforge.net/ -- the home of FXSL


"Martin Buchan" <martinb@xxxxxxx> wrote in message
news:3FB4BA8A.8000101@xxxxxxxxxx
> Hi,
>
> Could anyone help me out with this. I have looked through the table
> generation stuff on the site but I think my problem is slightly different.
>
> I need to build a 2 column table from only some of the nodes in my node
> set. Here is an excerpt from my XML.
>
> <archive id="111111">
>    <product label="Product" group="1. Product" required="t">aaa</product>
>    <title label="Title" group="1. Product" required="t">bbb</title>
>    <client label="Client" group="1. Product" required="t">ccc</client>
>    <brand label="Brand" group="1. Product" required="t">ddd</brand>
>    <jobref label="Job Reference" group="2. TV" required="f">eee</jobref>
>    <tape_no label="Tape Number" group="2. TV" required="f">fff</tape_no>
>    <duration label="Duration" group="2. TV" required="t">ggg</duration>
>    <agency label="Agency" group="3. Agency" required="t">hhh</agency>
>    <producer label="Producer" group="3. Agency" required="t">i</producer>
>    <director label="Director" group="3. Agency" required="f">j</director>
> </archive>
>
> I dont know the element names in advance and I dont know how many
> elements there will be. Also, I need to lay them out in groups. I.e. I
> start a table for group "1. Product", lay out all the rows, finish the
> table and start a new table for group "2. TV" etc.
>
> I was doing this initially using a key to select all elements in a group
> and the looping thru them with a for-each and using position() mod 2 = 1
> and following-sibling::*[1] to lay out the first and second td.
>
> This worked fine but now I need to only pick elements where @required =
> 'f'. This causes problems with following-sibling because there is no
> gaurantee that the following-sibling will have @required = 'f'. I can
> test for this but if it doesnt, then all i can do is layout an empty
> cell - which I dont want. What I want is someway to move forward int eh
> loop sort of.
>
> Or, I want to build a new nodeset where @required = 'f'. I thought i was
> doing this with a key which had @required = 'f' in the predicate but It
> doesnt actually select only those nodes. It is selecing all the nodes.
>
> Here is my key.
>
> <xsl:key name="notreq1" match="/archive[1]/*[@required = 'f']"
> use="@group" />
>
> Here is my current for loop.
>
> <xsl:for-each select="/archive/*[count(. | key('notreq1', @group)[1]) =
1]">
>
> <xsl:if test="not(@required = 't'">
>    <table width="100%" align="center" border="0">
>      <tr><td>
>        <xsl:value-of select="substring-after(../*[@group =
> current()/@group]/@group, '. ')"/>
>      </td></tr>
>      <xsl:for-each select="key('notreq1', @group)[@required =
> 'f'][not(position() = 0)]">
>        <xsl:sort select="@group" />
>        <xsl:when test="(position() mod 2) = 1">
>          <td><xsl:value-of select="."/></td>
>            <xsl:choose>
>              <xsl:when test="following-sibling::*[1]/@group = @group">
>                <xsl:with-param name="content"
> select="following-sibling::*[1]"/>
>              </xsl:when>
>              <xsl:otherwise>
>                <td>EMPTY CELL</td>
>              </xsl:otherwise>
>            </xsl:choose>
>          </xsl:if>
>        </xsl:for-each>
>      </xsl:if>
> </xsl:for-each>
>
>
> Can anyone see what I am trying to do?
> Basically, my problem is with the node set i am looping through. Is
> there some way I can loop through only nodes with @required = 'f'. I
> thought the key did this. I.e. I thought I could build a new node set
> with the key???
>
> I will provide more info if needed. I am in very new territory here so
> any help would be muchly appreciated.
>
> Thanks
>
> Martin
>
>
>
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>




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