RE: Looping in XSLT(old question, but maybe new proble

Thank you Jeff, that was indeed very quick response.

Just one question, if foo-bar-temp is a global variable, is the nodeset
initialized
only once or is it generated every time foo-bar-temp is called?

Regards,
LS
-----Original Message-----
From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx
[mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of Jeff Beadle
Sent: Tuesday, June 24, 2003 9:40 AM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: RE:  Looping in XSLT(old question, but maybe new problem)


Hey Liu,

Just replace the xsl:variable[@name='foo-bar'] definition with this:

<!-- temp table ... er, excuse me ... I mean, temp node list -->
<xsl:variable name="foo-bar-temp">
<xsl:for-each select="foo/bar">
<xsl:sort select="@id" data-type="number"/>
<xsl:copy-of select="."/>
</xsl:for-each>
</xsl:variable>
<!--
Now, construct the foo-bar node list from the sorted temp list;
Doing this instead of rolling the for-each directly within the foo-bar will
prevent you from
having to use your parsers node-set/node-list adapter function to turn the
RTF into a node list.
 -->
<xsl:variable name="foo-bar" select="msxsl:node-set($foo-bar-temp)/bar"/>

Sorry, for the delay ... just logged into my mail server.

-Jeff

-----Original Message-----
From: Liu Shuai [mailto:shuai@xxxxxxxxxxxxxxxxx]
Sent: Tuesday, June 24, 2003 12:52 AM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: RE:  Looping in XSLT(old question, but maybe new problem)


Jeff,

Thank you for your response.
One concern, though. If I sort foo/bar, will foo/bar[position()] return
nodes based in my sorted order
or original order in the source file? I used following-sibling in a template
I wrote and it return nodes
in the orginal order.

Thanks!
LS


-----Original Message-----
From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx
[mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of Jeff Beadle
Sent: Tuesday, June 24, 2003 12:12 AM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: RE:  Looping in XSLT(old question, but maybe new problem)


Hey Liu,

Fun one!  I had one of those rare moments where I actually had a bit of time
to work on one of the problems/questions raised on this list.

Also, please be advised ... other members on this list--the experts--will
most likely provide direction and/or solutions well beyond what I'm
providing, so you may want to wait a bit to see what they contribute.

But hey, this works!

<xsl:param name="N" select="5"/>
<xsl:template match="/">
  <table>
    <xsl:variable name="foo-bar" select="foo/bar"/>
    <xsl:variable name="foo-bar-cnt" select="count($foo-bar)"/>
    <xsl:for-each select="$foo-bar">
      <xsl:variable name="pos" select="position()"/>
      <!-- if at the Nth column, then create a new row -->
      <xsl:if test="not(($pos - 1) mod $N)">
        <tr>
          <!-- grab the splice ... all bars that are within the range
[$pos,$pos+$N] -->
          <xsl:for-each select="$foo-bar[position()&gt;=$pos and
position()&lt;($pos + $N)]">
            <td><xsl:value-of select="someElement"/></td>
          </xsl:for-each>
          <!-- if within the last row, construct enough empty cells to
"square" the table -->
          <xsl:if test="$foo-bar-cnt mod $N &gt; $foo-bar-cnt - $pos">
            <xsl:for-each select="$foo-bar[position() &gt;= 1 and
position()&lt;=($N - ($foo-bar-cnt mod $N))]">
              <td>&#160;</td>
            </xsl:for-each>
          </xsl:if>
        </tr>
      </xsl:if>
    </xsl:for-each>
  </table>
</xsl:template>

Anyhow, it was fun.

-Jeff

-----Original Message-----
From: Liu Shuai [mailto:shuai@xxxxxxxxxxxxxxxxx]
Sent: Monday, June 23, 2003 8:32 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject:  Looping in XSLT(old question, but maybe new problem)


Hi, all

I am trying to write a template that will generate a table based on a xml
file and right now I don't have any
clue how to do it.

If I have a source file looks like this

<foo>
	<bar id='1'>
		<someElement>a</someElement>
	</bar>
	<bar id='2'>
		<someElement>b</someElement>
	</bar>
	<bar id='3'>
		<someElement>c</someElement>
	</bar>
	<bar id='4'>
		<someElement>d</someElement>
	</bar>
	<bar id='5'>
		<someElement>e</someElement>
	</bar>
	...
</foo>

Can I write a style sheet that will transform the source file above to a
html table like this?

<table>
	<tr>
		<td>a</td>
		<td>b</td>
		<td>c</td>
	</tr>
	<tr>
		<td>d</td>
		<td>e</td>
		<td>&nbsp;</td>
	</tr>
	...
</table>

Basicly, I want to generate N columns per row but I don't know how many
"bar"s I have in the source file.

Any hint or suggestion will be greatly appreciated.

Shuai


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