Re: [xsl] testing for position of an element and displaying it accordingly

Subject: Re: [xsl] testing for position of an element and displaying it accordingly
From: Mark Carlson <carlsonm@xxxxxxxxxxxxxxxx>
Date: Mon, 22 Jan 2007 11:12:10 -0800
For one, your <fo:inline> elements are not closed in the bottom 3 template rules.


ms wrote:
Hello:

I have tried to implement this solution but I dont
know where I am going wrong. Here is my complete code:

This template is used for formatting.
<xsl:template name="lists">
	<xsl:param name="format"/>
	<fo:list-block>
		<fo:list-item>
			<fo:list-item-label>
				<xsl:choose>
					<xsl:when test="$format = '1'">
						<fo:block>
							<fo:inline>
								<xsl:number format="1"/>.</fo:inline>
						</fo:block>
					</xsl:when>
				</xsl:choose>
			</fo:list-item-label>
			<fo:list-item-body start-indent="body-start()"
end-indent="0pt">
				<fo:block>
					<xsl:apply-templates>
				
						</xsl:apply-templates>
				</fo:block>
			</fo:list-item-body>
		</fo:list-item>
	</fo:list-block>
</xsl:template>


This template calls the above template for formatting:


    <xsl:template match="r1">
<xsl:if test="child::a">
<xsl:apply-templates select="a" mode="t"/>
</xsl:if>

        <xsl:call-template name="lists">
            <xsl:with-param
name="format">1</xsl:with-param>
        </xsl:call-template>
    </xsl:template>

This is the template for element a which is a child of
<r1>.

<xsl:template match="a" mode="t">
		<fo:block space-before="6pt">
	<fo:inline><xsl:value-of select="."/>
		</fo:block>
	</xsl:template>

<xsl:template match="a">
		<fo:block space-before="6pt">
	<fo:inline><xsl:value-of select="."/>
		</fo:block>
	</xsl:template>

<xsl:template match="text">
		<fo:block space-before="6pt">
	<fo:inline><xsl:apply-templates/>
		</fo:block>
	</xsl:template>


And my Input is:
<r1>
<a>test</a>
<a>test2</a>
<text>test test</text>
<a>test3</a>
</r1>
My desired output is:


test
test2
1. test test
test3


The reason for checking if <a> is a child of <r1> and applying the mode is so that this <a> appears above <r1> without getting formatted as 1. test

So even though <r1> element is formatted to be 1., any
<a> element which appear immediately after <r1> should
not be formatted. If there is an <a> element after
some other element like in this example <text>, then
it should be displayed as it is in the document tree.


Please let me know if this is clear and if there is a
way to do it and if there is amistake in my code.


--- ms <mina_hurray@xxxxxxxxx> wrote:

Hello:

I did see the axis specifier thread on this list,
unfortunately it is not me, but I am facing a
similar
issue. I tried to implement the solutions given in
those, but I was not successful and thats te reason
for posting this issue.



--- Michael Kay <mike@xxxxxxxxxxxx> wrote:


I assume you're the same person as xslt.new who
posted "axis specifiers"
last week (if not, it's a remarkable that you have
exactly the same problem
and the same misunderstandings).

We need to go back a step. You haven't understood
a
word of the explanations
we gave you last week. Giving you the same
explanations again isn't going to
do much good. I'm really trying hard to read your
postings and work out
which step in the learning process you've got
stuck
on, but I'm struggling.

The normal way of processing an XML document is to
work top-down. First, you
have to think of the document as a tree, with the
root at the top (strange
in botany, but not in computing). The document
node
itself is at the top,
then the r1 element, and (a,a,b,a) are below the
r1
element, at the next
level down (I stress this because you talk of
test3
as being "below" b,
which means you're not yet thinking in tree
terms).
So top down processing
means you write a template rule for r1, which
typically applies templates to
each of its children. You then define template
rules
for each kind of child.

This works when the rules for an element are
independent of where it appears
relative to other elements at the same level. For
example, if every a
element is processed by converting the <a> tag to
a
<p>, then you can write
a template rule

<xsl:template match="a">
  <p><xsl:apply-templates/></p>
</xsl:template>

Similarly, if every b is processed by outputting a
<p> but with a sequence
number attached, you can write

<xsl:template match="b">
  <p><xsl:number/>: <xsl:apply-templates/></p>
</xsl:template>

Now, your input and your processing seems to
follow
this conventional style.
As far as we can see, you've got a perfectly
straightforward easy
transformation to do here that yields to the basic
techniques in chapter 1
of any textbook, and we can't see why you're
making
it so difficult.

Michael Kay
http://www.saxonica.com/



-----Original Message-----
From: ms [mailto:mina_hurray@xxxxxxxxx] Sent: 19 January 2007 14:58
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: [xsl] testing for position of an
element
and
displaying it accordingly

Hi all:

My input XML looks like this;

<r1>
<a>test</a>
<a>test2</a>
<b>test test</b>
<a>test3</a>
</r1>

For this XML, I would like to write an XSLT
which
basically
achieves the following:

1) For the <a> elements which are above the <b>
element, they
should be displayed this way:

Test
Test2
1)test test

2) For the <a> element which appears after the
<b>
element, I
want it displayed after the <b> element:

Test
Test2
1)test test
test3

How can I achieve this in XSLT? I tried testing
the
position, using position()=1. This works
obviously
for the
first <a> element and fails for the second one.

Please let me know if this is possible?




______________________________________________________________
______________________
Do you Yahoo!?
Everyone is raving about the all-new Yahoo! Mail
beta.
http://new.mail.yahoo.com






____________________________________________________________________________________
Need a quick answer? Get one in minutes from people
who know.
Ask your question on www.Answers.yahoo.com





____________________________________________________________________________________
Yahoo! Music Unlimited
Access over 1 million songs.
http://music.yahoo.com/unlimited

Current Thread