That's why I called it "e1link". You could also have "e2" and "e2link",
etc. Just associate them by attribute name.
________________________________
From: Nancy Brandt [mailto:
nancy_brndt@yahoo.com]
Sent: Tuesday, October 20, 2009 12:29 PM
To:
docbook-apps@lists.oasis-open.org; Cavicchio, Rob
Subject: RE: [docbook-apps] RE: A tricky xref problem
Hi Rob,
Thanks a lot for your suggestions! The second option you describe could
be ideal, if we could somehow associate between the 'e1' and 'e1link'
attributes, so that the value of 'sectionID' could go to the xref in the
desired table entry (in this case, entry1). Is it possible? Otherwise,
how does the compiler know where to send sectionID - xref of entry1 or
entry 2?
Thank you a lot again!!!
My best wishes,
Nancy
--- On Tue, 10/20/09,
Cavicchio_Rob@emc.com <
Cavicchio_Rob@emc.com>
wrote:
From:
Cavicchio_Rob@emc.com <
Cavicchio_Rob@emc.com>
Subject: RE: [docbook-apps] RE: A tricky xref problem
To:
nancy_brndt@yahoo.com,
docbook-apps@lists.oasis-open.org Cc:
bobs@sagehill.net Date: Tuesday, October 20, 2009, 9:38 AM
I think I understand now. You are transforming *into* DocBook,
and you want each entry's attribute value to include not only the text
for the entry, but the value of the "linkend" attribute that will go
into the DocBook output.
You could certainly perform text string manipulations so that if
the "linkend" value were written in a predictable way, you could extract
its value. For example, if your source document had this:
<row e1="blue_sectionID">
then your transform could do this:
<entry>
<xsl:value-of select="substring-before( @e1, '_' )" />
<xsl:if test="string-length( substring-after(@e1, '_')
)">
<xref>
<xsl:attribute name="linkend" />
<xsl:value-of select="substring-after( @e1,
'_' )" />
</xsl:attribute>
</xref>
</xsl:if>
</entry>
which should produce this:
<entry>
blue
<xref linkend="sectionID" />
</entry>
You could also use a separate source attribute for the "linkend"
attribute value; for example:
<row e1="blue" e1link="sectionID" />
That would achieve the same effect, but avoids having to do all
the string manipulations, and also avoids having to designate a special
character that you then can't easily use in the actual content.
________________________________
From: Nancy Brandt [mailto:
nancy_brndt@yahoo.com]
Sent: Tuesday, October 20, 2009 8:55 AM
To:
docbook-apps@lists.oasis-open.org; Cavicchio, Rob
Cc:
bobs@sagehill.net Subject: Re: [docbook-apps] RE: A tricky xref problem
Hi Rob,
First of all, thanks a lot for your quick response!
I think you misunderstood me, and it's my fault. I'll try to provide
more details about my "design":
what i tried to do is to minimize the amount of tags you need to type
in, to create a table.
So, I defined a kind of a macro, that builds the table's look and feel,
headings, etc.
<xsl:template match="jstatuscodestable">
<xsl:variable name="e1">
<xsl:apply-templates/>
</xsl:variable>
<xsl:variable name="e2">
<xsl:apply-templates/>
</xsl:variable>
<xsl:element name="table">
<xsl:attribute name="frame">
<xsl:value-of select="'all'"/>
</xsl:attribute>
<xsl:attribute name="id">
<xsl:value-of select="@id"/>
</xsl:attribute>
<tgroup cols="2">
<colspec colnum="1" colname="col1" colwidth="*"/>
<colspec colnum="2" colname="col2" colwidth="1.5*"/>
<row>
...
</row>
<xsl:for-each select="row">
<row>
<xsl:if test="$e1">
<entry>
<xsl:value-of select="@e1"/>
<xsl:if test="contains('xref')">
<xref>
<xsl:copy-of select="@*"/>
<xsl:attribute name="linkend">
<xsl:copy-of select="@*"/>
<xsl:value-of select="<extracted_text>"/>
</xsl:attribute>
</xref>
</xsl:if>-->
</entry>
</xsl:if>
<xsl:if test="$e2">
<entry>
<xsl:value-of select="@e2"/>
</entry>
</xsl:if>
</row>
</xsl:otherwise>
</xsl:choose>
</xsl:for-each>
</tgroup>
</xsl:element>
</xsl:template>
So, to add such a table, I only need to enter:
<jstatuscodestable>
<row e1="blue" e2="red">
<row e1="black" e2="yellow">
...
</jstatuscodestable>
instead of entering dozens of table elements each time.
But then, I discovered that I can't add an <xref> within an entry.
I can't write:
<row e1="blue (refer to <xref linkend="sect_test"/>)" e2="red"/>
So I wonder, if somehow a piece of text within e1's value (e.x.
sect_test) can be extracted and passed as a value for linkend I added
(as an example) in the first <entry> tag in the template above. Of
course, I would like to be able to add an xref in any entry within the
table.
I hope this time, my explanation is clearer :-)
Thanks a lot in advance!
Best wishes,
Nancy
--- On Tue, 10/20/09,
Cavicchio_Rob@emc.com <
Cavicchio_Rob@emc.com>
wrote:
From:
Cavicchio_Rob@emc.com <
Cavicchio_Rob@emc.com>
Subject: [docbook-apps] RE: A tricky xref problem
To:
nancy_brndt@yahoo.com,
docbook-apps@lists.oasis-open.org Cc:
bobs@sagehill.net Date: Tuesday, October 20, 2009, 8:20 AM
Your example is confusing, because in standard DocBook, <row>
does not have "entry" attributes, only <entry> child elements.
But assuming you know that and this is really what you want, it
sounds like you are saying that you want to use the concatenated values
of the other attributes as the link ID. I think that you would do that
not by dynamically adding a "linkend" attribute, but by modifying the
"xref" template so that it retrieves the values of those other
attributes directly; something like this:
<xsl:variable name="linkID" select="@linkend" />
<xsl:choose>
<xsl:when test="id( $linkID )">
<xsl:call-template name="createURI">
<xsl:with-param name="element" select="id( $linkID
)" />
</xsl:call-template>
</xsl:when>
<xsl:otherwise>
<xsl:call-template name="createURI">
<xsl:with-param name="element"
select="*[concat(@entry1, @entry2, @entry3) = $linkID]" />
</xsl:call-template>
</xsl:otherwise>
</xsl:choose>
You would have to make the entry text retrieval match the exact
pattern you want, but that's generally how I'd approach it.
________________________________
From: Nancy Brandt [mailto:
nancy_brndt@yahoo.com]
Sent: Tuesday, October 20, 2009 5:30 AM
To:
docbook-apps@lists.oasis-open.org Cc:
bobs@sagehill.net Subject: A tricky xref problem
Hi all,
I wonder if there is a way to extract a text string and pass it as a
value for the <xref linkend=""/> expression. For example, when I write:
Some text text text (see ref=[sect_test]),
I would like to take the text within the [ ] delimiters and pass it as a
value for the linkend attribute of the xref element called in my
customized table template. The reason for this special workaround is
that I've created a custom template that enables you to add a table row
with its entries using one command, like so:
...
<row entry1="red" entry2="blue" entry3="black"/> -which saves a lot of
typing.
So, if I wish to add a reference to some section from within entry1, I
can't write:
<row entry1="red (refer to <xref linkend="sect_test"/>)" entry2="blue"
entry3="black"/>
I need somehow to customize my table template that says something like:
...
<xsl:for-each select="row">
<row>
<xsl:if test="$entry1">
<entry>
<xsl:value-of select="@entry1"/>
<xsl:if test="contains('xref')">
<xref>
<xsl:copy-of select="@*"/>
<xsl:attribute name="linkend">
<xsl:copy-of select="@*"/>
<xsl:value-of select="<extracted_text_string>"/>
</xsl:attribute>
</xref>
</xsl:if>
</entry>
</xsl:if>
...
</row>
</xsl:for-each>
...
Please, advise me how to extract the text string from within [ ]
delimiters to pass it as a value for the <xref> expression.
Thanks a lot in advance!!!
Best wishes,
Nancy