RE: [xsl] one line of xml to indented xml doc

[Astor Rivera <arivera at esri dot com>]

Thanks for the help, although I may not have explained what I really needed.
You did point out some redundancy I found later that evening.
What I was trying to transform is a single line of xml being output from a
C++ app,
I must then take it and make it a well formed document with a hierachical
structure
and the transform is being done using MSXSL.

It must remain generic as the output tags will always vary.

Here is what I have up to now,

<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes" />
<xsl:template match=" * | @* ">
<xsl:copy>
        <xsl:apply-templates select=" * | @* " />
</xsl:copy>
</xsl:template>
</xsl:stylesheet>

Although I am currently looking into using xsl:key to generate the
hierarchical structure.

Any more help is great.

Thanks,
Astor



-----Original Message-----
From: Peter Davis
To: xsl-list@lists.mulberrytech.com
Sent: 3/24/02 2:26 AM
Subject: Re: [xsl] one line of xml to indented xml doc

On Friday 22 March 2002 15:52, Astor Rivera wrote:
> This is what I have and would like to know how to indent the document.
> The xml is coming from an application, and it's url encoded.

So you have a source document that is not indented, and you want to
indent 
it?  Well, if that's the case, then you have the right idea with
<xsl:output 
indent="yes"/>.  The problem is that processors aren't required to
support 
that; I know that Xalan doesn't do a very good job of it.  Saxon does, 
however.  Your milage will probably vary, though.

BTW, your template has some very redundant parts:

> <?xml version="1.0" encoding="UTF-8"?>
> <xsl:stylesheet version="1.0"
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
> xmlns:fo="http://www.w3.org/1999/XSL/Format";>
>         <xsl:output method="xml" indent="yes" encoding="UTF-8"/>
>         <xsl:template match=" * | node() | text() | @* ">

"text()" and "*" are both redundant since they are matched by "node()".
So 
really, the only thing you need here is <xsl:template match="node() |
@*"/>.

>                         <xsl:copy>
>                                 <xsl:apply-templates  select=" * |
node() |
> @*"/>                            </xsl:copy>

Same thing with your apply-templates.  "*" is taken care of by "node()",
so 
all you need is <xsl:apply-templates select="@* | node()"/>.

>         </xsl:template>
> </xsl:stylesheet>

Your question was a little confusing, so if this is not the answer you
wanted 
you really should clarify what you mean by "indent".  I would also
suggest 
re-indenting your stylesheet, because it's pretty ugly with the
<xsl:copy> 
half way across the page :).  Or maybe that's where your having problems
with 
indenting?  If so, please clarify.

-- 
Peter Davis
Excess on occasion is exhilarating.  It prevents moderation from
acquiring the deadening effect of a habit.
		-- W. Somerset Maugham

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