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[barkeeper]

Hi,

I have a xml input file like this:

<root>
<element>
<name>elem1</name>
<child>
<name>1</name>
</child>
<child>
<name>2</name>
</child>
</element>
<element>
<name>elem2</name>
<child>
<name>2</name>
</child>
<child>
<name>3</name>
</child>
</element>

Now I want apply a template on every element element that has a child
with name="3" or name="1" for example.

I do this right now:
<xsl:for-each-group select="root/elment" group-by="name">
<xsl:if test="contains(current()/child/name, '3') or
contains(current()/child/name, '1')" >
<xsl:apply-templates select="current-group()"/>
</xsl:if>
</xsl:for-each-group>

But I get a "too many items" error. I guess because there are more
than one child nodes in a elment node.

Thanks
Peter

[Pavel Lepin]

ba*******@jubii.de <ba*******@jubii.dewrote in
<fe**********************************@q21g2000hsa. googlegroups.com>:

<root>
<element>
<name>elem1</name>
<child>
<name>1</name>
</child>
<child>
<name>2</name>
</child>
</element>
<element>
<name>elem2</name>
<child>
<name>2</name>
</child>
<child>
<name>3</name>
</child>
</element>

Not well-formed.

Now I want apply a template on every element element that
has a child with name="3" or name="1" for example.

<xsl:apply-templates
select="/root/element[child/name/text()='1' or
child/name/text()='3']"/>

<xsl:for-each-group select="root/elment" group-by="name">
<xsl:if test="contains(current()/child/name, '3') or
contains(current()/child/name, '1')" >
<xsl:apply-templates select="current-group()"/>
</xsl:if>
</xsl:for-each-group>

I think I'll just add xsl:for-each-group to my list of Evil
Stuff That Confuses Neophytes.

--
When all you have is a transformation engine, everything
looks like a tree.

[Joseph Kesselman]

ba*******@jubii.de wrote:

Now I want apply a template on every element element that has a child
with name="3" or name="1"

<xsl:apply-templates select="/root/element[child/name=3 or child/name=1]"/>

--
Joe Kesselman / Beware the fury of a patient man. -- John Dryden

[barkeeper]

Well this works, but not like I need it.
I should have mentioned what I do in my template that I apply.

I want to output the element names but only once the same.
With the current solution I get the element names many times, to be
exactly once for every child that's name is 1 or 3.

That is why i grouped the names in a for-each-group loop.

On 11 Feb., 16:22, Joseph Kesselman <[e-mail]>
wrote:

[e-mail] wrote:

Now I want apply a template on every element element that has a child
with name="3" or name="1"

<xsl:apply-templates select="/root/element[child/name=3 or child/name=1]"/>

--
Joe Kesselman / Beware the fury of a patient man. -- John Dryden

[Joseph Kesselman]

ba*******@jubii.de wrote:

I want to output the element names but only once the same.

See the XSLT FAQ website's "grouping" page; that should give you some
useful techniques.

--
Joe Kesselman / Beware the fury of a patient man. -- John Dryden

[Martin Honnen]

ba*******@jubii.de wrote:

Well this works, but not like I need it.
I should have mentioned what I do in my template that I apply.

I want to output the element names but only once the same.
With the current solution I get the element names many times, to be
exactly once for every child that's name is 1 or 3.

That is why i grouped the names in a for-each-group loop.

I think it is better you provide a sample of your XML that makes it
clear what you need to group on and then describe the result you want to
create with your stylesheet.
So far we can only guess what you want, here is my attempt at guessing

<xsl:for-each-group select="root/element[child/name = '1' or
child/name = '3']" group-by="name">
<xsl:value-of select="current-grouping-key()"/>
<xsl:if test="position() != last()">
<xsl:text</xsl:text>
</xsl:if>
</xsl:for-each-group>

It outputs the grouping key of each group.

--

Martin Honnen
http://JavaScript.FAQTs.com/

[barkeeper]

Hi folks,

that was the clue:
<xsl:for-each-group select="root/element[child/name = '1' ....

Now it works.
Thanks very much!

I really have my problems with this xslt stuff :-)

Bye Peter

On 11 Feb., 17:48, Martin Honnen <[e-mail]:

[e-mail] wrote:

Well this works, but not like I need it.
I should have mentioned what I do in my template that I apply.

I want to output the element names but only once the same.
With the current solution I get the element names many times, to be
exactly once for every child that's name is 1 or 3.

That is why i grouped the names in a for-each-group loop.

I think it is better you provide a sample of your XML that makes it
clear what you need to group on and then describe the result you want to
create with your stylesheet.
So far we can only guess what you want, here is my attempt at guessing

<xsl:for-each-group select="root/element[child/name = '1' or
child/name = '3']" group-by="name">
<xsl:value-of select="current-grouping-key()"/>
<xsl:if test="position() != last()">
<xsl:text</xsl:text>
</xsl:if>
</xsl:for-each-group>

It outputs the grouping key of each group.

--

Martin Honnen
http://JavaScript.FAQTs.com/