grouping elements

[ofuuzo1@yahoo.no]

Hi,
I have the following xml file which I would like to transform to
another structure:

<person>
<dept>
<gr_name>sale </gr_name>
<gr_id>2<gr_i d>
....
<dept>
<name>Peter Pan</name>
<dept>
</person>

<person>
<dept>
<gr_name>produc tion</gr_name>
<gr_id>3<gr_i d>
....
<dept>
<name>John John</name>
<dept>
</person>

<person>
<dept>
<gr_name>packag e</gr_name>
<gr_id>11<gr_id >
....
<dept>
<name>Roy Kemp</name>
<dept>
</person>
......

New xml would look like this:
<sale>Peter Pan</sale>
<production_pac kage_etc>John John<production _package_etc>
<production_pac kage_etc>Roy Kemp<production _package_etc>

I have written some thing like this and it is not working. What is
wrong her?


<xsl:for-each select="//person">
<xsl:if test="//person/dept/gr_id = 1">
<sale<xsl:val ue-of select="name" /><sale>
</xsl:if>
</xsl:for-each>

<xsl:for-each select="//person">
<xsl:if test="//person/dept/gr_id != 1">
< production_pack age_etc<xsl:val ue-of select="name" /

>< production_pack age_etc>

</xsl:if>
</xsl:for-each>

Thanks in advance.

[Martin Honnen]

Re: grouping elements

ofuuzo1@yahoo.n o wrote:

I have the following xml file which I would like to transform to
another structure:
>
<person>
<dept>

^^^^^^
So here is a 'dept' start tag

<gr_name>sale </gr_name>
<gr_id>2<gr_i d>
....
<dept>

^^^^^^
And here is another one

<name>Peter Pan</name>
<dept>

^^^^^^
And here is yet another one while there are no matching end tags.
Can you post a well-formed sample? Currently it is not clear how your
input XML looks.

<xsl:for-each select="//person">
<xsl:if test="//person/dept/gr_id = 1">

You need a relative XPath here e.g.
<xsl:if test="dept/gr_id = 1">
but it is even better to write templates e.g.
<xsl:template match="person[dept/gr_id = 1]">


--

Martin Honnen

http://JavaScript.FAQTs.com/