Extend XML file attributes using XSLT

[=?iso-8859-1?q?Christian_R=FChl?=]

Hi folks!

I have a little noob problem here with XSLT. I have a XML file that
looks like this:

<archiveFiles>
<module path="c:/temp/module_m17_blabla.tmp"/>
<module path="c:/temp/module_m34_blabla.tmp"/>
...
<moduleAddition path="c:/temp/moduleaddition_m17.tmp"/>
...
</archiveFiles>

What I want to do now is the following:

1.) Copy name part of filename in <module>
---------------------------------------------
I want to copy that module name part and add it to a new attribute
called "name". How should I do this? The name of each module comes
right after module_ in the path-attributes.

2.) Reference additional information to these modules
---------------------------------------------
Finally another attribute "addition" should be added to each <module>
when there is a <moduleAdditionnode with the same module name in its
"path" attribute. So that the outcome of the upper XML file looks like
this:

<archiveFiles>
<module name ="m17" path="c:/temp/module_m17_blabla.tmp"
additon="c:/temp/moduleaddition_m17.tmp"/>
<module name ="m34" path="c:/temp/module_m34_blabla.tmp"/>
...
</archiveFiles>

I really hope you can help me. I never worked with XSLT before and I
couldn't find the right hints and/or tutorials yet. :(

Thanks a lot in advance!

/Chris

[=?ISO-8859-1?Q?Christian_R=FChl?=]

What the hell makes you react like I just came through your door and
started dissing you? I didn't make any harm, did I? You wanted a valid
source file... Well, it is valid! And I don't give out any business
secrets by posting it. Maybe I just can't express myself in English,
but... I wish I could understand your reaction. We "know" each other
since three or four posts and all of a sudden you patronize me. I
wouldn't do that with a noob that politely asked a question!

I guess I won't get help here any longer thanks to this lame fight...

I surely understand your hints, but if you once checked out my profile
you'd see that I don't have too much experience to make it right for
everybody in a Usenet newsgroup! I just don't know it better and your
reaction makes me guess that this is more relevant to you than the
fact that I am willing to learn! It not that I ignore your words only
waiting for the solution of my problem... Aw, whatever!

[=?ISO-8859-1?Q?Christian_R=FChl?=]

OK, break. I guess I'm getting closer now...

There's still one problem: By now I just get the top-level-nodes
attributes by using something like this: <xsl:value-of
select="$archiveDoc//Product/@moduleName"/>
How do I get the right nodes attribute here? I have to retrieve the
right node's attribute depending on the cid-attribute. But how do I
put that into the pattern "$archiveDoc//Product/@moduleName"? Or si
there another way doing this? Maybe with a <xsl:whenpart or
something like that?

Here's my entry-template (the other ones look similar):

<xsl:template name="component" match="/">
<xsl:for-each select="$treeDoc//Component">
<Component topLevelModule="{@topLevelModule}"
description="{@description}" cid="{@cid}">
<xsl:if test="@cid = $archiveDoc//Product/@cid">
<xsl:attribute name="moduleName"><xsl:value-of
select="$archiveDoc//Product/@moduleName"/></xsl:attribute>
<xsl:attribute name="subsystem"><xsl:value-of
select="$archiveDoc//Product/@subsystem"/></xsl:attribute>
<xsl:attribute name="tokens"><xsl:value-of select="$archiveDoc//
Product/@tokens"/></xsl:attribute>
<xsl:attribute name="path"><xsl:value-of select="$archiveDoc//
Product/@path"/></xsl:attribute>
<xsl:attribute name="tc"><xsl:value-of select="$archiveDoc//
Product/@tc"/></xsl:attribute>
</xsl:if>
<xsl:if test="SubComponent">
<xsl:call-template name="sub-component"/>
</xsl:if>
</Component>
</xsl:for-each>
</xsl:template>

[=?ISO-8859-1?Q?Christian_R=FChl?=]

On 26 Nov., 14:52, "Christian Rühl" <[e-mail]:

OK, break. I guess I'm getting closer now...

There's still one problem: By now I just get the top-level-nodes
attributes by using something like this: <xsl:value-of
select="$archiveDoc//Product/@moduleName"/>
How do I get the right nodes attribute here? I have to retrieve the
right node's attribute depending on the cid-attribute. But how do I
put that into the pattern "$archiveDoc//Product/@moduleName"? Or si
there another way doing this? Maybe with a <xsl:whenpart or
something like that?

Here's my entry-template (the other ones look similar):

<xsl:template name="component" match="/">
<xsl:for-each select="$treeDoc//Component">
<Component topLevelModule="{@topLevelModule}"
description="{@description}" cid="{@cid}">
<xsl:if test="@cid = $archiveDoc//Product/@cid">
<xsl:attribute name="moduleName"><xsl:value-of
select="$archiveDoc//Product/@moduleName"/></xsl:attribute>
<xsl:attribute name="subsystem"><xsl:value-of
select="$archiveDoc//Product/@subsystem"/></xsl:attribute>
<xsl:attribute name="tokens"><xsl:value-of select="$archiveDoc//
Product/@tokens"/></xsl:attribute>
<xsl:attribute name="path"><xsl:value-of select="$archiveDoc//
Product/@path"/></xsl:attribute>
<xsl:attribute name="tc"><xsl:value-of select="$archiveDoc//
Product/@tc"/></xsl:attribute>
</xsl:if>
<xsl:if test="SubComponent">
<xsl:call-template name="sub-component"/>
</xsl:if>
</Component>
</xsl:for-each>
</xsl:template>

Solution found: <xsl:attribute name="moduleName"><xsl:value-of
select="$archiveDoc//Product[attribute::cid=$compID]/@moduleName"/></
xsl:attribute>

Thank you and sorry for taxing your patience.