XSLT with input parameter

[bbembi_de@lycos.de]

Hello everyone,

is it possible to give a transformation a input parameter?
I want to use this parameter as a variable in my transformation.

thanks.

bye bembi

[Martin Honnen]

Re: XSLT with input parameter

bbembi_de@lycos .de wrote:

is it possible to give a transformation a input parameter?
I want to use this parameter as a variable in my transformation.

Yes, that is possible, you define a top-level xsl:param element with the
name of the parameter and then you use the API of your XSLT processor to
set that parameter before running the transformation.

<xsl:styleshe et
xmlns:xsl="http ://www.w3.org/1999/XSL/Transform"
version="1.0">

<xsl:param name="p1"/>

<xsl:template match="/">
<xsl:value-of select="$p1"/>
</xsl:template>
</xsl:stylesheet>





--

Martin Honnen

http://JavaScript.FAQTs.com/

[bbembi_de@lycos.de]

Re: XSLT with input parameter

Thank you very much.

bembi


On 16 Mai, 17:04, Martin Honnen <mahotr...@yaho o.dewrote:

bbembi...@lycos .de wrote:

is it possible to give a transformation a input parameter?
I want to use this parameter as a variable in my transformation.

>
Yes, that is possible, you define a top-level xsl:param element with the
name of the parameter and then you use the API of your XSLT processor to
set that parameter before running the transformation.
>
<xsl:styleshe et
xmlns:xsl="http ://www.w3.org/1999/XSL/Transform"
version="1.0">
>
<xsl:param name="p1"/>
>
<xsl:template match="/">
<xsl:value-of select="$p1"/>
</xsl:template>
</xsl:stylesheet>
>
--
>
Martin Honnen
http://JavaScript.FAQTs.com/

[bbembi_de@lycos.de]

Re: XSLT with input parameter

Hello again,

is it possible to give the XSLT a variable number of parameter and use
it in a array?


Thanks.
bye bembi



On 18 Mai, 18:28, "bbembi...@lyco s.de" <bbembi...@lyco s.dewrote:

Thank you very much.
>
bembi
>
On 16 Mai, 17:04, Martin Honnen <mahotr...@yaho o.dewrote:
>
>
>

bbembi...@lycos .de wrote:

is it possible to give a transformation a inputparameter?
I want to use thisparameteras a variable in my transformation.

>

Yes, that is possible, you define a top-level xsl:param element with the
name of theparameterand then you use the API of yourXSLTprocess or to
set thatparameterbe fore running the transformation.

>

   <xsl:styleshe et
     xmlns:xsl="http ://www.w3.org/1999/XSL/Transform"
     version="1.0">

>

     <xsl:param name="p1"/>

>

     <xsl:template match="/">
       <xsl:value-of select="$p1"/>
     </xsl:template>
   </xsl:stylesheet>

>

--

>

        Martin Honnen
       http://JavaScript.FAQTs.com/- Zitierten Text ausblenden -

>
- Zitierten Text anzeigen -

[Joseph J. Kesselman]

Re: XSLT with input parameter

bbembi_de@lycos .de wrote:

is it possible to give the XSLT a variable number of parameter and use
it in a array?

For the first: Parameters can have default values. Or you could pass a
parameter which your stylesheet then parses data out of.

XSLT has no concept of "arrays". Some processors will let you pass in
non-XSLT objects but the only thing you can do with them, really, is to
pass them along to extension functions.

You might get more useful answers if you told us what you're trying to
do rather than asking whether a particular approach is possible.

[Martin Honnen]

Re: XSLT with input parameter

bbembi_de@lycos .de wrote:

is it possible to give the XSLT a variable number of parameter and use
it in a array?

Depending on the XSLT processor you use it might be possible to pass in
an XML document exposed as a node-set to XSLT/XPath. For instance MSXML
allows you to build and pass in an XML DOM document.
XslCompiledTran sform allows that too. See

http://msdn.microsoft.com/en-us/library/system.xml.xsl.xsltargumentlist.addparam.aspx

for the .NET types you can pass in and how the translate into XPath/XSLT
types.


--

Martin Honnen

http://JavaScript.FAQTs.com/

[bbembi_de@lycos.de]

Re: XSLT with input parameter

Thanks for the answers so far.

To give you the whole story:
I want to do this: I have a XML like:
<root>
<entry type='a' />
<entry type='b' />
<entry type='c' />
</root>

Now I want to pass a variable number of types. All these will be
exported:
Example:
parameters: a, b
Output:
<root>
<entry type='a' />
<entry type='b' />
</root>

bye bembi



On 21 Mai, 15:47, "Joseph J. Kesselman" <keshlam-nos...@comcast. net>
wrote:

bbembi...@lycos .de wrote:

is it possible to give theXSLTa variable number ofparameterand use
it in a array?

>
For the first: Parameters can have default values. Or you could pass aparameterwhich your stylesheet then parses data out of.
>
XSLThas no concept of "arrays". Some processors will let you pass in
non-XSLTobjects but the only thing you can do with them, really, is to
pass them along to extension functions.
>
You might get more useful answers if you told us what you're trying to
do rather than asking whether a particular approach is possible.

[Joseph J. Kesselman]

Re: XSLT with input parameter

bbembi_de@lycos .de wrote:

parameters: a, b
Output:
<root>
<entry type='a' />
<entry type='b' />
</root>

I'd suggest passing a single parameter which contains a delimited string
of the values you want to match, and writing the test to use it in that
form... exactly as you've shown it here.

I did a simple and sloppy version of that in part 2 of my "styling
stylesheets" article on DeveloperWorks (http://www.ibm.com/xml).

[Martin Honnen]

Re: XSLT with input parameter

bbembi_de@lycos .de wrote:

To give you the whole story:
I want to do this: I have a XML like:
<root>
<entry type='a' />
<entry type='b' />
<entry type='c' />
</root>
>
Now I want to pass a variable number of types. All these will be
exported:
Example:
parameters: a, b
Output:
<root>
<entry type='a' />
<entry type='b' />
</root>

Which XSLT processor do you use?


--

Martin Honnen

http://JavaScript.FAQTs.com/

[bbembi_de@lycos.de]

Re: XSLT with input parameter

I am not sure which XSLT processor will be used.
The transformation will be integreated in a existing system. I think
it is saxon 8.?

bye bembi


On 21 Mai, 18:25, Martin Honnen <mahotr...@yaho o.dewrote:

bbembi...@lycos .de wrote:

To give you the whole story:
I want to do this: I have a XML like:
<root>
  <entry type='a' />
  <entry type='b' />
  <entry type='c' />
</root>

>

Now I want to pass a variable number of types. All these will be
exported:
Example:
parameters: a, b
Output:
<root>
  <entry type='a' />
  <entry type='b' />
</root>

>
WhichXSLTproces sor do you use?
>
--
>
        Martin Honnen
       http://JavaScript.FAQTs.com/- Zitierten Text ausblenden -
>
- Zitierten Text anzeigen -

[Martin Honnen]

Re: XSLT with input parameter

bbembi_de@lycos .de wrote:

I am not sure which XSLT processor will be used.
The transformation will be integreated in a existing system. I think
it is saxon 8.?

Well there are different versions of Saxon 8 supporting different
version of the XSLT 2.0 draft but with Saxon 8.9 or 9.0 which support
the final XSLT 2.0 specification you could easily use the tokenize
function to split up a string value passed in into a sequence of strings:

<xsl:styleshe et
xmlns:xsl="http ://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="xs"
version="2.0">

<xsl:output method="xml" indent="yes"/>

<xsl:param name="types" as="xs:string"/>

<xsl:variable name="type-list" as="xs:string* "
select="tokeniz e($types, ',')"/>

<xsl:template match="root">
<xsl:copy>
<xsl:apply-templates select="@*"/>
<xsl:apply-templates select="entry[@type = $type-list]"/>
</xsl:copy>
</xsl:template>

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

</xsl:stylesheet>

Then you could use Saxon 9 from the command line as follows:

java -jar saxon9.jar -s:input.xml -xsl:stylesheet. xsl types=a,b


--

Martin Honnen

http://JavaScript.FAQTs.com/