Hi,

Front End - XSLT/Java

I am trying to create a Spanish screen. Say i have a non ascii character ¿Continuar to be displayed as the button name.

Once i run the screen, i button name changes to %bfContinuar. Please help me.

It's easier to debug people's code when they show you the code...

My guess is you are including Javascript code within an HTML attribute that's designed to hold a URI. Instead of putting code in "href" attributes, put it in "onClick" where it belongs. Non-ascii characters in URI values will always be escaped using the %HH convention, which you don't want in Javascript code.

Michael Kay
http://www.saxonica.com/
Author, XSLT 2.0 and XPath 2.0 Programmer's Reference

Thanks.
I dont use a javascript. I just use a form element. Here is the code:

<input type ="button" name = "test" value = "¿Confirmar"/>

When this is run, the screen displays it as %bfConfirmar

I also tried an alternate:

Screen1.xsl: i defined the label
<xsl:template name="labelConfirm">¿Confirmar</xsl:template>

Screen2.xsl: I called the label name from Screen1.xsl
<xsl:with-param name="text"><xsl:call-template name="labelConfirm"/></xsl:with-param>

Still i get the same result.

What does the actual HTML generated by your XSLT look like? (viewed as source with markup, not viewed in the browser)

How are you running the transformation: client side or server side? Which browser, which XSLT processor?

I can't see any reason why %HH escaping should be applied in the cases you show.

Michael Kay
http://www.saxonica.com/
Author, XSLT 2.0 and XPath 2.0 Programmer's Reference

Many Thanks.

This is the code in the xsl file:
<input type ="button" name = "test" value = "¿Confirmar"/>

This is the Page Source:
<input type="button" name="test" value="%HH%bfConfirmar">

Transformation:
The Transformation is done at the Servlet -
transformer = templates.newTransformer()

Am using Both firefox and IE. I dont know which XSLT processor.

Few more information:
When i try this spain character (¿Confirmar) with a normal html page with out transformer = templates.newTransformer(), it works. This does not work when the xsl is transformed through transformer = templates.newTransformer().

Thanks
Venkat

Looks weird: I'd say at first sight it's a bug in your XSLT processor. If you're running in a servlet and you don't know what processor you are using then it is probably the version of Xalan that comes bundled with the JDK. Report a bug, and try switching to the Apache version of Xalan (which is much more reliable) or to Saxon.

Michael Kay
http://www.saxonica.com/
Author, XSLT 2.0 and XPath 2.0 Programmer's Reference