In Javascript (like Java and C) a backslash in a string literal must always be escaped as "\\".

However, a legal URI never contains a backslash. URI's use forwards slash as the separator character.

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

Thanks Michael,
One more query,
Please tell me whether there is any mistake in following xslt code lines

<a>
    <xsl:attribute name="href"><xsl:value-of select="NavigateUrl"></xsl:value-of></xsl:attribute>
    <xsl:attribute name="target">DetailsPage</xsl:attribute>
    <xsl:attribute name="onmouseover">doButtons('<xsl:value-of select="ID"></xsl:value-of>', '<xsl:value-of select="ImgUrl1"></xsl:value-of>')</xsl:attribute>
    <xsl:attribute name="onmouseout">doButtons('<xsl:value-of select="ID"></xsl:value-of>', '<xsl:value-of select="ImgUrl2"></xsl:value-of>')</xsl:attribute>
    <img><xsl:attribute name="name"><xsl:value-of select="ID"></xsl:value-of></xsl:attribute>
    <xsl:attribute name="src"><xsl:value-of select="ImgUrl1"></xsl:value-of></xsl:attribute></img>
    </a>

Thanks and regards,

I'm not into quiz questions. If you think there is an error, tell me why you think there is an error. If you don't, then why ask the question?

I'd suggest you put your complete stylesheet through the Saxon processor. Saxon will tell you whether there are any static errors in it.

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

Hi Michael,

It is giving me error:

"Oject does't support this property or method."

Swapnil.

That looks to me like a Javascript error, not an XSLT error.

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

Let's get this straight, this is entirely a JavaScript problem.
We need to back to basics, what is your XML, what HTML do you want to output?
Write the HTML first by hand and test it, if it works then it should be relatively simple to create from your XML.

If you can't write the HTML by hand then you need to post that as a problem on the Web => Javascript How-To forum.



--

Joe (Microsoft MVP - XML)

Thanks to all,

I got the solution,

Problem was in the function--

function doButtons(name, pic)
{
    document.getElementById(name).src = pic; //Here was the problem
}

follwing is the right answer:

function doButtons(name, pic)
{
    document[name].src=pic;
}

Regards,