Concatenate with escaped chars in xslt? - xslt

Question

I'm writing xslt code which concatenates some string:
<xsl:attribute name='src'>
<xsl:value-of select="concat('url(&apos;', $imgSrc, '&apos;)')" />
</xsl:attribute>
For some reason I can't use it, I keep getting this error:
Unknown function - Name and number of arguments do not match any function signature in the static context - 'http://www.w3.org/2005/xpath-functions:concat'
while evaluating the expression:
select="concat('url(&apos;', $imgSrc, '&apos;)')"
Any idea?
thx
====================
EDIT
I'm trying to get:
url('some_path')
Was having trouble with the apostrophes, but now it just doesn't work.

Answer 1

The ' references are resolved by the XML parser that parses your XSLT. Your XSLT processor never sees them. What your XSLT processor sees is:
concat('url('', $imgSrc, '')')
Which is not valid because the commas don't end up in the right place to separate the arguments. However, this might work for you, depending on the serializer your XSLT processor uses:
concat("url('", $imgSrc, "')")
This surrounds the arguments in double-quotes, so that your single-quotes do not conflict. The XSLT processor should see this:
concat("url('", $imgSrc, "')")
Another option is to define a variable:
<xsl:variable name="apos" select='"&apos;"'/>
Which can be used like this:
concat('url(', $apos, $imgSrc, $apos, ')')
More here:
When you apply an XSLT stylesheet to a
document, if entities are declared and
referenced in that document, your XSLT
processor won't even know about them.
An XSLT processor leaves the job of
parsing the input document (reading it
and figuring out what's what) to an
XML parser; that's why the
installation of some XSLT processors
requires you to identify the XML
parser you want them to use. (Others
include an XML parser as part of their
installation.) An important part of an
XML parser's job is to resolve all
entity references, so that if the
input document's DTD declares a cpdate
entity as having the value "2001" and
the document has the line "copyright
&cpdate; all rights reserved", the XML
parser will pass along the text node
"copyright 2001 all rights reserved"
to put on the XSLT source tree.

Answer 2

From http://www.w3.org/TR/xpath/#NT-Literal
[29] Literal ::= '"' [^"]* '"' | "'" [^']* "'"
Meaning that an XPath literal string value can't have the delimiter as also part of the content.
For this you should use the host language. In XSLT:
<xsl:variable name="$vPrefix">url('</xsl:variable>
<xsl:variable name="$vSufix">')</xsl:variable>
<xsl:attribute name="src">
<xsl:value-of select="concat($vPrefix, $imgSrc, $vSufix)" />
</xsl:attribute>
Or more proper:
<xsl:attribute name="src">
<xsl:text>url('</xsl:text>
<xsl:value-of select="$imgSrc"/>
<xsl:text>')</xsl:text>
</xsl:attribute>