regular expression in replace()

3zor · May 19th, 2009, 01:40 PM

For solution of the task with search and keyword replacement under the dictionary it is necessary to take advantage of following regular expression which selects a word (for example JAVA) having spaces on both sides:

Code:

(?<=\s*)JAVA(?=\s*)

It is impossible to register this expression in function analyze-string, gives out an error. Prompt, please, in what a problem and how it can be solved?

Code:

<xsl:analyze-string select="replace($text, concat('((?<=\s*)', $keyphrase, '(?=\s*))'), '[$1]', 'i')" flags="i" regex="\[(.*?)\]"> 
    <xsl:matching-substring>
        <xsl:element name="a">
            <xsl:attribute name="href" select="$href" />
            <xsl:value-of select="regex-group(1)"/>
        </xsl:element>
    </xsl:matching-substring>
    <xsl:non-matching-substring>
        <xsl:value-of select="." />
    </xsl:non-matching-substring>
</xsl:analyze-string>

mhkay · May 20th, 2009, 02:07 AM

I'm sorry, I don't know Java regular expressions well enough to know what this one is intended to do, but I can tell you that it is not a legal XPath expression. You will find a good description of what is allowed in XPath regular expressions in my book - there is a chapter devoted to the subject.

mhkay · May 20th, 2009, 04:22 AM

The book title is in my signature, and more details are here:

http://www.wrox.com/WileyCDA/WroxTit...470192747.html

Regular expressions are covered in Chapter 14. (You can buy the chapter online if you want).