A serious string problem. Urgent solution needed.

asfak · November 28th, 2007, 05:04 AM

I am new in XSLT.

I need a solution in XSLT from the experts. Say I have a string
with the following values (on the left). I need to extract a substring with the value (on the right) which does not have any trailing dots.

Given sub string I need to extract
------------ ------------------------------
"1.2.1..." "1.2.1"
"1." "1"
"1.2.1.1.1." "1.2.1.1.1"
"1.2.1.2.1...." "1.2.1.2.1"

I think I can make you understand my problem. XSLT has so limited functions for string. I have coded like the following. I had a label attribute. I need to print the value of label attribute which is free of trailing dots. But surely this (my following code) does not work. It works only for the case "1." but not any of the above except "1.". For the case "1.2.1..." my code gives the output "1"

Code:

<xsl:variable name="labelValue" select="@label"/>
<xsl:choose>
    <xsl:when test="ends-with($labelValue,'.')">
        <xsl:value-of select="substring-before($labelValue,'.')"/>    
    </xsl:when>
    <xsl:otherwise>
        <xsl:value-of select="@label"/>
    </xsl:otherwise>
</xsl:choose>

Please give me a solution as quick as possible. I need it urgent.

regards,
Asfak

asfak

samjudson · November 28th, 2007, 05:27 AM

If you are using XSLT 2.0 then there is probably a way to do this using tokenize() function. Alternatively you are looking at a call to a recursive template.

/- Sam Judson : Wrox Technical Editor -/

mhkay · November 28th, 2007, 06:08 AM

You need to be aware that when you say a problem is urgent, many people will avoid responding, because they prefer helping people who have time to study the answers carefully and learn from them.

The first thing to ask is: can you use XSLT 2.0 for this? You're using ends-with(), which is a 2.0 function, and that suggests you could use regular expressions, in which case you can use

replace($labelValue, '(^.*?)\.*$', '$1')

However, creative solutions are often possible in 1.0. I think you could solve this one by using translate() to translate "." into space, then using normalize-space(), then using translate() to translate spaces back into ".".

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