Format xml using xslt and regex

t_herbert47 · January 18th, 2010, 03:07 PM

Hallo Leute,
I have a XML which I want to format by using XSLT and Regular Expressions.

I would like to transform this

Code:

<Mutter>
	<Kind>
		<Name>Jonas Maier</Name>
		<Geburtstag>12-3-2001</Geburtstag>
	</Kind>
	<Kind>
		<Name>Leon Jung</Name>
		<Geburtstag>7-30-1981</Geburtstag>
	</Kind>
	<Kind>
		<Name>Anna Krause</Name>
		<Geburtstag>4-14-1995</Geburtstag>
	</Kind>
</Mutter>

into something like this

Code:

<Mutter>
	<Kind>
		<Name>Maier, Jonas</Name>
		<Geburtstag>3-December-2001</Geburtstag>
	</Kind>
	<Kind>
		<Name>Jung, Leon</Name>
		<Geburtstag>30-July-1981</Geburtstag>
	</Kind>
	<Kind>
		<Name>Krause, Anna</Name>
		<Geburtstag>14-April-1995</Geburtstag>
	</Kind>
</Mutter>

So:

The name is restructured, by replacing the places and by adding a "," in the middle
Example: Name Surname --> Surname, Name
The date is also reformated
Example: MM-dd-yyyy --> dd-MMMM-yyyy

Can you offer me some help?

mhkay · January 18th, 2010, 05:05 PM

As regards the names, you don't really have enough information to do the job properly: given the name Norman St John Stevas, you need some pretty sophisticated logic to turn that into St John Stevas, Norman. Ditto with names like de Klerk or van Dijk. But assuming you just want to blindly split it at the last space, it's

Code:

<xsl:variable name="tokens" select="tokenize($in, '\s+')"/>
<xsl:value-of select="$tokens[last()]"/>
<xsl:text>, </xsl:text>
<xsl:value-of select="$tokens[position() lt last()]"/>

For the dates: split it into three tokens using tokenize(), expand the day to two digits using format-number(), convert the month to a string by doing ("January", "February"...)[$month], and then use concat() to put it back together again.

t_herbert47 · January 18th, 2010, 05:17 PM

Thank you very mych. Could you just show me an example how to reformat the date with regex?

mhkay · January 18th, 2010, 06:21 PM

No, I'm not going to write the code for you. You've got to learn to do it for yourself. If there's part of my explanation you don't understand, please explain what you don't understand.

CHRD · January 19th, 2010, 05:53 AM

What is the best way to handle this issue (datetime-data)? Or: How should I store/format date-fields within a XML-file to avoid this kind of prehistoric format transformation? Sorry, but this technique reminds me on handling date 'format' at the DOS-prompt. What I am looking for is a mechanism like a date-format within a database and a convert-function. Is there something similar within XLST?

mhkay · January 19th, 2010, 06:06 AM

It's always best to store dates in ISO format (yyyy-mm-dd) and then you can use the XSLT format-date() function to display them using local conventions for your country or language.

CHRD · January 19th, 2010, 07:28 AM

Still don't get it. With the given data I would transform this to ISO and use format-date to any representation I like. Like that?

Code:

<?xml version="1.0" encoding="iso-8859-1"?>
<xsl:stylesheet 
	version="2.0" 
	xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
>
<xsl:output 
	method="xml" 
	encoding="iso-8859-1" 
	indent="yes" 
/>

<xsl:template match="/Mutter/Kind">

<xsl:variable name="tokens" select="tokenize(Geburtstag, '-')"/>
<xsl:variable name="demo-data">
	<xsl:value-of select="$tokens[3]"/>-<xsl:value-of select="substring(string(100+number($tokens[1])), 1, 2)"/>-<xsl:value-of select="substring(string(100+number($tokens[2])), 1, 2)"/>
</xsl:variable>
<xsl:value-of select="format-date($demo-data, '[D]. [MNn] [Y]')"/>
</xsl:template>

</xsl:stylesheet>

Result is closed to what is required, but there is still some remarable difference (Saxon 9.2.0.3):

Code:

[Language: en]10. November 2001
[Language: en]13. October 1981
[Language: en]11. October 1995

Regards

Christian

samjudson · January 19th, 2010, 07:51 AM

It would help if you code was correct.

substring(x, 1, 2) returns the first two characters - you want the second two, so substring(x, 2, 2).

Alternatively:

<xsl:variable name="demo-data">
<xsl:value-of select="$tokens[3]"/>-<xsl:value-of select="if (number($tokens[1]) < 10) then concat('0',$tokens[1]) else $tokens[1]"/>-<xsl:value-of select="if (number($tokens[2]) < 10) then concat('0',$tokens[2]) else $tokens[2]"/>
</xsl:variable>

While its slightly more verbose it explains what you are trying to do better and is easier to understand.

mhkay · January 19th, 2010, 08:13 AM

There are a couple of known problems with localization of dates in Saxon 9.2

This one concerns installation of localization code for languages other than English:

https://sourceforge.net/tracker/?fun...72&atid=397617

This one concerns the spurious [Language:en]:

https://sourceforge.net/tracker/?fun...72&atid=397617

I would recommend that if you want English format dates and your machine is configured to a language other than English, then you use the language argument of format-date() to say what language you want.

CHRD · January 19th, 2010, 02:42 PM

Ad 1 (Sam):

Of course the substring argument was wrong and should be 2,2

The alternative code was no real alternative, as it will only work when there is always no leading '0'. To make it equal you have to add another conversion. So the code gets even a bit more longer.

Code:

<xsl:value-of select="$tokens[3]"/>-<xsl:value-of select="if (number($tokens[1]) &lt; 10) then concat('0',number($tokens[1])) else $tokens[1]"/>-<xsl:value-of select="if (number($tokens[2]) &lt; 10) then concat('0',number($tokens[2])) else $tokens[2]"/>

I though that the if-statement alternative is much slower. But it figured out to perform the same way. I tested this by transforming some million elements.

So in the end it is maybe only a matter of taste.

Ad 2 (Mike):
I have to use the binaries 'as is'. There is no way to use a customized/localized version at my customer's site. When I try to use the language argument like

Code:

format-date($demo-data, '[D]. [MNn] [Y]', 'de', (), ())

there is no effect. The result has still a leading [Language: en] and it is not localized/translated.

Best regards

Christian