How sort elements and store them in a variable, XSLT

问题

I was wondering whether it's possible to sort some elements first and store them (already sorted) in a variable. I would need to refer to them thought XSLT that's why I'd like to store them in a variable.

I was trying to do the following, but it doesn't seem to work

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
 version="1.0">

<xsl:variable name="deposits">
  <xsl:for-each select="/BookingCostings/MultiDeposits">
    <xsl:sort select="substring(@DepositDate, 1, 4)" />
    <xsl:sort select="substring(@DepositDate, 6, 2)" />
    <xsl:sort select="substring(@DepositDate, 9, 2)" />
 </xsl:for-each>
</xsl:variable>

I was trying to sort the elements by @DepositDate in the format 'yyyy-mm-dd' and store them all in the $deposits variable. So that later, I could access them using $deposits[1].

I would appreciate any help and tips!

Thanks a lot!

回答1:

Firstly, in your variable declaration, you do need to do something to create new nodes. Strictly speaking, you are not sorting them, but just reading through them in a given order. I think you need to add some sort of xsl:copy command.

<xsl:variable name="deposits"> 
  <xsl:for-each select="/BookingCostings/MultiDeposits"> 
    <xsl:sort select="substring(@DepositDate, 1, 4)" /> 
    <xsl:sort select="substring(@DepositDate, 6, 2)" /> 
    <xsl:sort select="substring(@DepositDate, 9, 2)" /> 
    <xsl:copy-of select=".|@*" />
 </xsl:for-each> 
</xsl:variable>

What this creates is a 'node-set', but to access it you will need to make use of an extension function in XSLT. Which one you use depends on the XSLT processor you are using. In the example I am about to give, I am using the Microsoft one.

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:ms="urn:schemas-microsoft-com:xslt" version="1.0">

Then, to access the nodes in your variable, you can do something like this

<xsl:value-of select="ms:node-set($deposits)/MultiDeposits[1]/@DepositDate" />

Here is a good article to read up on node-sets

Xml.com article on Node-Sets

回答2:

Using XSLT version 2.0 you could use perform-sort and tell that your variable is of type of a sequence of MultiDeposits using the as keyword (as="element(MultiDeposits)+")
Since your data is already as yyyy-mm-dd you can avoid to use the subtring to get each part of the date and use the sort directly on the field

with this sample xml:

<?xml version="1.0" encoding="ISO-8859-1"?>
<BookingCostings>
  <MultiDeposits depositDate="2001-10-09">1</MultiDeposits>
  <MultiDeposits depositDate="1999-10-09">2</MultiDeposits>
  <MultiDeposits depositDate="2010-08-09">3</MultiDeposits>
  <MultiDeposits depositDate="2010-07-09">4</MultiDeposits>
  <MultiDeposits depositDate="1998-01-01">5</MultiDeposits>
</BookingCostings>

and using the XSLT version 2.0 sheet:

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

 <xsl:template match="/">
 <html>
  <body>

  <xsl:variable name="deposits" as="element(MultiDeposits)+">
   <xsl:perform-sort select="BookingCostings/MultiDeposits">
    <xsl:sort select="@depositDate"/>
   </xsl:perform-sort>
  </xsl:variable>

  first date:<xsl:value-of select="$deposits[1]/@depositDate"/>,
  last date:<xsl:value-of select="$deposits[last()]/@depositDate"/>

  </body>
 </html>
 </xsl:template>

</xsl:stylesheet>

the ouput will be:

first date:1998-01-01, last date:2010-08-09

回答3:

Guess (don't have dev env to hand):

Add <xsl:value-of select="." />

Before the closing </xsl:for-each>

来源：https://stackoverflow.com/questions/2272123/how-sort-elements-and-store-them-in-a-variable-xslt

标签

xslt

date

sorting

variables

store