Re: [xsl] shortest way to write this xsl:if statement

[henry human <henry_human@xxxxxxxx>]

Hi ken 

fine! thanks, I will try it.
k regards
hh



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Gesendet: 13:59 Freitag, 15.Juni 2012
Betreff: Re: [xsl] shortest way to write this xsl:if statement


At 2012-06-15
12:54 +0100, henry human wrote:
> Hello
> The following if statement is too
long if try it as in the sample bellow.
> How could be shorter?
> The if -
logic to create a for-each loop:
> ----------------
> xsl:if
> D4/G100/6id  or
D4/G100/9id or D4/G100/12id or D4/G100/6id15 or D4/G100/45id or D4/G100/22id,
D4/G100/10id
> or D4/G100/19id is 'A' or 'B' or 'C', or 'D' or 'F'
> 
> The
Sample:
> <xsl:if test="D4/G100/6id = 'A' OR test="D4/G100/6id = 'B' OR
test="D4/G100/6id = 'C' OR test="D4/G100/9id = 'A' OR test="D4/G100/9id = 'B'
.....>
> 
> <xsl:for-each select=" ......">
> ...
> </xsl:for-each>
> 
>
</xsl:if>

Element names cannot begin with digits, so I'm unclear how you are
going to be testing elements such as <6id>.

But, assuming you had elements
D4/G100/X and D4/G100/Y and D4/G100/Z, you could have in XSLT2 the following:
<xsl:if test="D4/G100/(X,Y,Z) = ('A','B','C')">

... which is equivalent to:
  D4/G100/X = 'A' or
  D4/G100/X = 'B' or
  D4/G100/X = 'C' or
  D4/G100/Y =
'A' or
  D4/G100/Y = 'B' or
  D4/G100/Y = 'C' or
  D4/G100/Z = 'A' or
 
D4/G100/Z = 'B' or
  D4/G100/Z = 'C'

When using the "=" comparison operator,
either operand can be a set.  The processor walks through the comparisons in
an arbitrary order eventually testing each of the left operand with each of
the right operand and stops when it hits a true() result and returns true(). 
If you get a false() returned, you know the processor has checked every
possible combination and every combination has returned false().

I hope this
helps.

. . . . . . . . . . . Ken

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