Question

XSLT. IF statement, or something more powerful

Asked by: silentz

Hello all.

I'm using XSLT to display data from an xml file.
Here's a very simplified idea of the format of the XML file

<Practice>
   <PracticeName>Practice One</PracticeName>
   <Doctor>
      <sex>M</sex>
   </Doctor>
   <Doctor>
      <sex>F</sex>
   </Doctor>
</Practice>

So a practice can have one or more Doctors, each of which has a sex node. That sounds dirty :)

Anyway. I need to ONLY SHOW A PRACTICE IF IT HAS AT LEAST 1 FEMALE DOCTOR.

I can filter out the Male doctors in a practice, but if there are only male doctors, it still shows the PracticeName. I would like it to completely skip the practice if there are no Females.

Here's pseudocode for what i'm trying to do (and how i'd do it in vb)

for each Practice
   for each Doctor
      if sex = 'M' then foundmale = true
   next
   if foundmale = false then
      Print Practice
   end if
next

Comprendez?  I think either a) XSLT cant do what i'm trying to do in which case its crap. or more likely b) I just dont know enough XSLT.

Something along the lines of <xsl:if test="count(sex='M')>0">  is what i'm looking for. But that obviously doesnt work!

Any help is appreciated greatly.

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Asked On
2002-09-18 at 07:41:32ID20360511
Tags

count

,

statement

,

xslt

,

xsl

Topics

Extensible Markup Language (XML)

,

Extensible Stylesheet Language Transformation (XSLT)

Participating Experts
2
Points
200
Comments
8

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Answers

 

by: raizonPosted on 2002-09-18 at 08:04:19ID: 7288492

I havn't tested it but take a look at this and see if this gets what you need.  Noticed I changed the sex node to an attribute.

<Practice>
  <PracticeName>Practice One</PracticeName>
  <Doctor sex="M">

  </Doctor>
  <Doctor sex="F">
     
  </Doctor>
</Practice>



<xsl:for-each select="/Practice/PracticeName/Doctor[@sex='F']>
<!--put your code here -->
</xsl:for-each>

 

by: silentzPosted on 2002-09-18 at 08:10:48ID: 7288516

Sorry i should have said - i can't change the format of the XML... it's coming to me from a webservice on a remote machine. i have to work with what i've got.

Also, your code would create too many repeats i think. it looks like it would show this:

Practice One
Doctor One - F

Practice One
Doctor Two - F

Practice One
Doctor Three - F

... see what i mean? it repeats the whole practicename.

its damn hard to explain, all this XML stuff.

 

by: raizonPosted on 2002-09-18 at 08:32:30ID: 7288586

Depends on what you do with the code inside the for-each loop.  I could show only what I needed for each practice that had a female doctor.


<xsl:for-each select="/Practice/PracticeName/Doctor/name(Sex)='F'>
<p>
  <xsl:value-of select="//PracticeName" />
  <xsl:for-each select="//Doctor">
    <xsl:value-of select="//sex"/>
  </xsl:for-each>
</p>
</xsl:for-each>

well thats the general Idea.  

 

by: monasPosted on 2002-09-18 at 08:46:29ID: 7288627

<xsl:for-each select="/Practice[count(Doctor/sex='F')>0]">
 ...
</xsl:for-each>

 

by: silentzPosted on 2002-09-18 at 09:43:23ID: 7288791

oooh. i thought you'd solved it there, Monas.

Thing is that looks right, and I was confident it would work, but it returns an error:

"Argument 1 must return a node-set. Practice[-->count(Doctor/sex='F')<-->0]"

The search continues.  I think that's close to being right, but...

 

by: monasPosted on 2002-09-18 at 23:17:32ID: 7290469

oops :-)

<xsl:for-each select="/Practice[count(Doctor/sex[string(.)='F'])>0]">
...
</xsl:for-each>

I just checked and it works.

 

by: silentzPosted on 2002-09-19 at 01:18:53ID: 7290665

Gentlemen we have a winner. Thanks very much, monas. Dont really understand the syntax, but it works perfectly.

Thanks to raizon too for your help.

 

by: monasPosted on 2002-09-19 at 01:49:26ID: 7290720

string(.) gives string representation of current (specified by ".") node's content.
This is what you want to compare with 'F'.

Doctor/sex - selects all "sex" nodes. Square brackets contains condition what subset to take into consideration.

count() counts.

Outer square brackets contains condition which "Practice" elements to choose.

Thant all!

20120131-EE-VQP-002

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