CODEWITHSUNDEEP
xmlxquery
lisha
lisha

Reputation: 11

xquery: how to change the order of display

<base>
    <restaurant nom="la tour d'argent" etoile="3" ville="Paris">
        <fermeture>dimanche et lundi</fermeture>  
        <menu nom="buffet" prix="200"/>  
        <menu nom="gourmet" prix="300"/> 
      </restaurant>  
      <ville nom="Paris" departement="75"> 
        <plusBeauMonument nom="tour Eiffel" tarif="30"/> 
      </ville>
</base>

I am using XQuery to do some search for this xml file.
I want to display the "nom" and the "departement" together. For example, for this xml file above, I want to show like this:

<restaurant nom="la tour d'argent" departement="75"/>

As you see, the "ville" of "la tour d'argent" is "Paris". If there is a whose "nom" is "Paris" too, we will take its "departement" and display it with the "nom" of restaurant.

Here is my xquery code:

for $r in //restaurant
for $v in //ville

return if($v/@nom=$r/@ville)
then <restaurant nom="{data($r/@nom)}" departement="{data($v/@departement)}"/>
else()

But this code gives me a result below:

<restaurant departement="75" nom="la tour d'argent"/>

you see, the order is not what I want. I want to make sure that "nom" is shown first, however the "departement" is first now.

Could you help me?

Upvotes: 1

Views: 170

Answers (1)

Nick
Nick

Reputation: 2514

It's counter intuitive, however when I switch the attributes in your then statement, I get the correct result.

for $r in //restaurant
for $v in //ville

return if($v/@nom=$r/@ville)
then <restaurant departement="{data($v/@departement)}" nom="{data($r/@nom)}" />
else()

Upvotes: 0

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