<xforms:repeat> and indexing

Todd,

The value of position() in this case is always 1. Instead you can use:

<xforms:group ref=".[empty(preceding-sibling::log)]">
   This will run only on the first log

<xforms:group ref=".[exists(preceding-sibling::log)]">
   This will run on all but the first log

Alex

On 11/14/06, Todd Gochenour <[hidden email]> wrote:

> Iterating through a series of <log> elements, I want to render the first one
> differently than the rest.  I have code that looks like:
>
> <xforms:repeat nodeset="log" id="log">
>  <xforms:group ref=".[position()>1]">
>   <xforms:group ref=".[by!='']">
>    <xforms:output ref="by"/>
>   </xforms:group>
>  </xforms:group>
>  <xforms:group ref=".[position()=1]">
>   <xforms:input ref="by" class="task-description-input">
>    <xforms:label>User </xforms:label>
>   </xforms:input>
>  </xforms:group>
> </xforms:repeat>
>
> When it renders, the second group (the input, not the output) is used in all
> cases.  Ideas?
>
> Todd
>
> --

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Todd,

The value of position() in this case is always 1. Instead you can use:

<xforms:group ref=".[empty(preceding-sibling::log)]">
   This will run only on the first log

<xforms:group ref=".[exists(preceding-sibling::log)]">
   This will run on all but the first log

Alex

On 11/14/06, Todd Gochenour <[hidden email]> wrote:

> Iterating through a series of <log> elements, I want to render the first one
> differently than the rest.  I have code that looks like:
>
> <xforms:repeat nodeset="log" id="log">
>  <xforms:group ref=".[position()>1]">
>   <xforms:group ref=".[by!='']">
>    <xforms:output ref="by"/>
>   </xforms:group>
>  </xforms:group>
>  <xforms:group ref=".[position()=1]">
>   <xforms:input ref="by" class="task-description-input">
>    <xforms:label>User </xforms:label>
>   </xforms:input>
>  </xforms:group>
> </xforms:repeat>
>
> When it renders, the second group (the input, not the output) is used in all
> cases.  Ideas?
>
> Todd
>
> --
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I hope I can explain this: it's clear in my own mind but a little tricky to put into words.

 

The reason that position() is always 1 in the case you had is that the meaning of position() is always in the current context, whatever that may be.

 

When you write <xforms:group ref=".[position()=1]">, the "." becomes the current context for the [position()=1] predicate, which means that it looks at the position of the current element among the list ".".  This list has only one element in it, which is why the position is 1.

 

On the other hand, the preceding-sibling:: axis isn't restricted to the current context so it can "see" the previous elements.

 

HTH,

 

    Dave McIntyre

 

 

 

Dave McIntyre Software Developer
2nd Floor, Orion House, cnr Enfield & Mary St, Mt Eden, PO Box 8273, Auckland, New Zealand M.+64 21 212 8087 P.+64 9 638 0600 F.+64 9 638 0699 S.dave.mcintyre E.[hidden email] W.<a onclick="return top.js.OpenExtLink(window,event,this)" href="http://www.orionhealth.com/" target="_blank">www.orionhealth.com

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>>> "Todd Gochenour" <[hidden email]> 16/11/2006 9:30 a.m. >>>

If the value of position() is always 1, then how can there be a preceding-sibling?

On 11/14/06, Alessandro Vernet <[hidden email]> wrote:

Todd,

The value of position() in this case is always 1. Instead you can use:

<xforms:group ref=".[empty(preceding-sibling::log)]">
   This will run only on the first log

<xforms:group ref=".[exists(preceding-sibling::log)]">
   This will run on all but the first log

Alex

On 11/14/06, Todd Gochenour <[hidden email]> wrote:

> Iterating through a series of <log> elements, I want to render the first one
> differently than the rest.  I have code that looks like:
>
> <xforms:repeat nodeset="log" id="log">
>  <xforms:group ref=".[position()>1]">
>   <xforms:group ref=".[by!='']">
>    <xforms:output ref="by"/>
>   </xforms:group>
>  </xforms:group>
>  <xforms:group ref=".[position()=1]">
>   <xforms:input ref="by" class="task-description-input">
>    <xforms:label>User </xforms:label>
>   </xforms:input>
>  </xforms:group>
> </xforms:repeat>
>
> When it renders, the second group (the input, not the output) is used in all
> cases.  Ideas?
>
> Todd
>
> --

> You receive this message as a subscriber of the [hidden email]
> mailing list.
> To unsubscribe: mailto: [hidden email]
> For general help: mailto:[hidden email]?subject=help
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>
>
>


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