[rules-users] Accumulate / collect
Esteban Aliverti
esteban.aliverti at gmail.com
Wed Nov 16 04:34:57 EST 2011
You already have 2 accumulate functions to do what you need:
- collectList
- collectSet
So, your rule will look like this:
$countries : HashSet(empty == false) from accumulate (City($name matches
"X.*", $country : country), collectSet($country))
Best Regards,
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Esteban Aliverti
- Developer @ http://www.plugtree.com
- Blog @ http://ilesteban.wordpress.com
2011/11/15 Wolfgang Laun <wolfgang.laun at gmail.com>
> This is possible with an accumulate where you code the init/action/result
> explicily. Perhaps not fully "out of the box" but a box that holds
> everything would be rather big, wouldn't it? ;-)
>
> -W
>
> 2011/11/15 Bruno Freudensprung <bruno.freudensprung at temis.com>
>
>> **
>> Hi all,
>>
>> There is something I can't express using "collect" or "accumulate" and I
>> would like to have your opinion.
>> Let's imagine I have the following types :
>>
>> # a country type
>> * declare Country
>> name : String
>> end
>> *
>> # a city type holding a reference to its country
>> * declare City
>> name : String
>> country : Country
>> end
>> *
>> Let's imagine I have all Country and City objects into the working memory.
>> I want to get the set of Countries corresponding to Cities whose name
>> starts with "X".
>>
>> I have the impression that I need a kind of (nonexistent right?)
>> "collect" syntax that would look like the "accumulate" syntax (a kind of
>> "anonymous" accumulate function):
>>
>> # meaning I want to collect $country objects and not City objects
>> $countries : HashSet() from collect (City($name matches "X.*",
>> $country : country)*, $country*)
>>
>> Or a home made accumulate function that builds a set of countries:
>>
>> # custom "buildset" accumulate function
>> $countries : HashSet() from accumulate (City($name matches "X.*",
>> $country : country), *buildset*($country))
>>
>> Do you see any other (possibly out of the box) solution?
>>
>> Many thanks in advance for your answers,
>> Best regards,
>>
>> Bruno.
>>
>>
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>>
>>
>
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