Dev - DISTINCT parameter in performFind

[Alexandre Gomes-8]

Hi all,

Is there a way to do a distinct performFind without changing the service
code?
(The reason behind this is that I'm using a view for searching
PartyClassifications with certain Parties but I just want the unique
PartyClassifications' details to show up in the result).

Thank you for your time,

Alexandre Gomes

 
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[David E. Jones]

Use the service to create the find condition and then do the find  
using an EE call (which you can even do in screen defs or whatever).

-David


On Jan 16, 2006, at 5:25 PM, Alexandre Gomes wrote:

> Hi all,
>
> Is there a way to do a distinct performFind without changing the  
> service
> code?
> (The reason behind this is that I'm using a view for searching
> PartyClassifications with certain Parties but I just want the unique
> PartyClassifications' details to show up in the result).
>
> Thank you for your time,
>
> Alexandre Gomes
>
>
> _______________________________________________
> Dev mailing list
> [hidden email]
> http://lists.ofbiz.org/mailman/listinfo/dev

 
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[Alexandre Gomes-8]

Thanks David, that's a brilliant idea.

Could you just help me with the syntax? I'm trying the following and it
isn't working:

<set field="entityName"
value="PartyClassificationGroupAndPartyAndPersonAndContactMech"/>
<service service-name="performFind" result-map-name="result">
   <field-map field-name="inputFields" env-name="requestParameters"/>
   <field-map field-name="entityName" env-name="entityName"/>
</service>        

<entity-condition
entity-name="PartyClassificationGroupAndPartyAndPersonAndContactMech"
list-name="partyClassificationGroupAndPartyAndPersonAndContactMechs"
distinct="true">
   <condition-object field-name="result.entityConditionList"/>
<order-by field-name="description"/>
</entity-condition>


Alexandre

On Mon, 2006-01-16 at 18:22 -0700, David E. Jones wrote:

> Use the service to create the find condition and then do the find  
> using an EE call (which you can even do in screen defs or whatever).
>
> -David
>
>
> On Jan 16, 2006, at 5:25 PM, Alexandre Gomes wrote:
>
> > Hi all,
> >
> > Is there a way to do a distinct performFind without changing the  
> > service
> > code?
> > (The reason behind this is that I'm using a view for searching
> > PartyClassifications with certain Parties but I just want the unique
> > PartyClassifications' details to show up in the result).
> >
> > Thank you for your time,
> >
> > Alexandre Gomes
> >
> >
> > _______________________________________________
> > Dev mailing list
> > [hidden email]
> > http://lists.ofbiz.org/mailman/listinfo/dev
>
>  _______________________________________________
> Dev mailing list
> [hidden email]
> http://lists.ofbiz.org/mailman/listinfo/dev

 
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[David E. Jones]

That's about the idea, but you'll want to use the "prepareFind"  
service. There are some examples, just search for prepareFind in the  
xml files in the project.

-David


On Jan 16, 2006, at 6:52 PM, Alexandre Gomes wrote:

> Thanks David, that's a brilliant idea.
>
> Could you just help me with the syntax? I'm trying the following  
> and it
> isn't working:
>
> <set field="entityName"
> value="PartyClassificationGroupAndPartyAndPersonAndContactMech"/>
> <service service-name="performFind" result-map-name="result">
>    <field-map field-name="inputFields" env-name="requestParameters"/>
>    <field-map field-name="entityName" env-name="entityName"/>
> </service>
>
> <entity-condition
> entity-name="PartyClassificationGroupAndPartyAndPersonAndContactMech"
> list-name="partyClassificationGroupAndPartyAndPersonAndContactMechs"
> distinct="true">
>    <condition-object field-name="result.entityConditionList"/>
> <order-by field-name="description"/>
> </entity-condition>
>
>
> Alexandre
>
> On Mon, 2006-01-16 at 18:22 -0700, David E. Jones wrote:
>> Use the service to create the find condition and then do the find
>> using an EE call (which you can even do in screen defs or whatever).
>>
>> -David
>>
>>
>> On Jan 16, 2006, at 5:25 PM, Alexandre Gomes wrote:
>>
>>> Hi all,
>>>
>>> Is there a way to do a distinct performFind without changing the
>>> service
>>> code?
>>> (The reason behind this is that I'm using a view for searching
>>> PartyClassifications with certain Parties but I just want the unique
>>> PartyClassifications' details to show up in the result).
>>>
>>> Thank you for your time,
>>>
>>> Alexandre Gomes
>>>
>>>
>>> _______________________________________________
>>> Dev mailing list
>>> [hidden email]
>>> http://lists.ofbiz.org/mailman/listinfo/dev
>>
>>  _______________________________________________
>> Dev mailing list
>> [hidden email]
>> http://lists.ofbiz.org/mailman/listinfo/dev
>
>
> _______________________________________________
> Dev mailing list
> [hidden email]
> http://lists.ofbiz.org/mailman/listinfo/dev

 
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