SELECT
DriverID, DriverName, list(CarCost)
FROM ( ... )
GROUP BY  DriverID, DriverName

Il giorno gio 7 mar 2019 alle ore 12:27 Luigi Dell'Aquila <
[email protected]> ha scritto:

> oh, sorry, I didn't scroll the screen :D
>
> You have to use a set() or list() aggregation on CarCost and do a GROUP BY
> on the other columns
>
> Thanks
>
> Luigi
>
> Il giorno gio 7 mar 2019 alle ore 12:26 Luigi Dell'Aquila <
> [email protected]> ha scritto:
>
>> Hi Michael,
>>
>> Just add a DISTINCT to the RETURN, ie
>>
>> RETURN DISTINCT driver.@rid AS DriverID, driver.name AS DriverName,
>> car.cost AS CarCost
>>
>> Thanks
>>
>> Luigi
>>
>> Il giorno mer 6 mar 2019 alle ore 22:46 Michael Faughn <
>> [email protected]> ha scritto:
>>
>>> SELECT * FROM (MATCH {Class: Person, as: driver, where:(name='Bob')}-
>>> Driving->{Class: Vehicle, as:car} RETURN driver.@rid AS DriverID, driver
>>> .name AS DriverName, car.cost AS CarCost)
>>>
>>> results in the following:
>>>
>>> DriverID
>>> DriverName
>>> CarCost
>>> #332:0
>>> <http://localhost:2480/studio/index.html#/database/test/browse/edit/332:0>
>>> Bob
>>> 222
>>> #332:0
>>> <http://localhost:2480/studio/index.html#/database/test/browse/edit/332:0>
>>> Bob
>>> 111
>>>
>>> I would instead like the following:
>>>
>>> DriverID
>>> DriverName
>>> CarCost
>>> #332:0
>>> <http://localhost:2480/studio/index.html#/database/test/browse/edit/332:0>
>>> Bob
>>> [222, 111]
>>>
>>> Is there a way to write the query so that this can happen?
>>>
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>>

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