Hello, thanks for the help.
This makes sense but it looks like that the value is calculated twice, one
in the query and the other accessing the property.
My idea is to have the result on the query return itself.
2017-03-06 18:31 GMT-03:00 mike bayer <mike...@zzzcomputing.com>:
>
>
> On 03/06/2017 04:16 PM, Leonardo Mata wrote:
>
>> Hello, My applications does some ordering using the distance from
>> latitude and longitude haversine distance, i was able to calculate this
>> using @hybrid.method and @.*expression, but i can't output the
>> calculated distance:
>>
>> /class PartnerAddress(db.Model, WithTimestampsModel, SerializeMixin):/
>> / /
>> / # Columns/
>> / id = db.Column(db.Integer(), primary_key=True, nullable=False)/
>> /
>> /
>> / partner_id = db.Column(/
>> / db.Integer(),/
>> / db.ForeignKey('partner.id'),/
>> / nullable=False/
>> / )/
>> /
>> /
>> / latitude = db.Column(db.Numeric(precision=9, scale=7),
>> nullable=False)/
>> / longitude = db.Column(db.Numeric(precision=10, scale=7),
>> nullable=False)/
>> / /
>> /
>> /
>> / @hybrid_method/
>> / def distance(self, lat, lng):/
>> / /
>> / return math.acos(math.cos(math.radians(self.latitude)) *
>> math.cos(math.radians(lat)) */
>> / math.cos(math.radians(self.longitude) -
>> math.radians(lng)) +/
>> / math.sin(math.radians(self.latitude)) *
>> math.sin(math.radians(lat))) * 6371/
>> /
>> /
>> / @distance.expression/
>> / def distance(cls, lat, lng):/
>> / return func.acos(func.cos(func.radians(cls.latitude)) *
>> func.cos(func.radians(lat)) */
>> / func.cos(func.radians(lng) -
>> func.radians(cls.longitude)) +/
>> / func.sin(func.radians(cls.latitude)) *
>> func.sin(func.radians(lat))) * 6371/
>> /
>> /
>> /
>> /
>> /class PartnerV3(db.Model, WithTimestampsModel,/
>> / SoftDeletableModel, SerializeMixin):/
>> / /
>> /
>> /
>> / name = db.Column(db.String(128), nullable=False)/
>> / /
>> /
>> /
>> / address = db.relationship(/
>> / 'mustafar.partner.v3.models.address.PartnerAddress',/
>> / backref='partner',/
>> / uselist=False/
>> / )/
>>
>>
>> when querying like this:
>>
>> query =
>> PartnerV3.query.join(PartnerAddress).order_by(PartnerAddress
>> .distance(lat,
>> lng)).paginate(1,10, False)
>>
>> I can paginate the result, but it returns only the PartnerV3 object, i
>> can't access the distance.
>>
>> How do I access this distance property?
>>
>
> you have a PartnerV3 there so the PartnerAddress is on the .address
> property:
>
> query =
> PartnerV3.query.join(PartnerAddress).order_by(PartnerAddress.distance(lat,
> lng))
>
>
> some_partner = query.first()
>
> partner_address = some_partner.address
> distance = partner_address.distance(lat, lng)
>
>
>
>
>
>
>>
>>
>>
>>
>> --
>> SQLAlchemy -
>> The Python SQL Toolkit and Object Relational Mapper
>>
>> http://www.sqlalchemy.org/
>>
>> To post example code, please provide an MCVE: Minimal, Complete, and
>> Verifiable Example. See http://stackoverflow.com/help/mcve for a full
>> description.
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>>
>
> --
> SQLAlchemy - The Python SQL Toolkit and Object Relational Mapper
>
> http://www.sqlalchemy.org/
>
> To post example code, please provide an MCVE: Minimal, Complete, and
> Verifiable Example. See http://stackoverflow.com/help/mcve for a full
> description.
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--
Leonardo Luiz Padovani da Mata
barr...@gmail.com
"May the force be with you, always"
"Nerd Pride... eu tenho. Voce tem?"
--
SQLAlchemy -
The Python SQL Toolkit and Object Relational Mapper
http://www.sqlalchemy.org/
To post example code, please provide an MCVE: Minimal, Complete, and Verifiable
Example. See http://stackoverflow.com/help/mcve for a full description.
---
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