How to filter by a function in sequelize.js

Question

I'm trying to filter results by a function and then also order the resultset with the function column. Here is my code so far:

db.trip.findAll({
  attributes: [
    [
      db.sequelize.fn('SUM', 
        db.sequelize.col('departure_time'),
        db.sequelize.col('arrival_time')),
      'summation'
    ]
  ],
  where: {
    summation: {
      gte: 500
    }
  },
  order: [['summation', 'ASC']],
  logging: console.log
}).then(function(results) {

});

The SQL that gets generated makes it hard to re-reference the alias that gets created in the attributes:

SELECT SUM("departure_time", "arrival_time") AS "summation"
FROM "trip" AS "trip" 
WHERE "trip"."summation" >= 500 
ORDER BY "trip"."summation" ASC;

Any ideas on how to make this work without making the whole query literal?

Answer 1

I found a way by making a literal statement act as an alias. I'm using Postgres, so I'm not sure if the quoting works in other dialects:

var summationAlias = db.sequelize.literal('"summation"');

db.trip.findAll({
  attributes: [
    [
      db.sequelize.fn('SUM', 
        db.sequelize.col('departure_time'),
        db.sequelize.col('arrival_time')),
      'summation'
    ]
  ],
  where: db.sequelize.where(summationAlias, '>=', 500),
  order: [[summationAlias, 'ASC']],
  logging: console.log
}).then(function(results) {

});

The above code generates the following SQL:

SELECT SUM("departure_time", "arrival_time") AS "summation"
FROM "trip" AS "trip" 
WHERE "summation" >= 500 
ORDER BY "summation" ASC;