Sequelize function on where - node.js

Question

I using sequelize 5.21.2 on Node 12.
I want to get like this:
SELECT `id`
FROM `order`
WHERE (substr(`order`.`code`,1,8) >= '20200101'
AND substr(`order`.`code`,1,8) <= '20201230')
So I wrote it like this:
return Model.order.findAll({
attributes: [
'id',
],
where: {
[sequelize.fn('substr', Model.sequelize.col('code'), 1, 8)]: {
[Op.gte]: params.start_date,
[Op.lte]: params.end_date,
},
},
});
But It returns:
SELECT `order`.`id`
FROM `order`
WHERE (`order`.`[object Object]` >= '20200101'
AND `order`.`[object Object]` <= '20201230')
I tried
[sequelize.literal('STR_TO_DATE(substr(code,1,8),\'%Y %m %d\')'), 'order_date']
[sequelize.literal('STR_TO_DATE(substr(code,1,8),\'%Y %m %d\')')]
[sequelize.fn('substr', Model.sequelize.col('code'), 1, 8)]
but they work fine when I SELECT, but not WHERE.
How do I get the results I want? I couldn't find any relevant information in the official manual.. (https://sequelize.org/v5/manual/querying.html#where)

Answer 1

Use sequelize.where and Op.and to generate the select statement using a function and not a column. See this page in the Sequelize documentation - Advanced queries with functions (not just columns)
const { sequelize, order } = Model;
return order.findAll({
attributes: [
'id',
],
where: {
[Op.and]: [
sequelize.where(sequelize.fn('substr', sequelize.col('code'), 1, 8), {
[Op.gte]: params.start_date,
}),
sequelize.where(sequelize.fn('substr', sequelize.col('code'), 1, 8), {
[Op.lte]: params.end_date,
}),
],
},
});