Sequelize (or clear SQL) query for selecting rows what includes value in JSON field? - mysql

Question

I have rows in my MYSQL and I Need Sequelize.js query.
Every row have col of type JSON what include this for example:
[
{id: 1234, blah: "test"},
{id: 3210, blah: "test"},
{id: 5897, blah: "test"}
]
I have id and I need to select row what include this id in at least one object in array.

Answer 1

Raw mysql query will be like this:
SELECT * FROM `user` WHERE JSON_CONTAINS(`comments`, '{"id": 1234}');
Simple sequelize example:
const { fn, col, cast } = this.sequelize;
const User = this.sequelize.define('user', {
id: {
type: Sequelize.INTEGER,
primaryKey: true,
autoIncrement: true
},
comments: DataTypes.JSON,
defaultValue: [],
})
User.findAll({
where: fn('JSON_CONTAINS', col('comments'), cast('{"id": 1234}', 'CHAR CHARACTER SET utf8')),
})
.then(users => console.log('result', users.map(u => u.get())))
.catch(err => console.log('error', err));
Cast function is used to unescape double quotes around "id" to avoid wrong query string like this:
SELECT * FROM `user` WHERE JSON_CONTAINS(`comments`, '{\"id\": 1234}');
There is one more dirty mysql query example (don't use it):
SELECT * FROM `user` WHERE `comments` LIKE '%"id": 1234%';