Sequelizejs, how sort field values when creating an index - mysql

Question

I have some 'invite' model:
'use strict';
export default function (sequelize, DataTypes) {
return sequelize.define('Invite', {
userFromId: {
type: DataTypes.INTEGER,
modal: 'User',
key: '_id',
onDelete: 'cascade'
},
userToId: {
type: DataTypes.INTEGER,
modal: 'User',
key: '_id',
onDelete: 'cascade'
},
accept1: {
type: DataTypes.BOOLEAN,
defaultValue: false
},
accept2: {
type: DataTypes.BOOLEAN,
defaultValue: false
}
},{
indexes: [
{
name:'id',
unique: true,
fields: ['userFromId','userToId'],
}]
}
);
}
I need that when i to create a new 'Invite', a combination of 'userFromId' and 'userToId' created one index regardless of who gets invited. For example invite {userFromId:1,userToId:2} and {userFromId:2,userToId:1} must have the same index.
I use sequelize: 3.29.0 and mysql.

Answer 1

This is not possible using a database index because [1, 2] != [2, 1]. If you want to enforce this type of uniqueness one option is to create a new field with a unique index and set it's value to be lower + : + higher so that in the example above you always end up with 1:2.
// Create a field called 'uniqueKey' set to the sorted IDs
export default function (sequelize, DataTypes) {
return sequelize.define('Invite', {
// ... other fields
uniqueKey: {
type: DataTypes.String,
allowNull: false,
defaultValue: function() {
// create an array of the IDs, sort, then join with :
return [
this.getDataValue('userFromId'),
this.getDataValue('userToId'),
].sort().join(':');
},
},
// ... other fields
},{
indexes: [
{
name:'uniqueKey',
unique: true,
fields: ['uniqueKey'],
}]
}
);
}
You can see how the defualtValue is calculated here, the key/result is the same regardless of the order of IDs:
[1,2].sort().join(':');
"1:2"
[2,1].sort().join(':');
"1:2"