Sequelize error when using sequelize.fn in where clause

Question

The Sequelize docs show:

Post.findAll({
  where: sequelize.where(sequelize.fn('char_length', sequelize.col('status')), 6)
});

My query:

models.appointment.findAll({
    where: Sequelize.where(Sequelize.fn('char_length', Sequelize.col('department')), 6)
})

Is throwing an error:

Unhandled rejection Error: Invalid value Where {
  attribute: Fn { fn: 'lower', args: [ [Object] ] },
  comparator: '=',
  logic: 6 }

With Sequelize v4.42.x (newest) and mysql. Why would this query be throwing an error like this?

Chuong Tran · Accepted Answer · 2019-02-27 02:58:02Z

1

Can you try it? and let me know. I found a solution at this post

models.appointment.findAll({
    where: Sequelize.where(Sequelize.fn('char_length', Sequelize.col('department')), '=',6)
})

answered Feb 27, 2019 at 2:58

Chuong Tran

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3 Comments

nice! I feel like that needs to be spelled out better in the docs. Thanks much

I agree. Currently, the document of sequelize is very simple, not cover all case. We need more case study

You can also use Raw where query like - where: { $and: Sequelize.literal("(char_length(department) = 6)") },