Sequelize.js delete query?


Question

Is there a way to write a delete/deleteAll query like findAll?

For example I want to do something like this (assuming MyModel is a Sequelize model...):

MyModel.deleteAll({ where: ['some_field != ?', something] })
    .on('success', function() { /* ... */ });
1
77
12/6/2011 3:40:58 PM

Accepted Answer

For anyone using Sequelize version 3 and above, use:

Model.destroy({
    where: {
        // criteria
    }
})

Sequelize Documentation - Sequelize Tutorial

179
8/16/2017 11:23:44 AM

I've searched deep into the code, step by step into the following files:

https://github.com/sdepold/sequelize/blob/master/test/Model/destroy.js

https://github.com/sdepold/sequelize/blob/master/lib/model.js#L140

https://github.com/sdepold/sequelize/blob/master/lib/query-interface.js#L207-217

https://github.com/sdepold/sequelize/blob/master/lib/connectors/mysql/query-generator.js

What I found:

There isn't a deleteAll method, there's a destroy() method you can call on a record, for example:

Project.find(123).on('success', function(project) {
  project.destroy().on('success', function(u) {
    if (u && u.deletedAt) {
      // successfully deleted the project
    }
  })
})

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