Using dynamic search parameters with Sequelize.js

Question

I'm trying to follow the Sequelize tutorial on their website.

I have reached the following line of code.

Project.findAll({where: ["id > ?", 25]}).success(function(projects) {
  // projects will be an array of Projects having a greater id than 25
})

If I tweak it slightly as follows

Project.findAll({where: ["title like '%awe%'"]}).success(function(projects) {
    for (var i=0; i<projects.length; i++) {
        console.log(projects[i].title + " " + projects[i].description);
    }
});

everything works fine. However when I try to make the search parameter dynamic as follows

Project.findAll({where: ["title like '%?%'", 'awe']}).success(function(projects) {
    for (var i=0; i<projects.length; i++) {
        console.log(projects[i].title + " " + projects[i].description);
    }
});

It no longer returns any results. How can I fix this?

Answer 1

Now on Sequelize you can try this

{ where: { columnName: { $like: '%awe%' } } }

See http://docs.sequelizejs.com/en/latest/docs/querying/#operators for updated syntax

Answer 2

I think you would do that like this:

where: ["title like ?", '%' + 'awe' + '%']

So if you were doing this with an actual variable you'd use:

Project.findAll({where: ["title like ?", '%' + x + '%']}).success(function(projects) {
    for (var i=0; i<projects.length; i++) {
        console.log(projects[i].title + " " + projects[i].description);
    }
});

Answer 3

Please try this code

const Sequelize = require('sequelize');
const Op = Sequelize.Op;
{ where: { columnName: { [Op.like]: '%awe%' } } }

Answer 4

I would do it in this way:

Project.findAll({where: {title: {like: '%' + x + '%'}, id: {gt: 10}}).success(function(projects) {
  for (var i=0; i<projects.length; i++) {
    console.log(projects[i].title + " " + projects[i].description);
  }
});

In this way you can have nicely more WHERE clausas