Specifying specific fields with Sequelize (NodeJS) instead of *
You have to specify the attributes as a property in the object that you pass to findAll():
Project.findAll({attributes: ['name', 'age']}).on('success', function (projects) { console.log(projects);});How I found this:
The query is first called here: https://github.com/sdepold/sequelize/blob/master/lib/model-definition.js#L131
Then gets constructed here: https://github.com/sdepold/sequelize/blob/master/lib/connectors/mysql/query-generator.js#L56-59
Try this in new version
template.findAll({ where: { user_id: req.params.user_id }, attributes: ['id', 'template_name'], }).then(function (list) { res.status(200).json(list);})
Use the arrays in the attribute key. You can do nested arrays for aliases.
Project.findAll({ attributes: ['id', ['name', 'project_name']], where: {id: req.params.id}}).then(function(projects) { res.json(projects);})Will yield:
SELECT id, name AS project_name FROM projects WHERE id = ...;