Postgresql – How to drop all tables with Sequelize.js using postgresql

postgresqlsequelize.js

I am trying:

if (process.NODE_ENV === 'test') {
  foreignKeyChecks = 0;
  forceSync = true;
} else {
  foreignKeyChecks = 1;
  forceSync = false;
}

global.db.sequelize.query("SET FOREIGN_KEY_CHECKS = " + foreignKeyChecks).then(function() {
  return global.db.sequelize.sync({
    force: forceSync
  });
}).then(function() {
  return global.db.sequelize.query('SET FOREIGN_KEY_CHECKS = 1');
}).then(function() {
  var server;
  console.log('Initialzed database on:');
  console.log(config.db);
  return server = app.listen(port, function() {
    return console.log("Server listening at http://" + (server.address().address) + ":" + (server.address().port));
  });
})["catch"](function(err) {
  return console.log('err', err);
});

module.exports = app;

But I get: SequelizeDatabaseError: unrecognized configuration parameter "foreign_key_checks"

I assume I can't have that keyword in postgres? But is there an equivalent way to drop all tables and recreate?

Best Solution

This is an updated answer, targeted at the googlers who wound up here like me.

Sequelize offers a drop function:

drop(options) => promise

Drop all tables defined through this sequelize instance. This is done by calling Model.drop on each model. Sequelize docs

Example

var sequelize = new Sequelize(config.database, config.username, config.password, config);

var someModel = sequelize.define('somemodel', {
  name: DataTypes.STRING
});

sequelize
  .sync() // create the database table for our model(s)
  .then(function(){
    // do some work
  })
  .then(function(){
    return sequelize.drop() // drop all tables in the db
  });