Created a model using:
sequelize-cli model:create --name User --attributes "dispName:string,email:string,phoneNum1:string,vendorId:integer"
Which resulted in the following migration:
'use strict';
module.exports = {
up: (queryInterface, Sequelize) => {
return queryInterface.createTable('Users', {
id: {
allowNull: false,
autoIncrement: true,
primaryKey: true,
type: Sequelize.INTEGER
},
dispName: {
type: Sequelize.STRING
},
// plus others...
});
},
down: (queryInterface, Sequelize) => {
return queryInterface.dropTable('Users');
}
};
I want to change the automatically defined ID to:
cognitoId: {
allowNull: false,
primaryKey: true,
type: Sequelize.STRING(100)
}
So:
Will sequelize be able to recognize this as the ID?
Where all do I need to make this change? I could only identify the migration file.
The model file doesn't have a definition for the cognitoId (or the original auto-generated id field): how will I be able to get the value of a User instance's cognitoId (in the data returned by queries)?
Will changing the auto-generated id field have repercussions down the line?
Is the field name id "magical"? I.e., does the primary key have to be named id?
Is there a better way to do this?
Will changing the types of the fields from Sequelize.STRING to Sequelize.STRING(100) create any issues down the line?
Why doesn't the models file generated by sequelize-cli have the id field defined?
When generating models+migrations from the command-line I couldn't find any syntax to specify the ID or any other customization for the fields.
Using:
[Node: 12.14.1, CLI: 5.5.1, ORM: 5.21.3]
PS: relatively new to NodeJS & completely new to Sequelize.
Yes
You should declare custom named PK in your model
see p.2. If you don't declare PK in your model then sequelize assumes you have id PK with an integer type, autoincremented. If you wish to assign your PK another name you should declare it in the model.
Depends on what changes you make
It is the default PK name in sequelize (see p.3). You can set different name to your PK manually declaring it in your model (see p.3)
Personally I prefer to declare all PKs in my models even if they have id name and default PK type and value.
No issues if all PK values do not exceed this length
see p.3
You can define names and types only for fields while generating models from the command line.
Related
I'm not very familiar with sequelize, but currently I'm working with Node.js and Sequelize and I need to create a new association one to many between two tables. I know the code to generate the association:
school.hasMany(student,{ foreignKey: 'school_id', as : 'studentSchool', sourceKey: 'school_id'});
student.belongsTo(school, {foreignKey: 'school_id', targetKey : 'school_id', as: 'studentSchool'});
My problem is that the application has been deployed and in use for at least 2 years now. So there is a lot of data already. I don't know how to introduce this new association without corrupting the current data or without having to reconstruct the database.
You will need to create a migration for that.
I am assuming you already use sequelize-cli (if you do not, install from npm)
In your terminal, run
npx sequelize-cli migration:generate --name added-association-to-school-and-student
This creates an empty migration file. Fill the file with the code below
module.exports = {
up: (queryInterface, Sequelize) => {
return queryInterface.addColumn("students", "school_id", {
type: Sequelize.DataTypes.INTEGER,
/*
The defaultValue below was assigned because by default constraints are set to true.
This means that all students must belong to a school
If no school_id is specified, mysql sees that this does not follow the constraints and will opt to delete all records from the database.
So assign a default value and after this, you can go ahead to manually assign the correct schools.
ENSURE THAT THE DEFAULT VALUE OF school_id PROVIDED HAS A CORRESPONDING EXISITING RECORD IN THE school TABLE
*/
defaultValue: 1, // or any other existing value. (This is very important!)
references: {
model: "schools",
key: "school_id",
},
});
},
down: (queryInterface, Sequelize) => {
return queryInterface.removeColumn("students", "school_id");
},
};
After creating the migration file, head over to your table definitions and add the associations for the respective tables
In student Table, add this association
school.hasMany(student,{ foreignKey: 'school_id', as : 'studentSchool', sourceKey: 'school_id'});
In the school Table, add this association
student.belongsTo(school, {foreignKey: 'school_id', targetKey : 'school_id', as: 'studentSchool'});
When this is done, run the migration file in your terminal
npx sequelize-cli db:migrate
ALSO, BACK THE DATA UP BEFORE DOING THIS (Just in case)
I'm using sequelize 6.5.0. I created a simple model to do two rudimentary things: a) find records, b) create records. I'm having trouble creating records; specifically, ones with primary key. If I designate the column as primaryKey like so:
const Table = sequelize.define('table', {
id: {
type: DataTypes.UUID,
primaryKey: true
},
datum: {...}
...
and try to create a record like so:
Table.create({datum: 'abc'})
then it will try (and fail) to set the primary key with:
INSERT INTO "table" ("id","datum") VALUES ($1,$2) RETURNING ...;
which is 50% what I did not ask it to do. Now, I don't need this to happen since default value for id is already handled at the database level. So, the next natural move was to not designate id as primaryKey:
const Table = sequelize.define('table', {
id: {
type: DataTypes.UUID,
// primaryKey: true
},
datum: {...}
...
But now sequelize attempts to get smart and throws a tantrum:
Uncaught Error: A column called 'id' was added to the attributes of 'table' but not marked with 'primaryKey: true'
Q) How do I get sequelize to NOT handle primary key on create?
I think you can skip the id field in the definition altogether, and PostgreSQL will still have one
I'm new to Sequelize.js and Databases in general, i haven't used migrations before, but i know that they can be used to make changes to the tables structure in a non-destructive way.
However i'm not sure where to declare column options (notNull, references, validate, ENUM values, etc...)
Should i declare such options in the the model file, or migration file? or both?
Wouldn't adding the options to both model and migration cause duplicate code?
(keep in mind that i'm talking about the initial migrations that create tables to the database, not the migrations that add columns and stuff...)
Any help would be appreciated!
I see three options you can take. The first two options might be edge cases but it helps to understand.
Destructive option
You want to prototype a project and you don't mind losing your data, then you could potentially do not care about migration files and synchronize your database according to your model with:
await sequelize.sync({ force: true });
It will execute on all your models:
DROP TABLE IF EXISTS "your_model" CASCADE;
CREATE TABLE IF NOT EXISTS "your_model" (...)
This command can be executed at the start of your application for example.
Static option
As you mentioned you don't want to add columns and stuff, it's probably a good option.
Now if you don't want to lose data, you could simply use the sync method without the force option:
await sequelize.sync({ });
It will only generate:
CREATE TABLE IF NOT EXISTS "your_model" (...)
Hence, your tables are created according to your models and you don't have to create migration files.
However, if you want to modify your model and it's the most frequent usecase, the new column will not be generated in the table dynamically that's why you need migration scripts.
Flexible option
You will have to define both migration file and your model. That's what the cli does. Here is an example:
# npx sequelize-cli init or create migrations and models folders
npx sequelize-cli model:generate --name User --attributes firstName:string,email:string
Now you will have two more files:
// migrations/<date>-create-user.js
module.exports = {
up: (queryInterface, Sequelize) => {
return queryInterface.createTable('Users', {
id: {
allowNull: false,
autoIncrement: true,
primaryKey: true,
type: Sequelize.INTEGER
},
firstName: {
type: Sequelize.STRING
},
email: {
type: Sequelize.STRING
},
createdAt: {
allowNull: false,
type: Sequelize.DATE
},
updatedAt: {
allowNull: false,
type: Sequelize.DATE
}
});
},
down: (queryInterface, Sequelize) => {
// I usually remove this and create the table only if not exists
return queryInterface.dropTable('Users');
}
};
// models/users.js
module.exports = (sequelize, DataTypes) => {
const User = sequelize.define('User', {
firstName: DataTypes.STRING,
email: DataTypes.STRING
}, {});
User.associate = function(models) {
// associations can be defined here
};
return User;
};
You could refactor the code from the migration and the model, however it will be rather cumbersome because some migration files will only add one column, so merging all of them into the model might probably be less clear.
You should do it in both because as time goes by your models and inital migration will differ from each other. So I suppose you should determine a final structure in models and after that create an initial migration.
Constraints are defined and run on SQL level while validations are run on application level. Sequelize supports having validations and constraints on Models, only constraints can be defined in migrations.
My opinion is to put all constraints in migrations, and validations in Models. That way you have some kind of separation of concern as validations are run before query is made to the database - where constraints are run. You can read more on Sequelize's Validations and Constraints Validations and Constraints
I'm building a Node/Express/Postgres version of an app that I already built in Rails. I'm learning Node, so I figured I'd rebuild something that I know works.
For now, I'm dumping everything in one file (set up my database, defined my models, etc.), just to make sure I have everything set up correctly before I divvy them up into different files.
I set up my postgres database at the very top of the file, like so:
var Sequelize = require('sequelize');
var db = new Sequelize('my_database_name', 'my_username', null, {
host: 'localhost',
dialect: 'postgres',
});
With regard to my models, I have a Politician model:
var Politician = db.define("politician", {
name: {
type: Sequelize.STRING,
},
politicalParty: {
type: Sequelize.STRING
}
});
A Category model:
var Category = db.define("category", {
name: {
type: Sequelize.STRING
},
keywords: {
type: Sequelize.ARRAY(Sequelize.TEXT)
},
});
And a join model of Politician and Category, called "Interest". Because Interest is a join model, it will have a "politicianId" and "categoryId" properties....but will those properties automatically generate in the database? And so, is this how I would define the Interest model, with no properties?
Interest Model:
var Interest = db.define("interest")
Or, will I have to be specific, and create "politicianId" and "categoryId" properties? Like so:
Interest Model:
var Interest = db.define("interest", {
id: {
type: Sequelize.INTEGER,
primaryKey: true,
autoIncrement: true
},
categoryId: {
type: Sequelize.INTEGER,
foreignKey: true
},
politicianId: {
type: Sequelize.INTEGER,
foreignKey: true
}
});
Also, do I need the "foreignKey: true" bit? Or will it automatically know that those properties are foreign keys? Also, do I need the "id" property? I know models automatically create their own primary key "id"...but again, I've been at this for hours, looking at docs, and trying everything.
I then defined my associations (again, all of this is the same file):
Politician.belongsToMany(Category, {through: "Interest"});
Category.belongsToMany(Politician, {through: "Interest"});
The Node/Sequelize docs seems to suggest that defining those 2 associations above will automatically "create a new model called Interest with the equivalent foreign keys politicianId and categoryId." So, do I even need to define a "Interest" model? Also, do I need the follow associations to describe that Interest belongs to Politician and Category?
Interest.belongsTo(Politician);
Interest.belongsTo(Category);
If I don't write the associations saying that Interest belongs to Politican and Catetory, I don't get the "politicianId" and "categoryId" columns in the Interest table. Just the "id" and createdAt/updatedAt columns.
I then created an instance of Politician, Category, and Interest, to persist everything to the database, to see if everything is there and set up correctly:
Politician Object:
var politician1 = Politician.sync({force: true}).then(function(){
return Politician.create(aPoliticianObjectDefinedInthisFile);
});
This works perfectly. I see this object in the politician table in the database.
Category Object:
var category1 = Category.sync({force: true}).then(function(){
return Category.create(aCategoryObjectDefinedInThisFile);
});
This works perfectly. I see this object in the category table in the database.
Here is what doesn't work. Creating an instance/object of Interest and synching it to the database. My thinking is, if I put integers as values, it will know that "politicianId: 1" means point to the politician object with an id of 1, and the same for "categoryId: 1". But when I write it as I have it below, the Interest table doesn't even show up in the Postgres database at all.
Interest Object:
Interest.sync({force: true}).then(function(){
return Interest.create(
{
politicianId: 1,
categoryId: 1
}
);
});
However, when I create the object of Interest like this, with no properties defined, the Interest table appears in the database, along with the "politicianId" and "categoryId" columns, however, those columns are empty. The object's primary id is in there at 1, and the "createdAt" and "updatedAt" columns have data too. But the foreign key columns are blank.
Interest Object:
Interest.sync({force: true}).then(function()
{
return Interest.create(
{
// No properties defined.
}
);
}
);
Sorry for this long post, lol, but, in all:
Am I creating the "Interest" model correctly?
Am I writing the associations for "Interest" correctly?
Do I even need to write associations for Interest, if I already have associations for its parent classes, Politican and Category defined?
In my Rails app, my associations for Politican and Category are like so:
Politician has_many interests
Politican has_many categories through interests
Category has_many interests
Category has_many politicians through interests
Interest belongs_to politician
Interest belongs_to category
But I use the "belongsToMay" association in Node because I got an error telling me to do so.
Basically, I need to create an instance of Politician, an instance of Category, and an instance of Interest that has "politicianId" and "categoryId" columns that point to those aforementioned instances of those classes.
politicanABC -- id: 1
categoryABC -- id: 1
instanceABC -- id: 1; politicanId: 1 (referring to politicanABC); categoryid: 1 (referring to categoryABC).
My app is set up like that in Rails and works wonderfully.
Help and thank you in advance :-)
You don't have to define the Interest model if you are not going to add any additional fields. Sequelize will internally define the model and add all required fields once you do following:
Politician.belongsToMany(Category, {through: "Interest"});
Category.belongsToMany(Politician, {through: "Interest"});
Sync needs to run on database level and not on tables since Interest model is implicit at this point.
db.sync({force: true});
Sequelize will add relationship build methods on both Politician and Category instances. Category will have methods addPolitician(), addPoliticians([]), setPoliticians([]), getPliticians(). Politician instances will have similar functions to associate categories to them. You can connect these after create option is performed on both objects successfully.
Politician.create({name: 'John Doe', politicalParty: 'Nice Party'})
.then(function(politician) {
Category.create({name: 'Nicers'})
.then(function(category) {
politician.addCategory(category);
});
});
You can also search and associate existing items using helper methods. Alternatively you can associate objects manually by accessing db.models.Interest model and running creates on it.
I'm new to sequelize and Node.js.
I coded for test sequelize, but error occured "ER_NO_SUCH_TABLE : Table 'db.node_tests' doesn't exist"
Error is very simple.
However, I want to get data from "node_test" table.
I think sequelize appends 's' character.
There is my source code.
var Sequelize = require('sequelize');
var sequelize = new Sequelize('db', 'user', 'pass');
var nodeTest = sequelize.define('node_test',
{ uid: Sequelize.INTEGER
, val: Sequelize.STRING} );
nodeTest.find({where:{uid:'1'}})
.success(function(tbl){
console.log(tbl);
});
I already create table "node_test", and inserted data using mysql client.
Does I misunderstood usage?
I found the answer my own question.
I appended Sequelize method option following. {define:{freezeTableName:true}}
Then sequelize not appends 's' character after table name.
Though the answer works nicely, I nowadays recommend the use of the tableName option when declaring the model:
sequelize.define('node_test', {
uid: Sequelize.INTEGER,
val: Sequelize.STRING
}, {
tableName: 'node_test'
});
http://docs.sequelizejs.com/manual/tutorial/models-definition.html
Sequelize is using by default the plural of the passed model name. So it will look for the table "node_tests" or "NodeTests". Also it can create the table for you if you want that.
nodeTest.sync().success(function() {
// here comes your find command.
})
Sync will try to create the table if it does not already exist. You can also drop the existing table and create a new one from scratch by using sync({ force: true }). Check the SQL commands on your command line for more details about what is going on.
When you define a model to an existing table, you need to set two options for sequelize to:
find your table name as-is and
not fret about sequelize's default columns updatedAt and createdAt that it expects.
Simply add both options like so:
var nodeTest = sequelize.define('node_test',
{ uid: Sequelize.INTEGER , val: Sequelize.STRING},
{ freezeTableName: true , timestamps: false} //add both options here
);
Note the options parameter:
sequelize.define('name_of_your_table',
{attributes_of_your_table_columns},
{options}
);
Missing either options triggers respective errors when using sequelize methods such as nodeTest.findAll().
> ER_NO_SUCH_TABLE //freezeTableName
> ER_BAD_FIELD_ERROR //timestamps
Alternatively, you can:
create a fresh table through sequelize. It will append "s" to the table name and create two timestamp columns as defaults or
use sequelize-auto, an awesome npm package to generate sequelize models from your existing database programmatically.
Here's the sequelize documentation for option configurations.
In my case, it was due to case. I was having:
sequelize.define('User', {
The correct way is to use lowercase:
sequelize.define('user', {