I have two tables: notes and tags, related by note_tags
.createTable('notes', function (table) {
table.increments('id').primary()
table.string('title', 100).notNullable()
table.string('content', 300).notNullable()
table.boolean('archived').notNullable().defaultTo(false)
table.timestamp('updated_at').defaultTo(knex.raw('CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP'));
})
.createTable('tags', function (table) {
table.increments('id').primary()
table.string('name', 20).notNullable()
})
.createTable('note_tags', function (table) {
table.increments('id').primary()
table.integer('note_id').unsigned().notNullable()
table.integer('tag_id').unsigned().notNullable()
table.foreign('note_id').references('notes.id')
table.foreign('tag_id').references('tags.id')
})
I want to get all the "used" tags, the ones that have at least one relation with any note
I don't know much about SQL, and with help of copilot i tried with:
async getUsedTags() {
return await Tag.query()
.select('tags.*')
.join('notes_tags', 'tags.id', 'notes_tags.tag_id')
.join('notes', 'notes.id', 'notes_tags.note_id')
.where('notes.archived', false)
}
but it returns "ER_NO_SUCH_TABLE: Table 'ensolvers-challenge.notes_tags' doesn't exist"
Related
I am quite new to sequelize and mySQL and feel like I have tried everything in order to pass a search term ('query') to both the books table (searching against titles) and the authors table (searching against first_name or last_name). In the event of matching any of those values substrings it is to return the whole book and author information as a JSON object. When I just have the query focused on book title, it returns everything just fine. The problem comes in when I try to pass in Author columns. I have tried aliasing, nesting where clauses, everything I can think of to do and nothing I come across on here or online seems to help me figure it out.
search: (req, res) => {
const { query } = req.query;
Book.findAll({
include: [Author],
where: {
[Op.or]: [
{ title: { [Op.substring]: query } },
]},
})
.then((Book) => res.json(Book))
.catch((err) => {
console.log(err);
res.status(500).json(err);
});
},
Here is the working code. In the where clause, I want to do { first_name: { [Op.substring]: query } }, for example but it isn't accessing the Author table. In the include statement I have tried aliasing and calling it in the where clause, but that throws a aliasing error saying I have already declared an alias (Author) but when I try to use that as { 'Author.first_name' { [Op.substring]: query } }, it returns that there is no Book.Author.first_name.
I am probably missing something simple, so anyone that might be able to help, let me know what I am doing wrong here!
Solved and it was super easy. I was missing the syntax for accessing the separate table which is '$Author.first_name$'.
I want to receive data like this:
categories
----category1
----category2
topProducts
----product1
--------photo1
--------photo2
----product2
--------photo1
--------photo2
I need get all categories and top x products.
Each product has two photos.
How can i do this by using yii2 restful?
Thanks.
the query shold look something like this
Category::find()
->with(['subcategories','topProducts', 'topProducts.images'])
->all();
you can use joinWith if you absolutely want a single query
if you retrieve your data with an ActiveController, you need to specify extraFields to the Category model. (here's a rest-specific usage example - rest of the guide should prove usefull as well)
Category model:
public function extraFields() {
return ['subcategories', 'topProducts'];
}
// product relation
public function getTopProducts(){
return $this->hasMany(Product::className(), ['category_id' => 'id'])
// ->order()->where() // your criterias
->limit(10);
}
// subcategories
public function getChildren(){
return $this->hasMany(Category::className(), ['id' => 'parent_id']);
}
Product model:
public function extraFields() {
return ['iamges'];
}
public function getImages(){
return $this->hasMany(Image::className(), ['product_id' => 'id'])
}
ps. since you haven't posed any code or table structure, all relations in my example are based on standard naiming convention
I am trying to save data with eloquent relationship.
I have following three tables: User Table, Category Table and Post Table.
Post Table
Schema::create('posts', function (Blueprint $table) {
$table->increments('id');
$table->integer('category_id')->unsigned();
$table->integer('user_id')->unsigned();
$table->string('heading');
$table->timestamps();
$table->foreign('category_id')->references('id')->on('categories');
$table->foreign('user_id')->references('id')->on('users');
});
Relations:
Category:
public function posts() {
return $this->hasMany('App\Post');
}
Post:
public function user() {
return $this->belongsTo('App\User');
}
public function category() {
return $this->belongsTo('App\Category');
}
User:
public function posts($category) {
return $this->hasMany('App\Post');
}
My Problem is that, how can I save post just by passing the Heading in create function. I want to use the relationship. As an example I want to use this kind of code:
$data = ['heading' => $heading];
$user->posts()->category()->create($data);
Is this possible to do this kind of stuff ?
Or any another simple way to achieve this.
EDIT
I need to create post by using this kind of relationship.
As per the process:
user will fill up the form from which I will get the data along with
the category id.
Now I need to create data for that user related with the given category id.
It's because after you call posts() method you won't get to the model's relation (only the query builder) so you will not access category() relation method. It's because posts are one-to-many relation and you don';t know exacly which record you refer to create data.
EDIT
If you want to create new post entry the the best way to sole this is:
$data = ['heading' => $heading, 'category_id' => $putHereCategoryId];
$user->posts()->create($data);
You'll need to obtain somehow the id of the desire category for the new post's entry.
Using Laravel 5.1: Given two related models, User and Item, with through table Item_User, how can I include a specific column from the through table, item_count using a with statement?
Item_User table:
Query:
$userGameProfile = User::with([
'items' => function($query) use ($profile_id) {
$query->with(['phrases'])->where('profile_id', '=', $profile_id);
}])->find($id);
Adding ->select('item_count') to this query only returns item count, and not the related item objects. Adding ->select('*') gets me the items and fields I want, but breaks the nested with(['phrases']) relationship I need to get related phrases to items.
I was told I should use withPivot, but I cannot find documentation or examples on how to use this.
Preferably, I'd like to add this to the relationship, rather than each query.
Models:
User:
public function items()
{
return $this->belongsToMany(Item::class, 'user_item');
}
Item:
public function users()
{
return $this->belongsToMany(User::class, 'user_item');
}
public function phrases()
{
return $this->belongsToMany(Phrase::class, 'item_phrase');
}
Phrase:
public function items()
{
return $this->belongsToMany(Item::class, 'item_phrase');
}
This article highlights how to include additional information from the pivot table.
In User model, I know I will always want to include item_count from user_items pivot table, so you can add withPivot to the relation to items:
public function items()
{
return $this->belongsToMany(Item::class, 'user_item')
->withPivot('item_count');
}
User now comes back with items and pivot data:
By below reference I understood how map many to many with a relationship table
http://sequelizejs.com/docs/latest/associations#many-to-many
User = sequelize.define('User', { user_name : Sequelize.STRING})
Project = sequelize.define('Project', { project_name : Sequelize.STRING })
UserProjects = sequelize.define('UserProjects', {
status: DataTypes.STRING
})
User.hasMany(Project, { through: UserProjects })
Project.hasMany(User, { through: UserProjects })
But how to query Project 's of a User
I Tried like
User.find({where:{id:1},include,[UserProjects]})
User.find({where:{id:1},include,[Projects]})
User.find({where:{id:1},include,[UserProjects]})
User.find({where:{id:1},include,[Projects]})
But i dont get results
Sequelize created table like below
users(id,name)
projects(id,project_name)
userprojects(id,UserId,ProjectId)
I tried https://github.com/sequelize/sequelize/wiki/API-Reference-Associations#hasmanytarget-options
User.find({where:{id:1}}).success(function(user){
user.getProjects().success(function (projects) {
var p1 = projects[0] // this works fine but 2 queries required. I expect in single find. without getProjects
p1.userprojects.started // Is this project started yet?
})
})
How to get all the projects of a USER ??
You should be able to get all of the properties of the user in two different ways: using includes and getting the projects from a user instance.
Using includes the code you submitted above is almost right. This method will only make one query to the database using the JOIN operation. If you want all of the users with their corresponding projects, try:
User.findAll({include: [Project]})
You can also get the projects directly from a user instance. This will take two queries to the database. The code for this looks like
User.find(1).then(function(user) {
user.getProjects().then(function(projects) {
// do stuff with projects
});
});
Does this work for you?