I'm trying to use leftJoin but I'm having problem with conflicts between equal names in both tables.
Tables
products: added_by - user_id - published - approved - featured
product_types: added_by - user_id - published - approved - featured
$products = ProductType::
leftJoin('products', 'products.product_type_id', '=', 'product_types.id')
->select('product_types.*')
How to solve this problem?
You should start by implementing the correct relationships, it will help you in the future.
class ProductType extends Model {
public function products(): HasMany
{
return $this->hasMany(Product::class);
}
// Add the other missing relations too (user, etc)
}
class Product extends Model {
public function productType(): BelongsTo
{
return $this->belongsTo(ProductType::class);
}
// Add the other missing relations too (user, etc)
}
You can now use:
// Query all Product Types with related products
$productTypes = ProductType::with('products')->get();
// Query all ProductTypes that have products
$productTypes = ProductType::withWhereHas('products')->get();
/** #var ProductType $productType */
foreach ($productTypes as $productType) {
// You have access to the collection of associated products, in memory
$productType->products;
}
More information on eloquent relationships in https://laravel.com/docs/9.x/eloquent-relationships
Related
Working on a search functionality on Laravel App(Blog/Posts).
There are multiple types of posts (each having a separate table in the database)
Like Business posts, Social Life posts etc..
Below is the search function on SearchController
class SearchController extends Controller
{
public function search(Request $request, $query = null)
{
if($query == null)
return redirect()->route('home');
$search = Business::where([['title','like','%'.$query.'%'],['status','=',1]])
->orWhere([['description','like','%'.$query.'%'],['status','=',1]])
->paginate(10);
return view('front.search',[
'results' => $search,
'query' => $query
]);
}
}
So basically my question is how to add other types of Post's table also?
My main motive is that when someone searches for anything, the result should be fetched from all types of posts table(business, nature, life & so on..).
You have to maintain common id in both the table
NOTE: Join is the preferable method
$querys = DB::table('Business')->where([['Business.title','like','%'.$query.'%'],['Business.status','=',1]])
->orWhere([['Business.description','like','%'.$query.'%'],['Business.status','=',1]]);
$querys->join('socialtable','socialtable.userid','=','Business.userid');
// Just join the social table
$querys->where('socialtable.title', 'like','%'.$query.'%');
$result = $querys->paginate(10);
If you have a model called Book, like this:
class Book extends Model
{
/**
* Get the author that wrote the book.
*/
public function author()
{
return $this->belongsTo('App\Author');
}
}
Then you can retrieve all of your books with authors like this:
$books = App\Book::with(['author'])->get();
Check out Eager loading from Laravel documentation.
Just add table name before every field
$querys = DB::table('Business')->where([['Business.title','like','%'.$query.'%'],['Business.status','=',1]])
->orWhere([['Business.description','like','%'.$query.'%'],['Business.status','=',1]]);
$querys->join('socialtable','socialtable.userid','=','Business.userid');
// Just join the social table
$querys->where('socialtable.title', 'like','%'.$query.'%');
$result = $querys->paginate(10);
I am trying to pull a list of characters that belong to a certain user. When I make the request I get an SQL Error. Reading through the error it is trying to us fields that don't exist.
Error:
SQLSTATE[42000]: Syntax error or access violation: 1066 Not unique table/alias: 'characters' (SQL: select `characters`.*, `characters`.`id` as `pivot_id`,
`characters`.`character_id` as `pivot_character_id`
from `characters` inner join `characters` on `characters`.`id` = `characters`.`character_id` where `characters`.`id` = 1)
"character_id" does not exist in my database. The problem is I can't find where Eloquent is making that field. I looked through the source code and there was a lot of "If this is not provided use $variable.'_id'. I could not find that code anywhere for this though.
Models are below.
class Character extends Eloquent {
protected $guarded = array('id');
protected $table = 'characters';
public function User ()
{
return $this->belongsTo('User', 'id');
}
}
class User extends Eloquent implements UserInterface, RemindableInterface {
use UserTrait, RemindableTrait;
protected $table = 'users';
protected $hidden = ['password', 'remember_token'];
protected $guarded = ['password'];
public function Character ()
{
return $this->belongsToMany('Character', 'characters', 'id');
}
}
There is a foreign key between user_id in the characters table, and id in the users table.
belongsToMany is for many-to-many relations. Laravel throws you an error because it expects third table - pivot table - containing both character_id and user_id.
If you dont want many-to-many but one-to-many then you should use hasMany and belongsTo.
How do I use join with Eloquent taking in consideration the following table structure:
I have a properies table
---------------------
ID | Name
---------------------
1 | Property Name
than I have rooms
----------------------
RoomID | Property
----------------------
A-212 | 1
----------------------
F-1231 | 1
here Property is the foreign key
than I want to get all Properties and count how many rooms do they have each
The query which retrives all looks like
class PropertiesRepository extends EloquentBaseRepository implements PropertiesInterface
{
use TaggableRepository;
/**
* Construct
* #param Properties $properties
*/
public function __construct( Properties $properties )
{
$this->model = $properties;
}
/**
* Get all properties joining rooms
* #return Properties
*/
public function getAll()
{
return $this->model->get();
}
}
How do I extend this query to get the desired result?
This is more of a MySQL join+group+select trick which includes following steps.
Join your relation table(use join if you want to exclude rows with RoomsCount=0, else use leftJoin)
Use groupBy by primaryKey to avoid duplicates of the join.
Select count of joined table
$this->model->leftJoin('Rooms', 'Properties.ID', '=', 'Rooms.Property')
->selectRaw('Properties.*, count(Rooms.RoomID) as RoomsCount')
->groupBy('Properties.ID')
->get();
Define the relationship on your Property model class:
<?php
namespace App\Models;
class Property extends Model {
public function rooms() {
return $this->hasMany(Room::class);
}
}
$properties = Property::withCount(['rooms'])->get();
This will add a rooms_count to the result.
im having problem with CRUd now that i filled the database. CRUD is taking ages to show, becouse it takes condition from M:M tables.
Tables:
Table USER. has many labels (hasMany)
Table LABLE, has many users (hasMany)
Intermidiate Table UserLabel, has two hasOne
I want to show all users from some label with CRUD like this:
MODEL USER:
class Model_User extends Model_Table {
public $table ='user';
function init(){
parent::init();
$this->addField('fbid')->mandatory('Facebook id required');
...
$this->hasOne('Application');
$this->hasMany('UserLabel');
$this->addExpression('ratio')->set(function($model,$select){
return $select->expr('ROUND(([f2] / [f1]) * 100,0)')
->setCustom('f1',$model->getElement('sends'))
->setCustom('f2',$model->getElement('clicked'));
});
$this->addHook('beforeSave',function($m){
$m['updated']=$m->dsql()->expr('now()');
});
}
MODEL LABEL:
class Model_Label extends Model_Table {
public $table ='label';
function init(){
parent::init();
$this->addField('name')->mandatory('Name required');
$this->addFIeld('application_id')->refModel('Model_Application')->defaultValue($this->api->recall('app'))->system(true);
$this->addField('active')->type('boolean')->defaultValue('true')->system(true);
$this->addField('created')->type('timestamp')->defaultValue($this->dsql()->expr('now()'))->system(true);
$this->addField('updated')->type('timestamp')->system(true);
$this->hasMany('UserLabel');
$m = $this->add("Model_UserLabel");
$this->addExpression("users", $m->dsql()
->field($m->dsql()->expr("count(*)"), "all users")
->where("label_id", $this->getField("id"))
);
MODEL USER LABEL
class Model_UserLabel extends Model_Table {
public $table ='userlabel';
function init(){
parent::init();
$this->hasOne('User');
$this->hasOne('Label');
}
}
CODE FOR CRUD
$c = $this->add('CRUD');
$c->setModel('User', array('name', 'gender','country','city'));
$c->model->addCondition('id','in',
$this->add('Model_UserLabel')->addCondition('label_id', $_GET['l'])->dsql()->field('user_id')
);
Is there any better way to do this?
ps. I tested this solution, it is a lot faster but still very slow at around > 5.000 users:
//get all users
$records = $this->api->db->dsql()->option('distinct')->table('user')->join('userlabel.user_id')->field('user.id')->where('userlabel.label_id',$_GET['l'])->do_getAll();
foreach($records as $record){
$users .= ','.$record['id'];
}
//create CRUD
$c = $this->add('CRUD');
$c->setModel('User', array('name', 'gender','country','city','sends','clicked','ratio'));
$c->model->addCondition("application_id", $this->api->recall('app'));
$c->model->addCondition('id','in',
'('.$users.')'
);
Source code express more than words, so you better add your model definition source code (maybe not full) in your question.
What should be one row in your CRUD/Grid? I guess it's not 1 user = 1 row, but 1 user_label should be one row in grid. So you should set UserLabel model as model for your grid.
And then define some additional fields in Model_UserLabel by joining them from user and/or label tables directly like this:
class Model_UserLabel extends SQL_Model {
function init() {
parent::init();
// ...
// fields from user table
$join_u = $this->join('user', 'user_id');
$join_u->addField('username'); // this adds fields in current model from joined table
$join_u->addField('email');
// fields from label table
$join_l = $this->join('label', 'label_id');
$join_l->addField('name');
}
}
Note: source code above is untested and put here only as example.
EDIT:
Try this solution - almost the same as I wrote earlier above:
MODEL USER LABEL
class Model_UserLabel extends Model_Table {
public $table ='userlabel';
function init(){
parent::init();
$this->hasOne('User');
$this->hasOne('Label');
// join user table and add fields to this model from joined user table
$j = $this->join('user', 'user_id');
$j->addField('name');
$j->addField('gender');
$j->addField('country');
$j->addField('city');
}
}
CODE FOR CRUD
$m = $this->add('Model_UserLabel'); // UserLabel here not User
$m->addCondition('label_id', $_GET['l']); // and then this is simple
$c = $this->add('CRUD');
$c->setModel($m, array('name', 'gender','country','city'));
Try this solution and as (almost) always - code is untested.
EDIT:
Please try this version - is it working faster? That's basically your P.S. example, but you shouldn't extract all user IDs, join them and then create huge select with a lot of 'in'.
Faster result should be if you could do all with just one DB request without any additional processing of data.
// parameters
$app_id = $this->api->recall('app);
$label_id = $_GET['l'];
// prepare model for grid
$m = $this->add('Model_User'); // default User model
$m->_dsql()->option('distinct') // add join to userlabel table + conditions
->join('userlabel.user_id')
->where('userlabel.label_id', $label_id)
->where($m->getField('application_id'), $app_id);
// create CRUD and set it's model. All conditions already set on model above
$c = $this->add('CRUD');
$c->setModel($m, array('name', 'gender','country','city','sends','clicked','ratio'));
NOTE: Source code above as often - untested :)
I am new to nhibernate and trying to create a query on a database with manytomany links between items and categories.
I have a database with 3 tables : items, categories and a lookup table categoryitem like this:
categorys - primary key categoryId
items - primary key itemId
categoryItem - categoryId column and itemId column
I want a query returning items for a particular category and have tried this and think i am along the right lines:
public IList<Item> GetItemsForCategory(Category category)
{
//detached criteria
DetachedCriteria itemIdsCriteria = DetachedCriteria.For(typeof(Category))
.SetProjection(Projections.Distinct(Projections.Property("Item.Id")))
.Add(Restrictions.Eq("Category.Id", category.Id));
criteria.Add(Subqueries.PropertyIn("Id", itemIdsCriteria));
return criteria.List<Item>() as List<Item>;
}
I only have business objects for category and item.
how do i create a repository method to find items for a particular category?
I assume that your classes look like this:
class Item
{
// id and stuff
IList<Category> Categories { get; private set; }
}
class Category
{
// id and stuff
}
query (HQL)
session.CreateQuery(#"select i
from Item i
inner join i.Categories c
where
c = :category")
.SetEntity("category", category)
Criteria
session
.CreateCriteria(typeof(Item))
.CreateCriteria("Categories", "c")
.Add(Restrictions.Eq("c.Id", category.Id))