i have two tables food table and restaurants table which are all related by restaurant id
This is the relationship in the food models
public function restaurant(){
return $this->belongsTo('App\Restaurant');
}
This is the food table
$table->bigIncrements('id');
$table->timestamps();
$table->integer('price');
$table->integer('food_item');
$table->integer('restaurant_id');
This is the relationship in the restaurants model
public function foods(){
return $this->hasMany('App\Food');
}
When the user puts input q which is the search term of food item i need it to produce all restaurants which food item according to the search term
Here is the query i wrote but returns 0 results when i do return json_encode($foodsAll);
public function search(Request $request){
$foodsAll = Restaurant::whereHas('foods',function($query) use ($request){
$query->where('food_item','like','%'.$request->q.'%');
});
}
You are missing get().
public function search(Request $request){
$foodsAll = Restaurant::whereHas('foods',function($query) use ($request){
$query->where('food_item','like','%'.$request->q.'%');
})->get();
}
Related
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
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 = $querys->paginate(10);
I am new to laravel and I experience some trouble. I try to obtain data stored in two different tables and display them:
News.php (model)
public static function Data($category) {
$perPage = config('var.news.perPage');
if ($category) {
$news = News::orderBy('id', 'desc')->where('category', $category)->SimplePaginate($perPage);
} else {
$news = News::orderBy('id', 'desc')->SimplePaginate($perPage);
}
return $news;
}
This is how I grab all data from News table which struct is:
id, title, body, created_at updated_at, created_by, updated_by, category
The category column contains values separated by comma, e.g. 1,2,3,4
Now, I have another table, News_Cat which has id, name columns.
In another method I try to grab the filters names against values stored in category column of News table
public static function getFilterNames($id) {
$filters = DB::table('News_Cat')
->select('News_Cat.name as name')
->leftJoin('News', DB::raw('CAST(News_Cat.id as nvarchar)'), DB::raw('ANY(SELECT(News.category))'))
->where('News.id', $id)
->get();
return $filters;
}
However, it completely does not work. What I try to achieve is to display filter name in view.blade as 'name' value for specified filter from News_Cat
#if($news->count())
#foreach($news as $article)
<a href="{{ route('news.show', $article->id) }}" class="item angled-bg" data-filters="{{ $filters }}">
<div class="row">
So as result I would get e.g. data-filters="news, update, hot, latest"> instead data-filters="1,2,3,4">
Thank you
You should use eloquent!
In your News Model
public function getFiltersAttribute(){
$categories = explode(',', $this->category);
return implode(', ', NewsCat::find($categories)->pluck('name')->toArray());
}
then in your view :
{{ $article->filters }}
will output news, update, hot, latest
BUT
You should use a pivot table between your categories and your news, it would be much easier.
This method can't allow you to eager load the relationship and make a request for each news
If you can't change your database structure, I can propose you this:
In the boot method of your AppServiceProvider:
Config::set('tags', NewsCat::all());
THEN
public function getFiltersAttribute(){
$categories = explode(',' $this->category);
return implode(', ', config('tags')->whereIn('id', $categories)->pluck('name')->toArray());
}
MANY TO MANY METHOD
I am using laravel naming convention for the table :
news, categories_news (the pivot), and categories
You will have 2 models : New and Category
In your New Model
public function categories(){
return $this->belongsToMany(Category::class)
}
in your Category Model :
public function news(){
return $this->belongsToMany(New::class);
}
if you are not using laravel naming conventions, you will have to customize these raltionship like this : https://laravel.com/docs/5.5/eloquent-relationships#many-to-many
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.
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 :)