//controller
$promotion = Promotion::findOrFail($id);
//return
Array
(
[id] => 2
[en_title] => promo1
[game_id] => Array
(
[0] => 3
[1] => 4
[2] => 5
)
[amount] => 100.00
[start_at] => 2021-02-22
[end_at] => 2222-02-22
[status] => 1
)
//model promotion
class Promotion extends Model
{
use HasFactory;
protected $guarded = [];
protected $casts = [
'game_id' => 'array'
];
public function getAllGames()
{
return $this->belongsTo(Game::class, 'game_id', 'id');
}
}
Question:
Currently, I have 2 tables which are games and promotion, but I get trouble when coming into a relationship because the column of game_id inside the promotion table is a JSON, so that is hard to join it. Is there any work around can easily join them together in order to retrieve games data?
make many-to-many relationship between your games and promotion tables,
put your game_ids in pivot table (game_promotion) instead of JSON field,
Many To Many Relationships
Related
I am using this Staudenmein package in Laravel and referring to this answer given by the package owner himself.
I have two tables, pets and notifications with models Pet and Notification.
The notifications table has a column called data which is of JSON dataType and stores JSON data
{"pet_id":"4","pet_type_id":1,"lost_report_id":3,"pet_sighting_id":21,"latitude":"22.676846","longitude":"88.338509"}
The key pet_id denotes the id column of the pets table. I need a relationship so that I can fetch the name of the pet from pets table.
By referring to the answer link given above, I wrote my Notification model like this:-
namespace App\Models;
use DB, App, Auth, Hash, Lang, Mail, Config, Exception, Validator, Globals;
use Illuminate\Database\Eloquent\Model;
use \Staudenmeir\EloquentJsonRelations\HasJsonRelationships;
use App\User;
use App\Models\AdminConfig;
use App\Models\ReceivedAppNotification;
class Notification extends Model
{
protected $casts = [
'data' => 'json'
];
public function notificationPet()
{
return $this->belongsTo('App\Models\Pet', 'data->pet_id');
}
}
When I am running the query like this:-
$notificationQuery = Notification::with('notificationPet')
->whereRaw("FIND_IN_SET('$userId', in_app_notification_receiver)")
->where(array(
'status' => Globals::SMALL_CHAR_ACTIVE,
'is_delete' => Globals::SMALL_CHAR_NO,
))->get()->toArray();
I get the notificationPet relationship as empty, ie. the data-set is like this:-
Array
(
[id] => 10
[title] =>
[message] =>
[data] => Array
(
[pet_id] => 4
[latitude] => 22.676846
[longitude] => 88.338509
[pet_type_id] => 1
[lost_report_id] => 3
[pet_sighting_id] => 34
)
[no_of_in_app_notification_receiver] => 2
[no_of_push_notification_receiver] => 1
[from_user] => 22
[in_app_notification_receiver] => 2,23
[push_notification_receiver] => 2
[status] => a
[is_delete] => n
[push_notification_sent] => n
[created_date] => 2020-03-19 13:23:17
[modified_date] => 2020-03-19 13:23:17
[created_at] => 2020-03-19 13:23:17
[updated_at] => 2020-03-19 13:23:17
[notification_pet] =>
)
However, I have a record in pets table with id = 4, already. So the relationship should not be empty.
What am I doing wrong?
Change:
protected $casts = [
'data' => 'json'
];
To:
protected $casts = [
'data' => 'array'
];
I have problem with my code. At this moment I have code view like this:
...Table.php
public function containBasic(){
return [
//Keywords
'CandidatesKeywords',
'CandidatesKeywords.Keywords',
//User
'CandidatesUser',
'CandidatesUser.User',
];
}
...Controller.php
$this->loadModel('Candidates');
$candidates = $this->Candidates->find()
->contain($this->Candidates->containBasic())
->where([
'CandidatesKeywords.Keywords.id'=>5
])
->all();
I include all other Models via contain. This method its works but when I try to search data in this code I have error like:
Unknown column 'CandidatesKeyword.id' in 'where clause
I don't know, how I can get column CandidatesKeywords -> Keywords -> Id from database :/
When I get (debug) all rows without where condition I get data like this:
'items' => [
(int) 0 => object(Cake\ORM\Entity) {
'id' => (int) 4,
.........
,
'candidates_user' => null,
'candidates_keyword' => object(App\Model\Entity\CandidatesKeyword) {
'id' => (int) 1,
..................
'keyword' => object(App\Model\Entity\Keyword) {
'id' => (int) 5,
.............
And I need to get rows with only keyword.id = 5.
I have defined 2 databases , for example
return [
'components' => [
'db1' => [
'class' => 'yii\db\Connection',
'dsn' => 'mysql:host=localhost;dbname=db1name',
'username' => 'db1username',
'password' => 'db1password',
],
'db2' => [
'class' => 'yii\db\Connection',
'dsn' => 'mysql:host=localhost;dbname=db2name',
'username' => 'db2username',
'password' => 'db2password',
],
],
];
Now i have a table as 'users' in 'db1' and table 'countries' in 'db2'
users
id , country_code , username , password
1 , DE , xyz , 12345
2 , FR , abc , 12345
countries
code , name
DE , Germany
FR , France
IN , India
I have defined the foreign key relation between users.country_code & countries.code
ISSUE
But when i try to create the model for 'users' table using gii it gives an error , possibly because the tables relation are from 2 different databases.
How to use tables from different databases in relations of a model.
Any suggestions are welcomed
This works in my case to list iten on GridView::widget
-> bd_sisarc is my secound data base
-> deposito_sondagem is a table from my first data base
public static function getDb() // on your model
{
return Yii::$app->get('db1');
}
public static function getDb() // on your model
{
return Yii::$app->get('db2');
}
public function getEmpresaSondagem() // Relation on you model
{
return $this->hasOne(EmpresaSondagem::className(), ['idEmpSondagem' => 'entidade_deposito']);
}
public function search($params)
{
$this->load($params);
$sql = "SELECT deposito_sondagem.*
FROM
deposito_sondagem,
`bd_sisarc`.`tbempresasondagem`
WHERE
`bd_sisarc`.`tbempresasondagem`.`idEmpSondagem`=`deposito_sondagem`.`entidade_deposito`
and deposito_sondagem.estado=1
and tbempresasondagem.estado=1
and numero_registo LIKE '%$this->numero_registo%'
and nomeempsondagem LIKE '%$this->nomeEntidade%'
and dono_sondagem LIKE '%$this->dono_sondagem%'
and data_deposito LIKE '%$this->data_deposito%'";
$query = DepositoSondagem::findBySql($sql);
$dataProvider = new ActiveDataProvider([
'query' => $query,
]);
if (!$this->validate()) {
// uncomment the following line if you do not want to return any records when validation fails
// $query->where('0=1');
return $dataProvider;
}
return $dataProvider;
}
Try this one
SELECT `users`.* FROM `users` LEFT JOIN `db2name`.`countries` ON `users`.`country_code` = `db2name`.`countries`.`code `
I'm using CakePHP 2.5.2 and having a bit of trouble searching for data efficiently.
In my application I've 3 tables, teams, players, skills... In teams there are 80 records, players 2400 records, skills 2400 records... I want to calculate the average skill of a team...
//Team model
public $actsAs = array('Containable');
public $hasMany = array('Player');
//Player model
public $actsAs = array('Containable');
public $hasOne = array('Skill');
public $belongsTo = array('Team');
//Skill model
public $actsAs = array('Containable');
public $belongsTo = array('Player');
My research is:
$team = $this->Team->find('all', array(
'contain' => array(
'Player' => array(
'Skill'
)
),
));
$this->set('team', $team);
that gives the expected result:
Array
(
[0] => Array
(
[Team] => Array
(
[id] => 1
[name] => my_team_name
)
[Player] => Array
(
[0] => Array
(
[id] => 000000419
[name] => Name
[surname] => Surname
[age] => 21
[team_id] => 1
[Team_id] => 1
[Skill] => Array
(
[id] => 20
[player_id] => 000000419
[skill] => 599
)
), ecc.....
This structure use at least 1680 queries... that are too much for me...
I've tried an other way, that involve just one query, returns a bad data structure but all the information that i need (also redundant). unfortunately follow this way i can not iterate in View to display what i need.
$player = $this->Team->Player->find('all', array(
'contains' => array(
'Skill',
),
that returns
Array
(
[0] => Array
(
[Player] => Array
(
[id] => 000000400
[nome] => my_player_name
[cognome] => my_player_surname
[nation_id] => 380
[age] => 29
[team_id] => 2
)
[Team] => Array
(
[id] => 2
[nome] => my_team_name
)
[Skill] => Array
(
[id] => 1
[player_id] => 000000400
[average] => 632
)
)
ecc.
Is there a way to iterate in VIEV to get the average skill of every team? Any other solutions?
Thanks!
You can use my plugin to solve this issue if you can upgrade CakePHP to 2.6 or later. The plugin has a high compatibility with ContainableBehavior, but generates better queries.
I think that the find operation will execute only 2 queries then.
I would be happy if you try it.
https://github.com/chinpei215/cakephp-eager-loader
Usage
1. Enable EagerLoader plugin
// In your model
$actsAs = ['EagerLoader.EagerLoader'];
If you are afraid that loading my plugin breaks something somewhere, you can also enable it on the fly.
// On the fly
$this->Team->Behaviors->load('EagerLoader.EagerLoader');
2. Execute the same find operation
$this->Team->find('all', ['contain' => ['Player' => ['Skill']]]);
3. See the query log
You will see the query log such as the following:
SELECT ... FROM teams AS Team WHERE 1 = 1;
SELECT ... FROM players AS Player LEFT JOIN skills AS Skill ON Player.id = Skill.player_id WHERE Player.id IN ( ... );
if you feeling that query searching so many tables (ie, models) then
you can unbind those model, before performing search with find()
if you want to fetch some particular column of a table, then remove
others column by selecting "fields" in find().
I am struck with one issue related to foreignKey concept.
I have two table users & user_relations.
users user_relations
id username id user_id friend_id status
1 abc 1 1 2 Accepted
2 def 2 2 3 Accepted
3 ghi 3 1 3 Accepted
here the friend_id is the foreign_key and my user_relations model looks like
public $belongsTo = array(
'Friend' => array(
'className' => 'User',
'foreignKey' => 'friend_id'
)
);
My conditions will be like
'conditions' => array(
"OR" => array(
'UserRelation.user_id' => $iLoggedUserId,
'UserRelation.friend_id' => $iLoggedUserId,
),
'ViewerRelationship.status' => 'Accepted'
)
Now when user "1" is logged in then he will get the info of users "2" & "3" based on foreignKey friend_id.
If user "2" gets logged in then, he is getting the info of users "2" & "3". But here I need to get the info of users "1" & "3". i.e., in one case it need to take the info based on the user_id not the friend_id
If I can keep the foreignKey in a condition I can get the result based on user_id or friend_id.
Any help please..!
I dont know how it is effective for you.
But it will solve your problem.
$respone = $this->UserRelation->find('all',
array('conditions'=> array(
'OR'=>array(
'UserRelation.user_id' => $iLoggedUserId,
'UserRelation.friend_id' => $iLoggedUserId
)
),
'fields' => array('UserRelation.*')
));
$reqFriendsId = array();
foreach($respone as $key => $val){
if($val['UserRelation']['friend_id'] == $iLoggedUserId){
array_push($reqFriendsId, $val['UserRelation']['user_id']);
}else{
array_push($reqFriendsId, $val['UserRelation']['friend_id']);
}
}
//For logged in user 1
Array
(
[0] => 2
[1] => 3
)
//For logged in user 2
Array
(
[0] => 1
[1] => 3
)