CAKEPHP question
I am querying my HABTM table successfully and returning the id of every student with the given group_id. This is my code for this.
$students = $this->GroupsStudents->find('list', array('conditions' => array('group_id' => $id)));
It works, no problem. I need to somehow use this info (namely the student's id), which is stored in $students, to query my students table and extract student's names based on their id.
If someone could give me some insight on this, that would be greatly appreciated.
if i'm understanding you right. as you can see from this if you have the id you can easily get the students name even though i'm not sure why you would do this and not just foreach the name.
foreach ($students as $id => $name) {
echo $students[$id]; // produces name
}
In Student model define relation with GroupStudent model as shown below:
var $hasMany = array(
'GroupStudent' => array(
'className' => 'GroupStudent',
'foreignKey' => 'student_id'
)
);
Then your write your query as
$students = $this->Student->find('all',
array(
'conditions' => array('GroupStudent.group_id' => $id)
'fields'=>array('Student.name','Student.id','GroupStudent.group_id'))
);
Note: Make sure your controller has $uses=>array('Student','GroupStudent'); defined!
and your are using plural names for model GroupStudents so correct them if possible
Related
I have this database structure:
I need the user to get only those reservations in ReservationsSeachModel that are in the warehouse belonging to the user. Model user_has_warehouse has a function getUserHasWarehouse($user_id, $warehouse_id) to check for relations. I am able to get the relations when viewing single, deleting and updating records in controllers. But I cannot make the SearchModel to return reservations that belong to current user.
I cannot depend on Reservation->user_id, because that value is not mandatory and reservations can be inserted publicly with no user_id. How to make the SearchModel work properly so it only shows reservations belonging to the warehouse where user has the warehouse?
ReservationsSearchModel:
$warehouse_id = User_Has_Warehouse::find()->select(['warehouse_id'])->where(['user_id' => Yii::$app->user->id]);
$freezer_type_id = Freezer_type::find()->select(['id'])->where(['warehouse_id' => $warehouse_id]);
$reservation_id = Reservation_freezer_type::find()->select(['reservation_id'])->where(['freezer_type_id'=>$freezer_type_id]);
...
$query->andFilterWhere([
'id' => $reservation_id,
'customer_id' => $this->customer_id,
'user_id' => $this->user_id,
]);
...
But the following did not work for me. What am I doing wrong?
Managed to solve it.
$h = User_Has_Warehouse::find()->select(['warehouse_id'])->where(['user_id' => Yii::$app->user->id]);
$rt = Freezer_type::find()->select(['id'])->where(['warehouse_id' => $h]);
$res = Reservation_freezer_type::find()->select(['reservation_id'])->where(['freezer_type_id' => $rt]);
// grid filtering conditions
$query->andFilterWhere(['id' => $res]);
please help I dont know how to get relation table when using sqldataprovider. Anyone understand how to use relation model?
$model = new Finalresult();
$searchModel = new FinalresultSearch();
$dataProvider = $searchModel->search(Yii::$app->request->queryParams);
$dataProvider = new SqlDataProvider([
'sql' => 'SELECT finalresult.bib,
finalresult.series_id,
finalresult.category_id,
GROUP_CONCAT(DISTINCT finalresult.point ORDER BY series.serie_seri_no DESC) AS seriPoint
FROM finalresult, series GROUP BY finalresult.bib',
'key' => 'bib',
]);
I'm trying to get relation table:
'attribute'=>'category_id',
'width'=>'300px',
'value'=>function ($model, $key, $index, $widget) {
return $model->category->category_name;
},
then getting trying to non-object
You can't use relations with SqlDataProvider, because each single result will be presented as array, for example:
[
'id' => 1,
'name' => 'Some name',
'category_id' => 1,
],
For example, you can access category_id as `$model['category_id'].
SqlDataProvider is for very very complex queries, your query can easily be written as ActiveQuery and you can use ActiveDataProvider and get all advantages of that (relations, etc.).
You can find category by id, but it will be lazily loaded that means amount of queries is multiplied by number of rows.
With ActiveDataProvider and relations you can use eager loading and reduce amount of queries. Read more in official docs.
Grid Columns example in documentation
try to change "value" to
'value'=> function($data) {
return $data['category']['category_name'];
}
I'm trying to create an AJAX form whereby the content of a select field populates based on the choice of a preceding select field (you see this a lot with 'country' populating 'state/province'). In my case, I want users to be able to choose their province only if active accounts exist in it.
The Javascript I can write no problem. Fetching the data is where I'm... not so much stuck as doing too much work. CakePHP likes to build select fields with options in an array of the form
$options = array(select_option_value => display_text)
My strategy, though functional, must be more convoluted than cake intended (this a is segment of a controller method).
$provinceData = $this->Account->find('all',array('recursive' => 0,
'joins' => array(
array(
'table' => 'provinces',
'type' => 'LEFT',
'conditions' => array('Account.province_id = provinces.id')
)),
'fields'=>array('provinces.id', 'provinces.name', 'provinces.abbrev'),
'conditions' => array('registration > 2')));
$provinces = array();
foreach($provinceData as $pd) {
/*note: lowercase, plural below b/c can't get 'alias' => 'Province'
to work in joins array above : ( */
$id = $pd['provinces']['id'];
$name = $pd['provinces']['name'];
$provinces[$id] = $name;
}
$this->set(compact('provinces'));
Can anyone point out a more appropriate way to do this? I assume there must be a MySQL query that can do this, but I'm pretty bad at writing elaborate MySQL queries in the first place, let alone via Cake's convention (and, for you MySQL gurus out there, I'm happy to do this from a Model->query(//MySQL code) call instead!
Any and all help truly appreciated.
Assuming the relationship Account belongsTo Province you can try this code:
$accounts = $this->Account->find(
'all',
array(
'fields' => array('Account.province_id', 'Province.name'),
'conditions' => array('Account.registration > 2'),
'group' => 'Account.province_id'
)
);
$provinces = Hash::combine($accounts, '{n}.Account.province_id', '{n}.Province.name');
$this->set(compact('provinces'));
edit: missed bracket and a period instead of an underscore . Now should work
I am not very good with DB queries. And with Yii it's more complicated, since I am not very used to it.
I need to optimize a simple query
$userCalendar = UserCalendar::model()->findByAttributes(array('user_id'=>$user->id));
$unplannedEvents = CalendarEvent::model()->findAllByAttributes(array('calendar_id'=> $userCalendar->calendar_id,'planned'=>0));
CalendarEvent table, i.e the second table from which I need records does not have an user_id but a calendar_id from which I could get user_id from UserCalendar, i.e. the first table hence I created a UserCalendar object which is not a very good way as far as I understand.
Q1. What could I do to make it into one.
Q2. Yii does this all internally but I want to know what query it built to try it seperately in MySQL(phpMyAdmin), is there a way to do that?
Thanks.
Q1: You need to have the relation between UserCalendar and CalendarEvent defined in both of your active record models (in the method "relations").
Based on your comments, it seems like you have the Calendar model that has CalendarEvent models and UserCalendar models.
Lets assume your relations in Calendar are:
relations() {
return array(
'userCalendar' => array(self::HAS_MANY, 'UserCalendar', 'calendar_id'),
'calendarEvent' => array(self::HAS_MANY, 'CalendarEvent', 'calendar_id'),
}
In CalendarEvent:
relations() {
return array( 'calendar' => array(self::BELONGS_TO, 'Calendar', 'calendar_id'), );
}
And in UserCalendar:
relations() {
return array( 'calendar' => array(self::BELONGS_TO, 'Calendar', 'calendar_id'), );
}
So to make the link between UserCalendar and CalendarEvent you'll need to use Calendar
$criteria = new CDbCriteria;
$criteria->with = array(
"calendarEvent"=>array('condition'=>'planned = 0'),
"userCalendar"=>array('condition'=> 'user_id =' . $user->id),
);
$calendar = Calendar::model()->find($criteria);
and $calendar->calendarEvent will return an array of calendarEvent belonging to the user
Q2: you can enable web logging so all the db request (and others stuffs) will appear at the end of your page:
Logging in Yii (see CWebLogging)
In your application configuration put
'components'=>array(
......
'log'=>array(
'class'=>'CLogRouter',
'routes'=>array(
array(
'class'=>'CWebLogRoute',
),
),
),
),
I am now discovering the Yii framework and doing it by trying to develop a simple application which allows to create Leagues and assign Players to it. The relationship between these is many to many (a player can subscribe to many leagues, and one league contains many players). So I have three db tables - tbl_league, tbl_player and tbl_league_player and I created foreign keys like this:
ALTER TABLE tbl_league_player ADD FOREIGN KEY league_id_idxfk (league_id) REFERENCES tbl_league (id);
ALTER TABLE tbl_league_player ADD FOREIGN KEY player_id_idxfk (player_id) REFERENCES tbl_player (id);
Now I am trying to display League with the list of all players subscribed to it, so my actionView in LeagueController says:
public function actionView($id)
{
$issueDataProvider = new CActiveDataProvider('Player', array(
'criteria' => array(
'condition' => 'league_id=:leagueId',
'params' => array(
':leagueId' => $this->loadModel($id)->id
),
),
'pagination' => array(
'pageSize' => 1,
),
));
$this->render('view',array(
'model'=>$this->loadModel($id),
'issueDataProvider' => $issueDataProvider,
));
}
What I am trying to do here is to get all players from the Player table subscribed to a particular league ID. For this I have the Player model with relation defined like this:
public function relations()
{
// NOTE: you may need to adjust the relation name and the related
// class name for the relations automatically generated below.
return array(
'League' => array(self::MANY_MANY, 'League', 'tbl_league_player(player_id, league_id)'),
);
}
The problem is that I go to, for example, league/view&id=2, I am getting this error: "SELECT COUNT(*) FROM tbl_player t WHERE league_id=:leagueId" so it looks like the relation does not work. What am I missing?
Thanks a lot!
So, first, the reason for the error is that you are adding a WHERE league_id = condition to your Player CActiveDataProvider, and your Player table does not have a league_id column.
What you should do to get a list of the players in your View is:
1) NOT make a CActiveDataProvider - just pass in the $model to your view
public function actionView($id)
{
$this->render('view',array(
'model'=>$this->loadModel($id),
));
}
2) Then, in your view, just lazy-load your MANY_MANY relation and loop through it like so:
<?php foreach($model->League as $player): ?>
<?php echo $player->name ?>
<?php endforeach; ?>
I would rename your "League" relation "players" or something.
If you really need to use a CActiveDataProvider... I can't remember how to (or if you can) pass in a a relation, but you can do it my manually adding join criteria to your CActiveDataProvider.
I hope this helps! Cheers
try this.
Inside LeagueController
public function actionView($id)
{
$criteria=new CDbCriteria;
$criteria->with='League';
$criteria->together=true;
$criteria->condition='league_id=:leagueId';
$criteria->params=array(':leagueId'=>$this->loadModel($id)->id);
$issueDataProvider = new CActiveDataProvider('Player',array(
'criteria'=>$criteria,
),
'pagination' => array(
'pageSize' => 1,
),
));