I have a case where I need to have a union query, and at the same time have the end result as ActiveRecord. To illustrate, I have two tables Products and Categories. Each has a model that extends from ActiveRecord. I'm trying to have a query as follows
$deletedProducts = Products::find()->select([$productsTable . '.id', $productsTable . '.selling_price'])
->where(['client' => $clientId])
->andWhere(['<>', 'hidden', 0]);
and
$deletedCats = Category::find()->select([$catTable . '.id', $catTable . '.selling_price'])
->where(['client' => $clientId])
->andWhere(['<>', 'hidden', 0]);
and
$archivedItemsQuery = $deletedProducts->union($deletedCats);
I then return the results through ActiveDataProvider
$dataProvider = new ActiveDataProvider(
[
'query' => $archivedItemsQuery,
]
]);
return $dataProvider->getModels();
The problem is that when I try to send pagination data in the request, or set it through the ActiveDataProvider, no pagination happens. I can change this query of course to
$unionQuery = (new \yii\db\Query())
->from(['counter_items' => $archivedItemsQuery]);
This results in correct pagination, but I dont get the result as ActiveRecords.
Any ideas on how to create a union query with 1. Pagination working and 2. The result is ActiveRecord.
Hi can you try the following, not very neatest of the ways but I hope it works for you I tested it on my end and if I understand correctly you want the result of $dataProvider->getModels() to return an ActiveRecord Models Object i.e Products model and want the GridView pagination to work correctly at the same time.
You need to change the ActiveDataProvider to use the union $archivedItemsQuery in the following way.
$dataProvider = new \yii\data\ActiveDataProvider([
'query' => Products::find()->from(['products' => $archivedItemsQuery]),
'pagination' => [
'pageSize' => 10,
],
]);
Related
I am using Yii 2.
I have model User with table in DB.
I tried link this table with table auth_assignment which has columns user_id, item_name and created_at using LEFT JOIN clause. I wrote this:
$model = User::find()->leftJoin('auth_assignment AS a', '`a`.`user_id` = `user`.`id`');
Then I attach $model to ActiveDataProvider:
$dataProvider = new ActiveDataProvider([
'query' => $model,
'pagination' => [
'pageSize' => 5,
],
]);
But when I run application it prints only Users columns: screenshot
What I do wrong?
In User model you can't get attributes of another table(auth_assignment ).
You can do it by two ways.
Define relationship between User and auth_assignment table in User model using hasOne()
Example : ( use your model name for auth_assignment table)
public function getAuth()
{
return $this->hasOne(Authassignment::className(), ['id' => 'user_id']);
}
Than you can use auth_assignment table attributes as $model->auth->user_id;
Generate query as array as below, in this scenario you don't need to define relationship in model.
$model = User::find()
->select('user.*,a*')
->leftJoin('auth_assignment AS a', 'a.user_id= user.id')
->asArray();
$dataProvider = new ActiveDataProvider([
'query' => $model,
'pagination' => [
'pageSize' => 5,
],
]);
Now you can use attributes as array variable i.e.
$model['user_id'] etc.
I have this form :
$builder
->add('restaurantsFilter1', 'entity', [
'label' => 'Commune',
'empty_value' => 'Dans toute la Narbonnaise',
'class' => 'AppBundle:City',
'choice_label' => 'name',
'query_builder' => function (EntityRepository $er) {
return $er
->createQueryBuilder('c')
->addSelect('d')
->leftJoin('c.documents', 'd')
->where('d.type = :type')
->orderBy('c.name')
->setParameter('type', Document::T_VILLAGE)
;
},
])
which is a select which displays a list of cities.
A client told me that he needed a field "Around me" which will display all cities around 20 km.
So to do so, I created a new city in my database with this name, but now I need to put it in the first position of my select.
In sql I would use something like ORDER BY (insee_code= '[specific_code_of_the_city]') but I dont know how I could that with the query builder.
Do you have an idea how I could do that with the symfony query builder ?
EDIT: That's the exact issue that How do I return rows with a specific value first?
You could create a hidden field and order by that.
return $er
->createQueryBuilder('c')
->addSelect('CASE
WHEN c.name = "specific_code_of_city"
THEN 0
ELSE 1
END as HIDDEN presetOrder')
->addSelect('d')
->leftJoin('c.documents', 'd')
->where('d.type = :type')
->orderBy('presetOrder', 'ASC')
->addOrderBy('c.name', 'ASC')
->setParameter('type', Document::T_VILLAGE)
;
My query gets the timeout error on each run. Its a pagination with joins.
I want to debug the SQL, but since I get a timeout, I can't see it.
How can I see the compiled SQL Query before execution?
Some cake code:
$this -> paginate = array(
'limit' => '16',
'joins' => array( array(
'table' => 'products',
'alias' => 'Product',
'type' => 'LEFT',
'conditions' => array('ProductModel.id = Product.product_model_id')
)),
'fields' => array(
'COUNT(Product.product_model_id) as Counter',
'ProductModel.name'
),
'conditions' => array(
'ProductModel.category_id' => $category_id,
),
'group' => array('ProductModel.id')
);
First off, set the debug variable to 2 in app/config/config.php.
Then add:
<?php echo $this->element('sql_dump');?>
at the end of your layout. This should actually be commented out in your default cake layout.
You will now be able see all SQL queries that go to the database.
Now copy the query and use the SQL EXPLAIN command (link is for MySQL) over the database to see what the query does in the DBMS. For more on CakePHP debugging check here.
Since your script doesn't even render you can try to get the latest log directly from the datasource with:
function getLastQuery()
{
$dbo = $this->getDatasource();
$logs = $dbo->getLog();
$lastLog = end($logs['log']);
return $lastLog['query'];
}
This needs to be in a model since the getDatasource() function is defined in a model.
Inspect the whole $logs variable and see what's in there.
One more thing you can do is ....
Go to Cake/Model/DataSource/DboSource.php and locate function execute() and print $sql variable.
That should print the sql.
This certainly is not be the cleanest way (as you are changing Cake directory) .. but certainly would be quickest just to debug if something is not working with sql.
Try...
function getLastQuery($model) {
$dbo = $model->getDatasource();
$logData = $dbo->getLog();
$getLog = end($logData['log']);
echo $getLog['query'];
}
Simple way to show all executed query of your given model:
$sqllog = $this->ModelName->getDataSource()->getLog(false, false);
debug($sqllog);
class YourController extends AppController {
function testfunc(){
$this->Model->find('all', $options);
echo 'SQL: '.$this->getLastQuery();
}
function getLastQuery()
{
$dbo = ConnectionManager::getDataSource('default');
$logs = $dbo->getLog();
$lastLog = end($logs['log']);
return $lastLog['query'];
}
}
or you can get all the query by adding following line in to the function execute() in lib/Cake/Model/DataSource.php
Debugger::dump($sql);
set the debug variable to 2 in app/config/config.php.
echo $this->Payment->save();
Out put like =>SQL Query: INSERT INTO photoora_photoorange.payments VALUES (*******)
[insert query][2]
set the debug variable to 2 in app/config/config.php.
And
My title will look like naive but I have to say I read/searched/tested everything possible, but my find() method don't implement the JOIN to related tables in the SQL query. I used it several times in other projects without problems but here...
Here my 2 models (nothing special but the manual definition of the related model) :
class Pflanzen extends AppModel {
public $useTable = 'pflanzen';
public $hasAndBelongsToMany = array(
'Herbar' => array(
'order'=>'Herbar.order ASC',
'joinTable' => 'herbar_pflanzen',
'foreignKey' => 'pflanzen_id',
'associationForeignKey' => 'herbar_id')
);
}
class Herbar extends AppModel {
public $useTable = 'herbar';
public $hasAndBelongsToMany = array(
'Pflanzen' => array('joinTable' => 'herbar_pflanzen',
'foreignKey' => 'herbar_id',
'associationForeignKey' => 'pflanzen_id')
)
}
Here my query in the "Herbar" controller (can't be more normal...) :
$pflanzen = $this->Herbar->Pflanzen->find('all',array(
'fields'=>array('Herbar.name','Pflanzen.linkplatter'),
'conditions' => array('Pflanzen.linkplatter' => true),
'order' => 'Herbar.name',
'limit' => 10,
'recursive'=>2)
);
$this->set('pflanzen',$pflanzen);
and the resulting error in the view :
Error: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'Herbar.name' in 'field list'
SQL Query: SELECT `Herbar`.`name`, `Pflanzen`.`linkplatter`, `Pflanzen`.`id` FROM `burgerbib`.`platter_pflanzen` AS `Pflanzen` WHERE `Pflanzen`.`linkplatter` = '1' ORDER BY `Herbar`.`name` ASC LIMIT 10
You can see that their is no JOIN in the SQL. Why ?? What do I wrong ?
I would really appreciate your help as I'm searching for hours and do no more see any solutions and didn't find nothing using google. Thanks in advance !!
HABTM doesn't make joined queries, it makes a query for all base records and more queries as needed for each relationship to fill the array. Your condition assumes a join, hence the error.
You can force a join using the 'joins' parameter. http://book.cakephp.org/1.2/en/view/872/Joining-tables
In the End, the better way of doing this is using the containable behaviour. Force Join is only useful when the containable behavior don't respond to the need :
http://book.cakephp.org/2.0/fr/core-libraries/behaviors/containable.html#using-containable
I have the following example code:
$dataProvider = new CActiveDataProvider('firstTable',
array('criteria' => array(
'select' => 't.firstfield,secondTable.secondfield',
'join' => 'join secondTable on secondTable.id=t.secondTable_id',
),
'pagination' => array(
'pageSize' => 10,
),
));
$results=$dataProvider->getData();
After running the code above, firstField (from the model table - firstTable) is available in the object, but secondField (from the joined table - secondTable) is not.
Can anyone provide assistance on what is wrong with the code or why the "select" option is not picking up the secondField?
it would be better if you use CDbCriteria, that has a better solution to join table with the help of relations. I can show the example with CDbCriteria.
$criteria = new CDbCriteria;
$criteria->select = 'firstfield';
$criteria->with = array('secondTable_relation'=>array('select'=>'secondfield'));
$dataProvider = new CActiveDataProvider('firstTable',
array('criteria' => $criteria,
'pagination' => array(
'pageSize' => 10,
),
));
$results=$dataProvider->getData();
secondTable_relation is a relation name with secondTable.
Can anyone provide assistance on what is wrong with the code or why the "select" option is not picking up the secondField?
Answer:
That is happening because you have not selected the field which relates the two tables, i.e the foreign key in firstTable : secondTable_id. So if you do:
'select' => 't.firstfield,t.secondTable_id,secondTable.secondfield',
you will be able to access the secondField of secondTable:
$singleresultrow->secondTableRelationName['secondField'];// a single result row can be obtained by foreach iteration over $results
However there will still be another query (Lazy Loading) when you access the secondField. And the initial(before lazy) object returned will not have the secondTable object filled.
The problem i think lies in the fact that by default yii accesses the related fields by lazy loading, and for that to happen the related foreign_key should be present in the model you are trying to make a dataprovider of, here it is firstTable, and the foreign_key secondTable_id.