There are two models: Auteurs and Books. Auteurs model contains following relations:
public function relations() {
return array(
'books' => array(self::MANY_MANY, 'Books','a_liens(id_auteur,id_book)'),
'booksCount' => array(self::STAT, 'Books', 'a_liens(id_auteur,id_book)'),
);
}
How to write criteria to get all Auteurs with zero booksCount?
Instead of counting books, you may just find Auteurs models without any book - result is the same, but question is different. You may achieve this by using LEFT JOIN and finding records with missing books:
$auteurs->with([
'books' => [
'together' => true,
'select' => false,
'joinType' => 'LEFT JOIN',
'condition' => 'books.id IS NULL',
],
]);
Related
hi i wrote this code and it works just fine but i think its not the best way to do it!
i want to get all the jobs for 1 company.
each company can have many addresses and each address can have many jobs
here is my code:
$company = Company::find($id)->with('addresses.jobDetails.job')->first();
$jobs = [];
foreach ($company->addresses as $address) {
foreach ($address->jobDetails as $detail) {
array_push($jobs, [
'id' => $detail->job->id,
'title' => $detail->job->title,
'country' => $detail->job->country,
'city' => $detail->job->city,
'type' => $detail->job->type,
'work_types' => JobType::where('job_id',$detail->job->id)->pluck('title'),
'income' => $detail->income,
]);
}
}
return $jobs;
can anyone help me to change this to better code please
thank you in advance
You do the opposite and start with JobDetails
$jobDetails = JobDetail::whereHas('address.company', function($companyQuery) use($id) {
$companyQuery->where('id', $id);
})->whereHas('jobs', function($jobQuery) {
$jobQuery->where('is_active', 1);
})->with('jobs')->get();
foreach ($jobDetails as $detail) {
array_push($jobs, [
'id' => $detail->job->id,
'title' => $detail->job->title,
'country' => $detail->job->country,
'city' => $detail->job->city,
'type' => $detail->job->type,
'work_types' => JobType::where('job_id',$detail->job->id)->pluck('title'),
'income' => $detail->income,
]);
}
return $jobs;
EDIT:
In your query
Company::find($id)->with('addresses.jobDetails.job')->first();
You run 4 queries with eager loading. one for each model. You can check in the result that you got that all the data is present in the variable $company.
The example I gave you it runs only two queries, the first one (job_details) will use joins to filter the Job results by the id of the companies table (you can make it faster by using the field company_id in the addresses table)
The second one is for the jobs relation using eager loading.
scenario is crm with tables account, account_contact, contact and account_contact_role. The latter contains roles like 'project lead' or 'account manager' for the combos defined in the junction table.
My challenge is the account view, that is listing also the connected persons with their roles. I want my grid to show: Doe | John | employee.
The problem is now when the contact has 2+ entries in the junction table. How can I print the correct role for the row? As you can see in the code I solved it the static way which shows only 1 out of n times the correct value. Tried it with inner join without success. Is it a problem of the search in the model or with the access in the view?
the relation from the account model to the contacts:
public function getContacts($role = null)
{
// many-to-many
return $this->hasMany(ContactRecord::className(), ['id' => 'contact_id'])
->via('accountContacts')
->innerJoinWith(['accountContacts.role'])
->andWhere(['account_contact.account_id' => $this->id])
->andWhere(['account_contact_role.type' => $role])
;
}
the view
<?= \yii\grid\GridView::widget([
'dataProvider' => new \yii\data\ActiveDataProvider([
'query' => $model->getContacts('internal'),
'pagination' => false
]),
'columns' => [
'lastname',
'firstname',
[
'label' => 'Role',
'attribute' => 'accountContacts.0.role.name',
],
[
'class' => \yii\grid\ActionColumn::className(),
'controller' => 'contacts',
'header' => Html::a('<i class="glyphicon glyphicon-plus"></i> Add New', ['contact-records/create', 'account_id' => $model->id]),
'template' => '{update}{delete}',
]
]
]); ?>
defined relations are:
account has many accountContacts has one contact
accountContacts has one accountContactRole
Many thanks in advance!
You are showing account's contacts, so you have to list from Contact model.
Inside Contact model (or Contact ActiveQuery file):
public static function queryContactsFromAccountAndRole($account, $role = null)
{
// many-to-many
return ContactRecord::find()->innerJoinWith(['accountContacts' => function($q) use($account, $role) {
$q->innerJoinWith(['accountContactsRole'])
->andWhere(['account_contact.account_id' => $account->id])
->andWhere(['account_contact_role.type' => $role]);
}])
->andWhere(['account_contact.account_id' => $account->id]);
}
Now you have one record for each contact and the gridview will show all contacts.
I can't get the values from a second level contain using different field name as link.
Asistencia model:
var $belongsTo = array('Employee');
Horario model:
var $belongsTo = array('Employee');
Employee model:
var $hasMany = array(
'Horario' => array(
'foreignKey' => 'employee_id',
'dependent' => true
),
'Asistencia' => array(
'foreignKey' => 'employee_id',
'dependent' => true
)
);
I'll explain using these values on my example record:
Asistencia: employee_id = 3701
Employee : id = 3701
In my find() from Asistencia, I get to contain Employee by switching Employee primaryKey just fine:
$this->Asistencia->Employee->primaryKey = 'id';
$this->paginate = array(
'contain' => array(
'Employee' => array(
'foreignKey' => 'employee_id',
//'fields' => array('id', h('emp_ape_pat'), h('emp_ape_mat'), h('name')),
'Horario' => array(
'foreignKey' => 'employee_id',
//'fields' => array('id' )
))
),
'conditions' => $conditions,
'order' => 'Asistencia.employee_id');
However, my Horario record is linked to Employees via another field: emp_appserial
Employee : emp_appserial = 373
Horario : employee_id = 373
How can my Employee array contain Horario array? (they do share the value just mentioned).
Currently, the Horario contain is using the value on Asistencia.employee_id and Employee.id (3701). (checked the sql_dump and is trying to get the Horario via
"WHERE `Horario`.`employee_id` = (3701)"
but for the Employee to contain Horario, it should use the value on Employee.emp_appserial and Horario.employee_id (373).
This is what i get (empty array at bottom)
array(
(int) 0 => array(
'Asistencia' => array(
'id' => '5',
'name' => null,
'employee_id' => '3701',
'as_date' => '2012-11-19',
),
'Employee' => array(
'id' => '3701',
'emp_appserial' => '373',
'emp_appstatus' => '8',
'AgentFullName' => '3701 PEREZ LOMELI JORGE LORENZO',
'FullNameNoId' => 'PEREZ LOMELI JORGE LORENZO',
'Horario' => array()
)))
Please notice:
'employee_id' => '3701', (Asistencia)
and
'emp_appserial' => '373', (Employee)
my Horario has 'employee_id' = 373.
How could I make the switch so the relation Employee<->Horario is based on emp_appserial?
Thank you in advance for any help.
Firstly you may be getting problems because you're using Spanish words for your Model names and I suspect you're also using them for Controllers.
This is a very bad practice since CakePHP's idea is:
Convention over configuration.
This is achieved through the Inflector class which "takes a string and can manipulate it to handle word variations such as pluralizations or camelizing and is normally accessed statically". But this works ONLY WITH English words. What this means for you is that you may be missing the data because Cake is not able to build the DB queries right, since you're using Spanish words. By doing so you're making the use of CakePHP's flexible persistence layer obsolete - you will have to make all the configurations by hand. Also most likely pagination will NOT WORK. Other parts of the framework may also not work properly:
HtmlHelper, FormHelper, Components, etc. ...
These problems will expand if you try to use more complex Model associations such as HABTM or "hasMany through the Join Model".
So I do not know why exactly you aren't seeing the Horario record, but what I just explained most likely is your problem.
What you're trying to do is against the core principles of CakePHP and you'd save yourself a lot of problems if you refactor a bit and use English words. Your other option is NOT TO use Cake.
Using Yii framework.
I have 3 models.
Articles - table articles(id, name)
ArticlesToAuthors - table articles_to_authors (id, article_id, author_id, type)
Authors - table authors (id, name)
I need to get authors.*, article_to_authors.type for specific article_id.
I was trying to make relation in ArticlesToAuthors model like that:
'authors' => array(self::HAS_MANY, 'Authors', array('id' => 'author_id'), ),
then i made query:
$authors = ArticlesToAuthors::model()->with('authors')->findAll(array('condition' => 'article_id=2217') );
foreach($authors as $a)
{
//this is not working
#var_dump($a->authors->name);
#var_dump($a->name);
//this works fine
foreach($a->authors as $aa)
{
var_dump($aa->name);
}
}
Can i get all active record object in one and not to do foreach in foreach?
I need an analogue to sql "SELECT atoa., a. FROM articles_to_authors atoa LEFT JOIN authors a ON atoa.author_id=a.id WHERE atoa.article_id=:article_id"
Am i doing right?
p.s. - sorry for my bad english.
It looks like you need the following relations:
in your ArticlesToAuthors table:
'author' => array(self::BELONGS_TO, 'Authors', 'author_id'),
'article' => array(self::BELONGS_TO, 'Articles', 'article_id'),
and, for completeness, in your Authors table:
'articlesToAuthors' => array(self::HAS_MANY, 'ArticlesToAuthors', array('id' => 'author_id'), ),
This should allow you to access $a->author->name. Not tested, that's just off the top of my head.
I am semi-frustrated with this Yii CGridView problem and any help or guidance would be highly appreciated.
I have two related tables shops (shop_id primary) and contacts (shop_id foreign) such that a single shop may have multiple contacts. I'm using CGridview for pulling records and sorting and my relation function in shops model is something like:
'shopscontact' => array(self::HAS_MANY, 'Shopsmodel', 'shop_id');
On the shop grid, I need to display the shop row with any one of the available contacts. My attempt to filter, search the Grid has worked pretty fine, but I'm stuck in one very strange problem. The respective grid column does not display the value that is intended.
On CGridview file, I'm doing something like
array(
'name' => 'shopscontact.contact_firstname',
'header' => 'First Name',
'value' => '$data->shopscontact->contact_firstname'
),
to display the contact's first name. However, even under circumstances that searching/sorting are both working (I found out by checking the db associations), the grid column comes out empty! :( And when I do a var_dump
array(
'name' => 'shopscontact.contact_firstname',
'header' => 'First Name',
'value' => 'var_dump($data->shopscontact)'
),
The dump shows record values in _private attributes as follows:
private '_attributes' (CActiveRecord) =>
array
'contact_firstname' => string 'rec1' (length=4)
'contact_lastname' => string 'rec1 lsname' (length=11)
'contact_id' => string '1' (length=1)
< Edit: >
My criteria code in the model is as follows:
$criteria->with = array(
'owner',
'states',
'shopscontacts' => array(
'alias' => 'shopscontacts',
'select' => 'shopscontacts.contact_firstname,shopscontacts.contact_lastname',
'together' => true
)
);
< / Edit >
How do I access the values in their respective columns? Please help! :(
Hmm, I have not used the with() and together() methods much. What's interesting is how in the 'value' part of the column, $data->shopscontacts loads up the relation fresh, based on the relations() definition (and is not based on the criteria you declared).
A cleaner way to handle the array output might be like this:
'value' => 'array_shift($data->shopscontacts)->contact_lastname'
Perhaps a better way to do this, though, would be to set up a new (additional) relation, like this in your shops model:
public function relations()
{
return array(
'shopscontacts' => array(self::HAS_MANY, 'Shopsmodel', 'shop_id'), // original
'firstShopscontact' => array(self::HAS_ONE, 'Shopsmodel', 'shop_id'), // the new relation
);
}
Then, in your CGridView you can just set up a column like so:
'columns'=>array(
'firstShopscontact.contact_lastname',
),
Cheers
Since 'shopscontact' is the name of the has-many relation, $data->shopscontact should be returning an array with all the shops related... did you modify the relation in order to return only one record (if I didn't get you wrong, you only need to display one, right?)? If you did it, may I see your filtering code?
P.S. A hunch to get a fast but temporal solution: have you tried 'value' => '$data->shopscontact['contact_firstname']'?