I have created two custom entities using hook_entity_info in drupal 7.The entities is created for the given database tables.
I am able to create a view for each of the entities separately.But want to create a view of both entities together.The "relationship" option in view displays "no relationship available". And add fields option displays only the field of the secleted entity.
How do I relate both entities?
I was able to come with two solutions:
1)using relation ,relation end field,relation UI
2)using hook_views_data_alter example from commerce module:
Function hook_views_data_alter(){
// Expose the uid as a relationship to users.
$data['users']['uc_orders'] = array(
'title' => t('Orders'),
'help' => t('Relate a user to the orders they have placed. This relationship will create one record for each order placed by the user.'),
'relationship' => array(
'base' => 'uc_orders',
'base field' => 'uid',
'relationship field' => 'uid',
'handler' => 'views_handler_relationship',
'label' => t('orders'),
),
);
}
Related
Here is Yet another cakePHP question! I have table called blood_groups which has blood_group_id and group fields.. Then I have another table called donors, which has several fields such as name, surname etc. Another field included inside this table is the foreign key 'blood_group_id' which will need to map to the blood_group table on retrieval. in the donor registration view, i want to be able to retrieve the values from the blood_groups table, and display them using the formHelper (with their respective id's).
I have gone through CAKE doc, and I understand that I would need to create the association between my models, but I am struggling to figure this one out. Should I create $hasOne association inside the Donor Model (considering that the Donor table has the fk of the other table). And how would I go about retrieving the options of blood_groups from the blood_groups Model?
Should It work like this?(and are any other prerequisites involved?) :
In my DonorController -
$this->set('blood_groups', $this->Donor->Blood_Group->find('all'));
in Views/Donor/add.ctp
echo $this->Form->input('blood_group_id');
Accessing data through associations is fine. But for radios or checkboxes you want to do a find('list). Your model and variable name does not match the CakePHP convention, there should be no underscore.
Properly named this should be already enough to populate the input.
// controller
$this->set('bloodGroups', $this->Donor->BloodGroup->find('list'));
// view
echo $this->Form->input('blood_group_id');
If you don't follow the conventions for some reason:
echo $this->Form->input('blood_group_id', array(
'options' => $bloodGroups
));
See:
Linking Models Together
The Form Helper
Create one function in BloodGroup Model
function getDonors(){
$options = array(
// 'conditions' => array('Donor.blood_group_id'=>$id),
'joins' => array(
array(
'alias' => 'Donor',
'table' => 'donors',
'type' => 'LEFT',
'conditions' => array(
'Donor.blood_group_id = BloodGroup.blood_group_id',
),
)
),
'fields' => array('Donor.name','Donor.surname','Donor.blood_group_id',
'BloodGroup.blood_group_id')
);
$returnData = $this->find('all',$options);
return $returnData;
}
Now from controller call this function
App::import('model','BloodGroup');
$BloodGroup = new BloodGroup;
$donorList = $BloodGroup->getDonors();
$this->set('donorList',$donorList);
In view file you will get list of donors in $donorList.
I am using CakePHP2.4 and the search plugin https://github.com/CakeDC/search
I have the following
Employee hasOne EmployeeProfile
Employee hasMany Qualification
So I have a single search bar.
the search bar will search using LIKE through the following fields
Employee.name
EmployeeProfile.email
Qualification.title
how do I configure the model Employee->filterArgs for this search?
This is a cross-posting of the original issue here
The documentation includes an example.
'username' => array('type' => 'like', 'field' => array('User.username', 'UserInfo.first_name')),
You just have to make sure the models you're calling are available in your find() call. In the example the find will do a like on the users table username and on the user_infos first_name field at the same time.
I'd like to expand on this as I've been trying to setup a search on a hasMany relationship for a few hours and couldn't find anything. Mark mentionned "custom bindModel (as hasOne) for hasMany relationship (Qualification)". Here's how it's done :
$this->Employee->bindModel(array(
'hasOne' => array(
'Qualification' => array(
'foreignKey' => false,
'conditions' => array('Employee.id = Qualification.employee_id')
)
)
), false);
Just bind it before your paginate and add Qualification.title in your field list in your filterArgs
I'm writing an application that uses Zend Framework 2 and Doctrine (both the latest stable version).
There is much documenation (mainly tutorials and blog posts) that deal with saving doctrine entities to the database in combination with Zend Form. Unfortunately they only deal with simple entities that do not have one-to-many or many-to-many relationships.
This is one of those examples that i have adopted into my own code.
http://www.jasongrimes.org/2012/01/using-doctrine-2-in-zend-framework-2/
I understand that in the Album Entity of this example, the artist is a string to keep the (already lengthy) tutorial as simple as possible. But in a real world situation this would of course be a one-to-many releationship with an Artist Entity (or even a many-to-many). In the view, a select-box could be displayed where the artist can be selected, listing all the artist-entities that could be found in the database, so the right one can be selected.
Following the example with the album, this is how i've set up an 'edit' Action in my controller:
public function editAction()
{
// determine the id of the album we're editing
$id = $this->params()->fromRoute("id", false);
$request = $this->getRequest();
// load and set up form
$form = new AlbumForm();
$form->prepareElements();
$form->get("submit")->setAttribute("label", "Edit");
// retrieve album from the service layer
$album = $this->getSl()->get("Application\Service\AlbumService")->findOneByAlbumId($id);
$form->setBindOnValidate(false);
$form->bind($album);
if ($request->isPost()) {
$form->setData($request->getPost());
if ($form->isValid()) {
// bind formvalues to entity and save it
$form->bindValues();
$this->getEm()->flush();
// redirect to album
return $this->redirect()->toRoute("index/album/view", array("id"=>$id));
}
}
$data = array(
"album" => $album,
"form" => $form
);
return new ViewModel($data);
}
How would this example need to be altered if the artist wasn't a string, but an Artist Entity?
And suppose the album also has multiple Track Entities, how would those be processed?
The example would not need to be changed at all, the changes will happen with your entity and your form.
This is a good reference: Doctrine Orm Mapping
So to save yourself a lot of extra work, your OnToMany relationship would use: cascade = persist:
/**
* #ORM\OneToMany(targetEntity="Artist" , mappedBy="album" , cascade={"persist"})
*/
private $artist;
When it comes to persisting the form object, the entity knows it must save the associated entity as well. If you did not include this, then you would have to do it manually using a collection.
To make like easier with your form, you can use Doctrines Object Select like this:
$this->add(
[
'type' => 'DoctrineModule\Form\Element\ObjectSelect',
'name' => 'artist',
'options' => [
'object_manager' => $this->objectManager,
'target_class' => 'Artist\Entity\Artist',
'property' => 'name', //the name field in Artist, can be any field
'label' => 'Artist',
'instructions' => 'Artists connected to this album'
],
'attributes' => [
'class' => '', //Add any classes you want in your form here
'multiple' => true, //You can select more than one artist
'required' => 'required',
]
]
);
So now your form takes care of the collection for you, the controller as per your example does not need to change since the entity will take care of the persisting...
Hope this gets you on track.
I have two tables QBQuestion(Questionid,Question,OptionId) and Option(OptionId,Option). I want to display option form on view form of QBQuestion? I want to create multiple choice question. i.e.for single question we can add multiple options.For such purpose i want to cretae option field with add button si that when we click add button,we can insert more options and also want to display that all inserted options in table using grid.
So what should i do? please help me....
1) Add relations in the model for this two stuff.
public function relations() {
return array(
'valOptions' => array(self::BELONGS_TO, 'Option', 'OptionId'),
);
}
2) Use lazy loading in CGridView.
$this->widget('zii.widgets.grid.CGridView', array(
'dataProvider' => new CActiveDataProvider('QBQuestion'),
'columns' => array(
'Questionid',
'Question',
'valOptions.Option',
),
));
I think that's what you need.
Hi I am building my User model to have only one profile picture, which is an example of an Image model. I want the conditions of the hasOne to only include the image where the Image.id is equal to the "picture" column in the user table. I know there is something wrong...here is the relevant part of code.
var $hasOne = array(
'ProfileImage' => array(
'className' => 'Image',
'conditions' => array('ProfileImage.id' => 'User.picture'),
'dependent' => false,
'foreignKey' => 'user_id',
)
);
What is wrong with my conditions? How do I specifc ProfileImage.id to be equal to the picture column in the User model? Thanks!
If the foreign key (picture) is in the User model, then your association should be User belongsTo Image. If the image can only be used by one user at a time, the reverse would be Image hasOne user. If an image can be used by multiple users, it would be Image hasMany User.
But, if the image can only be used by one user at a time, the better data structure would be to add a user_id column to the images table. Then you have a User hasOne Image/Image belongsTo User relationship.
For neither you need conditions, only the foreignKey.
After your clarification of the issue, your condition needs to look like:
'conditions' => array('ProfileImage.id = User.picture')
Otherwise Cake thinks ProfileImage.id should equal the string "User.picture".