Can someone tell me what does $this->alias do or mean in cakephp?
$this->alias is the name of your model. For instance, if you have a model named User, $this->alias will be equal to 'User'.
Further to the other answer, Cake models have the property $this->name, which is the name of the model. $this->alias is normally the same value, but in some contexts it is different, for example:
class MyModel extends AppModel {
$hasOne = [
'OtherModelAlias' => [
'className' => 'OtherModel'
]
];
}
When dealing with this associated model OtherModel from the class MyModel, you would refer to it by $this->OtherModelAlias. The value of $this->OtherModelAlias->alias is 'OtherModelAlias', but $this->OtherModelAlias->name is 'OtherModel'
We use $this->alias when we write some code within the model itself. For example, if we are trying to access a data after form submit,
Model: User.php
$this->data[$this->alias]["email"] --------------(1)
Controller: UsersController.php
$this->request->data["Users"]["email"] ----------------(2)
Both (1) and (2) implies the same email field submitted in the form.
N.B: $this->data[$this->alias]["email"] is same as $this->data["User"]["email"], but it's better if we follow the former and stick to the CakePHP conventions. Better safe than sorry! :)
Peace! xD
Related
I just confused because of a find() result. This is my configurations:
First, users can have different User.role values: student, admin, and some others.
// Book
public $belongsTo = array(
'Student' => array(
'className' => 'User',
'foreignKey' => 'student_id'
'conditions' => array('User.role' => 'student')
);
);
When I chain Models like $this->Book->Student->find('list'); I was expecting to get only users whose role are 'student', but instead, it gets all users. What is going on here, what is conditions for on association definition, where can it and cannot be used. Any lead would help, thanks.
PS: I am aware that I could put conditions on find(), that's not the issue
There is a difference between associated data and accessing an associated model object. If you access $this->Book->Student you're accessing the Student model and work in it's scope. The conditions in the defined associations work only in the context of the accessed object.
So if you do a find on the Book and list the students for that book:
$this->Book->find('first', array('contain' => array('Student'));
Your code will work correctly. It will find the book plus the first user who has the role stundent. BUT your association is wrong then: It should be hasMany. because why would you filter a book by role if the book just belongsTo one student?
If you want to filter users by their role you can implement a query param that is checked in beforeFind(), pseudocode: if isset roleFilter then add contention to filter by that role from roleFilter.
Or, if you don't need to paginate just create a getStudents() method in the user model that will return a find('list') that has the conditions.
Or Student extends User and put the filter in the beforeFind() and use that model instead of the User model in your Book association.
If you want to filter on model level or per model I think the last one is a good option. Don't forget to set $useTable and $name or the inherited model will cause problems.
you have miss , inside your model.
try this:
public $belongsTo = array(
'Student' => array(
'className' => 'User',
'foreignKey' => 'student_id', //<------ miss
'conditions' => array('User.role' => 'student')
);
);
Yoi can debug your query to check what is the real query that you make.
Personally I have never use this approach, I prefer to use foreign key with another table for examples Rolesand User.role_id.
Is better for me to use this approach to have more flexibility inside your app.
After I prefer to use a conditions where inside controller to check well the query, because in your way every query you search always for student role not for the other and can be a problem for the rest of role, because inside controller you see a find without conditions but it doesn't take right value because in your model there is a particular conditions.
For me the good way is to create a new table, use foreign key and where conditions inside action of the controller to view well what are you doing.
For default all relations are "left join", you must set the parameter "type" with "inner" value
// Book
public $belongsTo = array(
'Student' => array(
'className' => 'User',
'foreignKey' => 'student_id'
'conditions' => array('Student.role' => 'student'), // <-- Fix That (field name)
'type' => 'inner', // <-- add that
);
);
Is it possible for a virtualFields var to be the sum of a field from a linked table?
For example, in, say, an Invoice model, could you have
public $virtualFields = array(
'invoiceNett' => 'SUM(InvoiceLine.nett)'
);
but obviously only SUMming the lines that belong to that invoice?
Thanks.
== Using CakePHP 2.0
You could use the afterFind callback to get the sum. This avoids storing the calculated values, which should be avoided when possible.
function afterFind($results)
{
foreach($results as &$result)
{
/*
Use something like:
$this->InvoiceLine->find('all', array('fields' => array('SUM(InvoiceLine.nett) as total'),
'conditions' => array('invoice_id' => $result['Invoice']['id'])));
*/
}
unset($result);
}
As far as I know, the best way to do that would be to have an actual total field, and update it anytime the data is saved (likely with a afterSave callback method).
So - anytime an InvoiceLine is saved, you run some code to update it's associated Invoice with a new total.
//InvoiceLine model
public function beforeSave() {
//code to update Invoice's "total" field
}
Theoretically, yes, if that linked table is an associate that is joined (belongsTo and hasOne).
However this would be a poor idea because if you decide not to include that table you would generate a SQL error.
You'd be better off having a separate function grab the data or creating a virtual field that was a nested SQL query.
Defining your virtual field in the respective Model would make more sense.
If not done so, you will be breaking the MVC pattern.
You can use that virtual field from other related models.
If you don't want using them in all related models, you can always use the field
attribute whilst defining relations.
public $hasMany = array(
'IwantVirtualField' => array(
'className' => 'MyModel',
...
)
);
In a model where you don't want virtual field
public $belongsTo = array(
'IwantVirtualField' => array(
'className' => 'MyModel1',
'fields' => array('MyModel1.id', 'MyModel1.name')
...
)
);
Is it possible to get the Model that is trying to save data to a related model in a saveAll transaction. If forexample, User hasMany Goal, and I make a call;
$this->User->saveAll($data);
Can I get the Model User in the Goal Model class using some predefined attritbute, method or mechanism?
Thanks for your help in advance,
Roland.
Edit:
Assume a hypothetical situation where a user has pictures, with a Picture model and Posts( blogs) with a corresponding model. These both can be commented upon. So rather than creating a separate comment model for each of these models, I create one central Comment model. The association might look like this;
//In the Picture model
var $hasMany = array(
`` 'Comment' => array(
'className' => 'Comment',
'foreignKey' => 'picture_id',
'conditions' => array('Comment.objectType' => 'picture')
)`
);
The association for the Post model will be similar, the only difference being;
'conditions' => array('Comment.objectType' => 'post')
With this structure, if I were to query any of these Models, their corresponding comments will be retrieved from the DB table using the objectType field.
If I were to do something like;
$this->Picture->saveAll($data);
, or
$this->Post->saveAll($data);
with the $data array well structured and containing a Comments part to be saved, at some point in time during the saving transaction, data will be saved to the comments table through the Comment model.
What I want is to be able to know that it is the Picture model trying to save comment in the Comment model's beforeSave method, i.e;
//In the Comment mode
beforeSave() {
//Post model is trying to save a comment here
}
With CakePHP, if you associate your models using the hasMany, while you are in the controller you can access that model.
In UserModel
var $hasMany = array('Goal');
In GoalModel
function myCrazyFunc() {
//crazy stuff happens here
}
In UsersController
$this->Goal->myCrazyFunc();
//more stuff
$this->User->saveAll($data);
In my "Reports" controller, which is just a dummy controller without any actual database, I'm trying to generate a paginated view of other models. For example, to generate paginated view of "Transactions" model I'm doing the following:
$this->loadModel('Transactions');
$this->Transactions->bindModel(array('belongsTo'=>array('Item'=>array('className'=>'Item'),'Member'=>array('className'=>'Member'))));
$results = $this->paginate('Transactions',null,array('recursive'=>1));
But this is not giving me associated data from Items and Members. If I do a
$this->Transactions->find('all',array('recursive'=>1))
I get the associated data, but not paginated. How will I get paginated view which includes the associated data too?
Two things: First, even when plural model names can work for some odd reason, the convention is that model names are singular, like $this->loadModel('Transaction');. See the manual on naming conventions.
Second, forget about recursive and go for the Containable behavior. Frankly, it's so useful that I wonder why it isn't the default process (perhaps because Containable got created when the framework was very mature). Matt has a good book explaining why Containable is good (download it, really, it's almost mandatory :D ). But to help even more, I'm going to tell you exactly how you solve your issue with containable:
1) Define the associations in the models, like:
In Transaction model:
var $belongsTo = array(
'Item' => array(
'className' => 'Item',
'foreignKey' => 'item_id',
)
);
In Item model:
var $hasMany = array(
'Transaction' => array(
'className' => 'Transaction',
'foreignKey' => 'item_id',
'dependent' => true,
'exclusive' => true,
)
);
Do the same for the Member model.
2) Create an app_model.php file in /app/ with this code:
(The $actsAs variable here within the AppModel class tells all models to use Containable)
<?php
class AppModel extends Model {
var $recursive = -1;
var $actsAs = array('Containable');
}
?>
3) In the Reports Controller, change the code to something like this:
(The contain parameter is an array of all the associated models that you want to include. You can include only one assoc. model, or all, or whatever you want).
$this->loadModel('Transaction');
$this->paginate = array('Transaction' => array('contain' => array('Item', 'Member')));
$results = $this->paginate('Transaction');
And that's it!
In one of my models, I have a "LONGTEXT" field that has a big dump of a bunch of stuff that I never care to read, and it slows things down, since I'm moving much more data between the DB and the web app.
Is there a way to specify in the model that I want CakePHP to simply ignore that field, and never read it or do anything with it?
I really want to avoid the hassle of creating a separate table and a separate model, only for this field.
Thanks!
Daniel
As #SpawnCxy said, you'll need to use the 'fields' => array(...) option in a find to limit the data you want to retrieve. If you don't want to do this every time you write a find, you can add something like this to your models beforeFind() callback, which will automatically populate the fields options with all fields except the longtext field:
function beforeFind($query) {
if (!isset($query['fields'])) {
foreach ($this->_schema as $field => $foo) {
if ($field == 'longtextfield') {
continue;
}
$query['fields'][] = $this->alias . '.' . $field;
}
}
return $query;
}
Regarding comment:
That's true… The easiest way in this case is probably to unset the field from the schema.
unset($this->Model->_schema['longtextfield']);
I haven't tested it, but this should prevent the field from being included in the query. If you want to make this switchable for each query, you could move it to another variable like $Model->_schemaInactiveFields and move it back when needed. You could even make a Behavior for this.
The parameter fields may help you.It doesn't ignore fields but specifies fields you want:
array(
'conditions' => array('Model.field' => $thisValue), //array of conditions
'fields' => array('Model.field1', 'Model.field2'), //list columns you want
)
You can get more information of retrieving data in the cookbook .
Another idea:
Define your special query in the model:
function myfind($type,$params)
{
$params['fields'] = array('Model.field1','Model.field2',...);
return $this->find($type,$params);
}
Then use it in the controller
$this->Model->myfind($type,$params);
Also try containable behaviour will strip out all unwanted fields and works on model associations as well.
Containable
class Post extends AppModel { <br>
var $actsAs = array('Containable'); <br>
}
where Post is your model?
You can add a beforeFilter function in your Table and add a select to the query
Excample:
public function beforeFind(Event $event, Query $query){
$protected = $this->newEntity()->hidden;
$tableSchema = $event->subject()->schema();
$fields = $tableSchema->columns();
foreach($fields as $key => $name){
if(in_array($name,$protected)){
unset($fields[$key]);
}
}
$query->select($fields);
return $event;
}
In this excample I took the hidden fields from the ModelClass to exclude from result.
Took it from my answer to a simular question here : Hidden fields are still listed from database in cakephp 3