Hello there,
thanks in adv for help.
i have two table: named as options fields are (id,option_title,created,status) and second table as option values fields as (id,option value,options_id,created,status). THe option values table has a foreign key as options_id. Now when i created association between these two table . i didn't work.. Let me explain you how i am creating association.
In the Option model i am using following code.
var $hasMany = array(
'values' => array(
'className' => 'Optionvalue',
'foreignKey' => 'options_id'
));
But this is not working and i am not able to get data from options table. Can any please advice me.. where i am wrong. I just need options table data in the optionsvalue add view file.
Assuming you have a table called options and another called optionvalues. Here is how the Option model looks like
class Option extends AppModel {
public $hasMany = array(
'Optionvalue' => array(
'className' => 'Optionvalue',
'foreignKey' => 'option_id',
'dependent' => FALSE
)
);
}
In the Optionvalue model you then go like this
class Optionvalue extends AppModel {
public $belongsTo = array(
'Option' => array(
'className' => 'Option',
'foreignKey' => 'option_id',
'conditions' => '',
'fields' => '',
'order' => ''
)
);
}
This is how it should look like. In the Optionvalue Add method, you will need to write this in the view
echo $this->Form->create('Optionvalue');
echo $this->Form->input('option_id');
echo $this->Form->end('Save');
and the options will appear by themselves.
Very important though, you did not follow the naming conventions. options_id needs to be option_id among others.
Use
option_id in place of 'options_id'
Cakephp strongly follows naming conventions.
http://book.cakephp.org/2.0/en/getting-started/cakephp-conventions.html
Also looks like you are not using proper naming conventions for other places too.
Related
I try to code a order system and it is my first cake project.
i have a order, a user and a useraddress.
//Order model
public $belongsTo = array(
'User' => array(
'className' => 'User'
));
//User model
public $hasOne = array(
'Useradr' => array(
'className' => 'Useradr',
'foreignKey' => 'user_id',
'order' => ''
)
);
//maybe not necessary Useradr model
public $belongsTo = 'User';
Now i want to access the useraddress while i add/edit a voucher.
I can access the User in the voucher edit view:
echo $this->Form->input('User.email',array(
'label' => 'Email',
));
But not the Useraddress. What iam doing wrong ?
Thank you very much!
Julius
You have to use the correct model prefix.
echo $this->Form->input('Useradr.some_field',array(
'label' => 'Some Field',
));
I suggest you to do the blog tutorial instead of starting blindly coding. Also reading about the conventions is a good idea.
Further Useradr is a pretty bad name, better would be UserAddress. Abbreviations suck for multiple reason. First they're hard to guess if you don't know what they mean and the inflector very likey does not like them either and might screw up the singular/plural cases. Read Clean Code, very nice book about how to write clean code.
I have 2 models, User and Entity. I need to on my entities page, have pagination, and a simple search function. The search function will filter the entities, but the data it filters is a virtualField within the User model.
The Entity model:
public $belongsTo = array(
'User' => array(
'className' => 'User',
'foreignKey' => 'user_id',
'conditions' => '',
'fields' => '',
'order' => ''
)
};
The virtual field in the User model:
public $virtualFields = array("searchFor" => "CONCAT(User.first_name, ' ',User.last_name, ' ',User.email)");
And the condition within the entity controller:
$conditions["User.searchFor LIKE"] = "%".str_replace(" ","%",$this->passedArgs["searchFor"])."%";
$this->paginate = array(
'conditions' => $conditions,
'order' => array('Re.created' => 'DESC'),
'limit' => 20
);
From what I can read this is not possible because I cannot use virtualFields within a query of an associative model, and it says I must use it in "run time", but I really do not understand how to do that, or how to do that in this instance. Any ideas or a way I can get this to work?
You could try attaching the virtualField to the Entity model like so
$this->Entity->virtualFields['searchFor'] = $this->Entity->User->virtualFields['searchFor'];
But you have to make sure that a join is done and not 2 queries.
I believe its discussed in the book.
Edit: Book page
For this purpose I needed to do a search on a concatinated string from an associated model. Normally this can be done using Virtualfields in cake, but cake does not support using a virtualField from an associated model in the search.
Seeing that the 2 models were already linked with the belongsTo, I merely changed the condition to:
"conditions" => "CONCAT(User.first_name, ' ',User.last_name, ' ',User.email) LIKE" => "%".str_replace(" ","%",$this->passedArgs["searchFor"])."%"
Probably not the most elegant solution, but it works.
I want to create a binding from one model to my User model on the fly so that the JOIN is not called every time that I perform a find on that model. I am using the binding to perform a HABTM save. However, when I use the bindModel function, the HABTM data is not saved in the database.
What makes this odd is that if I move my binding to the User model, then the save works perfectly. I don't see any indication in the documentation that the save behavior would be different when the association is made in the model versus the bindModel function (though, I may have missed it if there is any).
Here is my bindModel code in my controller:
$this->User->bindModel(
array(
'hasAndBelongsToMany' => array(
'Othermodel' => array(
'className' => 'Othermodel',
'joinTable' => 'othermodels_users',
'foreignKey' => 'user_id',
'associationForeignKey' => 'othermodel_id',
'unique' => true,
)
)
)
);
if($res = $this->User->save($data)){
return true;
}
And this is my user model.
class User extends AppModel {
public $name = 'User';
public $belongsTo = array();
public $hasOne = array();
public $hasMany = array();
public $hasAndBelongsToMany = array(
'Othermodel' => array(
'className' => 'Othermodel',
'joinTable' => 'othermodels_users',
'foreignKey' => 'user_id',
'associationForeignKey' => 'othermodel_id',
'unique' => true
)
);
Again, I only have the relationship active in one place at one time, so I know the problem is not with the binding itself. It seems the problem is solely related to the fact that I have tried to use bindModel. Is this the intended behavior?
Based on the responses in this article, it looks like the bindModel function only exists for finds.
CakePHP: Bind Model not working
Though there's no information in the CakePHP 1.3 documentation that states that the bind is only for finds, though that would explain the behavior...
http://book.cakephp.org/1.3/en/The-Manual/Developing-with-CakePHP/Models.html
So, I am marking this as answered.
I'm not sure if this is relevant, but are you sure you are not doing any other find operations between the binding and the save. bindModel only binds the model for the next find operation so you may need to pass true through to bindModel to tell it to keep the binding around.
Relevant links: CakePHP: Bind Model not working and http://groups.google.com/group/cake-php/browse_thread/thread/316c9796603eac57?pli=1.
I hope this helps.
Try this,
$this->Message->bindModel(
array(
'belongsTo'=>array(
'User'=>array(
'foreignKey'=>false,
'conditions'=>array('Npl.to=User.id '),
'fields'=>array('recive')
),
'User_e'=>array(
'className'=>'User',
'foreignKey'=>false,
'alias'=>'User_e',
'conditions'=>array('Npl.from=User_e.id '),
'fields'=>array('recive')
)
)
)
);
I am using CakePHP 1.3.
I have a few questions regarding CakePHP Models:
1) Is there a way to define a Model with conditions, such that when invoked using the Containable, behavior, I do not need to define 'conditions' for it again. For example: I have two models - "Store" and "Deal" where "Store" that hasMany "Deal". I want to accomplish the following without having to define these conditions for "Deal" every time:
$this->Store('all', array('contain'=>array('Deal'=>array('isactive'=>1,'now() < expirydate', 'qty > 0')));
2) Also, is there a way to define relationships between Models differently based on conditions? In other words, how can I define "Store" hasMany "Deals", "ActiveDeal", "ExpiredDeal", etc.., all on the 'deals' table but differing based on conditions I set for each.
Much appreciate any help.
Thanks/Regards..
If you look at the first couple of code examples on the Associations: Linking Models Together page of the CakePHP Cookbook, you'll see that you can add conditions to your model joins that will work whenever you do a basic find.
Therefore you should be able to do something like the following:
class Store extends AppModel {
var $name = 'Store';
var $hasMany = array(
'Deal' => array(
'className' => 'Deal',
'foreignKey' => 'store_id',
'conditions' => array(
'Deal.isactive' => '1',
'now() < Deal.expirydate',
'Deal.qty > 0'
),
'order' => ''
),
'ExpiredDeal' => array(
'className' => 'Deal',
'foreignKey' => 'store_id',
'conditions' => array('now() >= ExpiredDeal.expirydate'),
'order' => ''
)
);
}
I have a segment which can have many comments and each comment can have many tags. I can bind the comments to the segments using code like the below which is a function in the segment model class.
function prepareForGettingSegmentsWithComments() {
$this->bindModel(
array('hasMany' => array(
'Comment' => array(
'className' => 'Comment',
'foreignKey' => 'segmentID'
)
)
)
);
}
However how can I bind in the Tags as well?
Yes, in this situation, I bind the Tags to the Comments with a belongsTo. Then filter the results with some query conditions.
Let me see if I can find an example snippet somewhere,
if(isset($this->params['named']['category'])){
$this->Link->bindModel(
array('belongsTo' => array(
'CategoriesLink' => array(
'className' => 'CategoriesLink',
'foreignKey' => 'id',
)
)),
array('belongsTo' => array(
'Category' => array(
'className' => 'Category',
'foreignKey' => 'categories_link_id',
)
))
);
$data = $this->paginate('Link', array('CategoriesLink.category_id'=>$this->params['named']['category']));
} else {
$data = $this->paginate('Link', array('Link.status_id'=>'1'));
}
$this->set('links', $data);
This is how I did it when I was trying to paginate my Link model by a related field. This was with Cake1.2 though, but I think the principle is the same.
I would also recommend installing DebugKit, http://www.ohloh.net/p/cakephp-debugkit , then tinker with the links and conditions until you get a query which works for you.
Sorry, not very technical ;) I'm sure someone can give you a more accurate answer.
PS, Having just reread the question, do you not have these Models linked already? Surely hooking them up in the Models through CakePHP relationships, you'd not need to bind the models and could just use Containable or unbindModel()