CakePHP - problem with HABTM paginate query - php

Tables
restaurants
cuisines
cuisines_restaurants
Both restaurant and cuisine model are set up to HABTM each other.
I'm trying to get a paginated list of restaurants where Cuisine.name = 'italian' (example), but keep getting this error:
1054: Unknown column 'Cuisine.name' in 'where clause'
Actual query it's building:
SELECT `Restaurant`.`id`, `Restaurant`.`type` .....
`Restaurant`.`modified`, `Restaurant`.`user_id`, `User`.`display_name`,
`User`.`username`, `User`.`id`, `City`.`id`,`City`.`lat` .....
FROM `restaurants` AS `Restaurant` LEFT JOIN `users` AS `User` ON
(`Restaurant`.`user_id` = `User`.`id`) LEFT JOIN `cities` AS `City` ON
(`Restaurant`.`city_id` = `City`.`id`) WHERE `Cuisine`.`name` = 'italian'
LIMIT 10
The "....." parts are just additional fields I removed to shorten the query to show you.
I'm no CakePHP pro, so hopefully there's some glaring error. I'm calling the paginate like this:
$this->paginate = array(
'conditions' => $opts,
'limit' => 10,
);
$data = $this->paginate('Restaurant');
$this->set('data', $data);
$opts is an array of options, one of which is 'Cuisine.name' => 'italian'
I also tried setting $this->Restaurant->recursive = 2; but that didn't seem to do anything (and I assume I shouldn't have to do that?)
Any help or direction is greatly appreciated.
EDIT
models/cuisine.php
var $hasAndBelongsToMany = array('Restaurant');
models/restaurant.php
var $hasAndBelongsToMany = array(
'Cuisine' => array(
'order' => 'Cuisine.name ASC'
),
'Feature' => array(
'order' => 'Feature.name ASC'
),
'Event' => array(
'order' => 'Event.start_date ASC'
)
);

As explained in this blogpost by me you have to put the condition of the related model in the contain option of your pagination array.
So something like this should work
# in your restaurant_controller.php
var $paginate = array(
'contain' => array(
'Cuisine' => array(
'conditions' => array('Cuisine.name' => 'italian')
)
),
'limit' => 10
);
# then, in your method (ie. index.php)
$this->set('restaurants', $this->paginate('Restaurant'));

This fails because Cake is actually using 2 different queries to generate your result set. As you've noticed, the first query doesn't even contain a reference to Cuisine.
As #vindia explained here, using the Containable behavior will usually fix this problem, but it doesn't work with Paginate.
Basically, you need a way to force Cake to look at Cuisine during the first query. This is not the way the framework usually does things, so it does, unfortunately, require constructing the join manually
. paginate takes the same options as Model->find('all'). Here, we need to use the joins option.
var $joins = array(
array(
'table' => '(SELECT cuisines.id, cuisines.name, cuisines_restaurants.restaurant_id
FROM cuisines_restaurants
JOIN cuisines ON cuisines_restaurants.cuisines_id = cuisines.id)',
'alias' => 'Cuisine',
'conditions' => array(
'Cuisine.restaurant_id = Restaurant.id',
'Cuisine.name = "italian"'
)
)
);
$this->paginate = array(
'conditions' => $opts,
'limit' => 10,
'joins' => $joins
);
This solution is a lot clunkier than the others, but has the advantage of working.

a few ideas on the top of my mind:
have you checked the model to see if the HABTM is well declared?
try using the containable behavior
in none of those work.. then you could always construct the joins for the paginator manually
good luck!

Cuisine must be a table (or alias) on the FROM clausule of your SELECT.
so the error:
1054: Unknown column 'Cuisine.name' in 'where clause'
Is just because it isn't referenced on the FROM clausule

If you remove the Feature and Event part of your HABTM link in the Restaurant model, does it work then?
Sounds to me like you've failed to define the right primary and foreing keys for the Cuisine model, as the HABTM model is not even including the Cuisine tabel in the query you posted here.

Related

How to do join with 3 tables in CakePHP 2.x [duplicate]

can anyone tell me, how to retrieve joined result from multiple tables in cakePHP ( using cakePHP mvc architecture). For example, I have three tables to join (tbl_topics, tbl_items, tbl_votes. Their relationship is defined as following: a topic can have many items and an item can have many votes. Now I want to retrieve a list of topics with the count of all votes on all items for each topic. The SQL query for this is written below:
SELECT Topic.*, count(Vote.id) voteCount
FROM
tbl_topics AS Topic
LEFT OUTER JOIN tbl_items AS Item
ON (Topic.id = Item.topic_id)
LEFT OUTER JOIN tbl_votes AS Vote
ON (Item.id = Vote.item_id);
My problem is I can do it easily using $this-><Model Name>->query function, but this requires sql code to be written in the controller which I don't want. I'm trying to find out any other way to do this (like find()).
$markers = $this->Marker->find('all', array('joins' => array(
array(
'table' => 'markers_tags',
'alias' => 'MarkersTag',
'type' => 'inner',
'foreignKey' => false,
'conditions'=> array('MarkersTag.marker_id = Marker.id')
),
array(
'table' => 'tags',
'alias' => 'Tag',
'type' => 'inner',
'foreignKey' => false,
'conditions'=> array(
'Tag.id = MarkersTag.tag_id',
'Tag.tag' => explode(' ', $this->params['url']['q'])
)
)
)));
as referred to in nate abele's article: link text
I'll be honest here and say that you'll probably be a lot happier if you just create a function in your model, something like getTopicVotes() and calling query() there. Every other solution I can think of will only make it more complicated and therefore uglier.
Edit:
Depending on the size of your data, and assuming you've set up your model relations properly (Topic hasMany Items hasMany Votes), you could do a simple find('all') containing all the items and votes, and then do something like this:
foreach ($this->data as &$topic)
{
$votes = Set::extract('/Topic/Item/Vote', $topic);
$topic['Topic']['vote_count'] = count($votes);
}
Two things are important here:
If you have a lot of data, you should probably forget about this approach, it will be slow as hell.
I've written this from my memory and it might not look like this in real life and/or it may not work at all :-)
You can easily set the "recursive" property on a find() query.
$result = $this->Topic->find('all', array('recursive' => 2));
Alternatively, you can use the Containable behavior in your model. Then you can use:
$this->Topic->contain(array(
'Item',
'Item.Vote',
));
$result = $this->Topic->find('all');
or
$result = $this->Topic->find('all', array(
'contain' => array(
'Item',
'Item.Vote',
),
));
What you need is recursive associations support, which is not possible with stock CakePHP currently.
Although it could be achieved using some bindModel trickery
or an experimental RecursiveAssociationBehavior.
Both of these solutions will either require you to use extra code or rely on a behaviour in your application but if you resist the temptation to write pure SQL code, you'll be rewarded with being able to use Cake`s pagination, auto conditions, model magic etc..
I think this answer is already submitted, but I am posting here for someone who seeks still for this.
The joins can be done with find() method can be like below
$result = $this->ModelName1->find("all",array(
'fields' => array('ModelName1.field_name','Table2.field_names'), // retrieving fileds
'joins' => array( // join array
array(
'table' => 'table_name',
'alias' => 'Table2',
'type' => 'inner',
'foreignKey' => false,
'conditions'=> array('ModelName1.id = Table2.id') // joins conditions array
),
array(
'table' => 'table_name3',
'alias' => 'Table3',
'type' => 'inner',
'foreignKey' => false,
'conditions'=> array('Table3.id = Table2.id')
)
)));
You should study HaBTM (Has and Belongs to Many)
http://book.cakephp.org/2.0/en/models/associations-linking-models-together.html

CakeDC search plugin using complex conditions in query with HABTM

I wrote 2 days ago to ask about andConditions and it appeared that I didn't understand the idea but the fact is that for two days now I am stuck with the next step using CakeDC:
How do I implement complex HABTM conditions in "query" methods for CakeDC search plugin?
I have Offer HABTM Feature (tables: offers, features, features_offers) and the below works just fine when used in controller:
debug($this->Offer->find('all', array('contain' => array(
'Feature' => array(
'conditions' => array(
'Feature.id in (8, 10)',
)
)
)
)
)
);
The problem comes when I want to use the same conditions in the search:
public $filterArgs = array(
array('name' => 'feature_id', 'type' => 'query', 'method' => 'findByFeatures'),
);
........
public function findByFeatures($data = array()) {
$conditions = '';
$featureID = $data['feature_id'];
if (isset($data['feature_id'])) {
$conditions = array('contain' => array(
'Feature' => array(
'conditions' => array(
'Feature.id' => $data['feature_id'],
)
)
)
);
}
return $conditions;
}
I get an error:
Error: SQLSTATE[42S22]: Column not found: 1054 Unknown column
'contain' in 'where clause'
which makes me think that I cannot perform this search and/or use containable behavior in searches at all.
Can someone with more experience in the field please let me know if I am missing something or point me to where exactly to find a solution for that - perhaps a section in the cookbook?
EDIT: Also tried the joins. This works perfectly fine in the controller, returning all the data I need:
$options['joins'] = array(
array('table' => 'features_offers',
'alias' => 'FeaturesOffers',
'type' => 'inner',
'conditions' => array(
'Offer.id = FeaturesOffers.offer_id'
),
array('table' => 'features',
'alias' => 'F',
'type' => 'inner',
'conditions' => array(
'F.id = FeaturesOffers.feature_id'
),
)
),
);
$options['conditions'] = array(
'feature_id in (13)' //. $data['feature_id']
);
debug($this->Offer->find('all', $options));
... and when I try to put in the search method I get the returned conditions only in the where clause of the SQL
WHERE ((joins = (Array)) AND (conditions = ('feature_id in Array')))
...resulting in error:
SQLSTATE[42S22]: Column not found: 1054 Unknown column 'joins' in 'where clause'
EDIT: Maybe I am stupid and sorry to say that but the documentation of the plugin sucks a ton.
I double, triple and quadruple checked (btw, have lost already 30 hours at least on 1 filed of the search form facepalm) and the stupid findByTags from the documentation still doesn't make any sense to me.
public function findByTags($data = array()) {
$this->Tagged->Behaviors->attach('Containable', array('autoFields' => false));
$this->Tagged->Behaviors->attach('Search.Searchable');
$query = $this->Tagged->getQuery('all', array(
'conditions' => array('Tag.name' => $data['tags']),
'fields' => array('foreign_key'),
'contain' => array('Tag')
));
return $query;
}
As I understand it
$this->Tagged
is supposed to be the name of the model of the HABTM association.
This is quite far from the standards of cakePHP though: http://book.cakephp.org/2.0/en/models/associations-linking-models-together.html#hasandbelongstomany-habtm
The way it is described here, says that you don't need another model but rather you associate Recipe with Ingredient as shown below:
class Recipe extends AppModel {
public $hasAndBelongsToMany = array(
'Ingredient' =>
array(
'className' => 'Ingredient',
'joinTable' => 'ingredients_recipes',
'foreignKey' => 'recipe_id',
'associationForeignKey' => 'ingredient_id',
'unique' => true,
'conditions' => '',
'fields' => '',
'order' => '',
'limit' => '',
'offset' => '',
'finderQuery' => '',
'deleteQuery' => '',
'insertQuery' => ''
)
);
}
meaning that you can access the HABTM assoc table data from Recipe without needing to define model "IngredientRecipe".
And according to cakeDC documentation the model you need is IngredientRecipe and that is not indicated as something obligatory in the cakePHP documentation. Even if this model is created the HABTM assoc doesn't work properly with it - I tried this as well.
And now I need to re-write the search functionality in my way, using only cakePHP even though I spent already 30 hours on it... so unhappy. :(
Every time I come to do this in a project I always spend hours figuring out how to do it using CakeDC search behavior so I wrote this to try and remind myself with simple language what I need to do. I've also noticed that although Using the CakeDC search plugin with associated models this is generally correct there is no explanation which makes it more difficult to modify it to one's own project.
When you have a "has and belongs to many" relationship and you are wanting to search the joining table i.e. the table that has the two fields in it that joins the tables on either side of it together in a many-to-many relationship you want to create a subquery with a list of IDs from one of the tables in the relationship. The IDs from the table on the other side of the relationship are going to be checked to see if they are in that record and if they are then the record in the main table is going to be selected.
In this following example
SELECT Handover.id, Handover.title, Handover.description
FROM handovers AS Handover
WHERE Handover.id in
(SELECT ArosHandover.handover_id
FROM aros_handovers AS ArosHandover
WHERE ArosHandover.aro_id IN (3) AND ArosHandover.deleted != '1')
LIMIT 20
all the records from ArosHandover will be selected if they have an aro_id of 3 then the Handover.id is used to decide which Handover records to select.
On to how to do this with the CakeDC search behaviour.
Firstly, place the field into the search form:
echo $this->Form->create('Handover', array('class' => 'form-horizontal'));?>
echo $this->Form->input('aro_id', array('options' => $roles, 'multiple' => true, 'label' => __('For', true), 'div' => false, true));
etc...
notice that I have not placed the form element in the ArosHandover data space; another way of saying this is that when the form request is sent the field aro_id will be placed under the array called Handover.
In the model under the variable $filterArgs:
'aro_id' => array('name' => 'aro_id', 'type' => 'subquery', 'method' => 'findByAros', 'field' => 'Handover.id')
notice that the type is 'subquery' as I mentioned above you need to create a subquery in order to be able to find the appropriate records and by setting the type to subquery you are telling CakeDC to create a subquery snippet of SQL. The method is the function name that are going to write the code under. The field element is the name of the field which is going to appear in this part of the example query above
WHERE Handover.id in
Then you write the function that will return the subquery:
function findByAros($data = array())
{
$ids = ''; //you need to make a comma separated list of the aro_ids that are going to be checked
foreach($data['aro_id'] as $k => $v)
{
$ids .= $v . ', ';
}
if($ids != '')
{
$ids = rtrim($ids, ', ');
}
//you only need to have these two lines in if you have not already attached the behaviours in the ArosHandover model file
$this->ArosHandover->Behaviors->attach('Containable', array('autoFields' => false));
$this->ArosHandover->Behaviors->attach('Search.Searchable');
$query = $this->ArosHandover->getQuery('all',
array(
'conditions' => array('ArosHandover.aro_id IN (' . $ids . ')'),
'fields' => array('handover_id'), //the other field that you need to check against, it's the other side of the many-to-many relationship
'contain' => false //place this in if you just want to have the ArosHandover table data included
)
);
return $query;
}
In the Handovers controller:
public $components = array('Search.Prg', 'Paginator'); //you can also place this into AppController
public $presetVars = true; //using $filterArgs in the model configuration
public $paginate = array(); //declare this so that you can change it
// this is the snippet of the search form processing
public function admin_find()
{
$this->set('title_for_layout','Find handovers');
$this->Prg->commonProcess();
if(isset($this->passedArgs) && !empty($this->passedArgs))
{//the following line passes the conditions into the Paginator component
$this->Paginator->settings = array('conditions' => $this->Handover->parseCriteria($this->passedArgs));
$handovers = $this->Paginator->paginate(); // this gets the data
$this->set('handovers', $handovers); // this passes it to the template
If you want any further explanation as to why I have done something, ask and if I get an email to tell me that you have asked I will give an answer if I am able to.
This is not an issue of the plugin but how you build the associations. You need to properly join them for a search across these three tables. Check how CakePHP is fetching the data from HABTM assocs by default.
http://book.cakephp.org/2.0/en/models/associations-linking-models-together.html#joining-tables
Suppose a Book hasAndBelongsToMany Tag association. This relation uses
a books_tags table as join table, so you need to join the books table
to the books_tags table, and this with the tags table:
$options['joins'] = array(
array('table' => 'books_tags',
'alias' => 'BooksTag',
'type' => 'inner',
'conditions' => array(
'Books.id = BooksTag.books_id'
)
),
array('table' => 'tags',
'alias' => 'Tag',
'type' => 'inner',
'conditions' => array(
'BooksTag.tag_id = Tag.id'
)
)
);
$options['conditions'] = array(
'Tag.tag' => 'Novel'
);
$books = $Book->find('all', $options); Using joins allows you to have
a maximum flexibility in how CakePHP handles associations and fetch
the data, however in most cases you can use other tools to achieve the
same results such as correctly defining associations, binding models
on the fly and using the Containable behavior. This feature should be
used with care because it could lead, in a few cases, into bad formed
SQL queries if combined with any of the former techniques described
for associating models.
Also your code is wrong somewhere.
Column not found: 1054 Unknown column 'contain' in 'where clause'
This means that $Model->contain() is somehow called. I don't see such a call in your code pasted here so it must be somewhere else. If a model method can not be found this error usually happens with the field name as column.
I want to share with everyone that the solution to working with HABTM searches with the plugin lies here: Using the CakeDC search plugin with associated models
#burzum, the documentation is far from ok man. Do you notice the use of 'type' => 'checkbox' and that it is not mentioned anywhere that it is a type?
Not to mention the total lack of grammar and the lots of typos and missing prepositions. I lost 2 days only to get a grasp of what the author had in mind and bind the words in there. No comment on that.
I am glad that after 5 days on the uphill work I made it. Thanks anyway for being helpful.

CakePHP returning multiple records from a habtm

I have a habtm relation between questions and categories. This habtm relation is standard and setup in the respective models. I now have the following issue. In a controller action I have an array of several question ids. I want to return all of the categories that are linked to these questions. The find I am using looks as followed.
$case_questions_categories = $this->Category->find('all', array('conditions' => array('OR' => $case_question_id_array)));
// Here the $case_question_id_array is generated as followed:
foreach($question['CaseQuestion'] as $case_question) {
$case_question_id_array[] = "Category.question_id = " . $case_question['id'];
}
Its clear that this will not work, and it throws the following error:
Error: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'Category.question_id' in 'where clause'
SQL Query: SELECT `Category`.`id`, `Category`.`name`, `Category`.`slug`, `Category`.`category_order`, `Category`.`created`, `Category`.`modified`, `Category`.`is_deleted` FROM `streetofwalls`.`categories` AS `Category` WHERE `Category`.`question_id` = 182
This makes sense because the join table from the habtm relation is not referenced in the query I have generated. How do I get habtm relation to be properly handled so that I can pass a group of question ids and be returned the relevant categories.
Thank you to anyone who offers time or help.
Use a join
The kind of sql query you need to generate is:
SELECT
categories.* from categories
JOIN
categories_questions ON (categories_questions.category_id = categories.id)
WHERE
categories_questions.question_id IN (list, of, question, ids)
The simplest way to do that is to just use the join key in your query - this is also in the documentation.
Applied to the example in the question that'd be:
$ids = array()
foreach($question['CaseQuestion'] as $q) {
$ids[] = $q['id'];
}
$this->Category->find('all', array(
'conditions' => array(
'CategoriesQuestion.id' => $ids // You do not need an OR
),
'joins' => array(
array(
'table' => 'categories_questions',
'alias' => 'CategoriesQuestion',
'type' => 'inner',
'conditions' => array(
'Category.id = CategoriesQuestion.category_id'
)
)
)
));

Query on Multiple Associations with CakePHP Models

I'm very new to CakePHP (started last night) and I'm having an issue with queries.
The way my model works is this:
Classgroups << hasAndBelongsTo >> Users << hasAndBelongsTo >> PermissionsGroups
Basically a Classgroup is made up of a load of users and each user belongs to a permission group. The group defines what type of user they are as well as their permissions.
Now I need to make a query where I get all Users that belong to a certain PermissionGroup but I'm doing it from the ClassgroupsController.
However whenever I try to do this:
$this->Classgroup->User->find('list', array(
'fields' => array('User.surname_firstname'),
'order' => array(
'User.surname_firstname'
),
'conditions' => array(
'PermissionGroup.permissions' => '10'
)
)
I get this error:
Column not found: 1054 Unknown column 'PermissionGroup.permissions' in 'where clause'
Because I'm in the ClassgroupController it's as if the Group fields can't be seen but I would have assumed because User is associated with PermissionGroup it should be ok. I have set up the associations between the 3 models but the problem seems to be that the query isn't looking "deep" enough into the model.
I've also tried setting the recursive value on the find to 2,3 and 4. From what I've read I think this probably isn't a great idea and Containable is better but surely it should at least work?
I should also point out that these associations are one way. I've only added them to the Models I needed. So Classgroups has an association to User and User has an association to PermissionGroup
Am I doing something obviously wrong? Or have I misunderstood a CakePHP concept?
Thanks
UPDATE: For the second time I've seen someone say I should use joins, like so:
'joins'=>array(
array(
'table'=>'rooms',
'alias'=>'Room',
'type'=>'inner',
'conditions'=>array(
'Room.hotel_id'=>'Hotel.id'
)
)
But I have the feeling that's an outdated way to do it... I've also tried using contain like so:
'contain' => array(
'Group' => array(
'conditions' => array(
'Group.name' => '10'
)
)
)
But it didn't work, I still got all of the Users.
You were on the right track with joins:
$joins = array(
array(
'table'=>'permission_groups',
'alias'=>'PemissionGroup',
'type'=>'inner',
'conditions'=>array(
'User.id = PermissionGroup.user_id',
'PermissionGroup.permissions = 10'
)
)
$this->Classgroup->User->find('list', array(
'fields' => array('User.surname_firstname'),
'joins' => $joins,
'order' => array(
'User.surname_firstname'
)
)
Try to work with that!

CakePHP order query results on more than 1 level

I'm using the Containable behavior to get a list of Comments (belongsTo Post, which belongs to Question; Question hasMany Post, and Post hasMany Comments; all of these belong to Users).
$data = $this->Question->find ( 'first',
array ('contain' =>
array ('User',
'Post' => array ('User', /* 'order' => 'User.created DESC'*/ )
)
)
);
It works, when I comment out the section in comments above. I suppose this is to be expected, but what I want is all of the Posts that are found, should be sorted in order of the 'created' field of the 'User' they belong to. How do I accomplish this deeper level sorting in CakePHP? I always get, "Warning (512): SQL Error: 1054: Unknown column 'User.created' in 'order clause'"
Thanks for your help!
Also, you might be trying to group on a related table from a find call that doesn't use joins.
Set your debug level to something greater than 1 so you can see the query log and make sure that Cake isn't doing two queries to fetch your data. If that is the case then the first query is not actually referencing the second table.
If you want to manually force a join in these situations you can use the Ad-Hoc joins method outlined by Nate at the following link.
http://bakery.cakephp.org/articles/view/quick-tip-doing-ad-hoc-joins-in-model-find
I have found two ways to get around this.
The first is to define the second level associacion directly in the model.
Now you will have access to this data everywhere.
It should look something like this.....
var $belongsTo = array(
'Foo' => array(
'className' => 'Foo', //unique name of 1st level join ( Model Name )
'foreignKey' => 'foo_id', //key to use for join
'conditions' => '',
'fields' => '',
'order' => ''
),
'Bar' => array(
'className' => 'Bar', //name of 2nd level join ( Model Name )
'foreignKey' => false,
'conditions' => array(
'Bar.id = Foo.bar_id' //id of 2nd lvl table = associated column in 1st level join
),
'fields' => '',
'order' => ''
)
);
The problem with this method is that it could make general queries more complex than they need be.
You can thus also add the second level queries directly into te find or paginate statement as follows: (Note: I found that for some reason you can't use the $belongsTo associations in the second level joins and will need to redefine them if they are already defined. eg if 'Foo' is already defined in $belongsTo, you need to create a duplicate 'Foo1' to make the association work, like the example below.)
$options['joins'] = array(
array('table' => 'foos',
'alias' => 'Foo1',
'type' => 'inner',
'conditions' => array(
'CurrentModel.foo_id = Foo1.id'
)
),
array('table' => 'bars',
'alias' => 'Bar',
'type' => 'inner',
'foreignKey' => false,
'conditions' => array(
'Bar.id = Foo1.bar_id'
)
)
);
$options['conditions'] = array('Bar.column' => "value");
$this->paginate = $options;
$[modelname] = $this->paginate();
$this->set(compact('[modelname]'));
I hope this is clear enough to understand and that it helps someone.
Check your recursive value. If it's too limiting, it will ignore the containable links, IIRC. I remember bumping into this a few times. I'd try containing multiple models, but my recursive option was set to 0 and nothing would get pulled. For your example, I'd think that a value of 1 (the default) would suffice, but maybe you've explicitly set it to 0 somewhere?
You can add before your call to find() the following:
$this->Question->order = 'Question.created DESC';
Yeah, I couldn't work out how to sort based on the related/associated model, so ended up using the Set::sort() method. Checkout this article for a good explanation.
// This finds all FAQ articles sorted by:
// Category.sortorder, then Category.id, then Faq.displaying_order
$faqs = $this->Faq->find('all', array('order' => 'displaying_order'));
$faqs = Set::sort($faqs, '{n}.Category.id', 'ASC');
$faqs = Set::sort($faqs, '{n}.Category.sortorder', 'ASC');
...And yes, it should probably be a Category->find() but unfortunately the original developer didn't code it that way, and I didn't wanna rework the views.

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