I have a model A that has a relationship of type HAS_MANY with model B.
B's attributes are:
id,
user_id,
message,
date,
parent_message_id
I need elements of model B to be ordered by date (descending), but in case the parent_message_id is different from null, the date to be taken into consideration should be the date corresponding to parent_message_id.
Is it possible to customize the criteria used to order the relation?
Ok, i solved this the following way: model A HAS_MANY model B, therefore, i redefined the relationships method to the following:
public function relations()
{
return array(
'messages' => array(self::HAS_MANY, 'WallMessages', 'liga_id',
'condition'=>'specific_post.parent_message_id IS NULL',
'order'=>'specific_post.date DESC',
'alias'=>'specific_post'),
);
}
Therefore, I only compare the date of those messages with no parent id. The downside is that I have to access each post's "child messages"... but well, couldn't find another workaround. Thanks all for your help!
I think it is only possible to sort elements according to the same column. However there might be a database ninja that could help you.
In an perfect world you could use the afterFind() method for any customization at any level, not only sorting, but also changing some values for others. It is not as fast as a plan order, maybe, but this way you release the MySQL server from a little load. Plus it is called automatically. Cheers
Related
In a legacy project we had issues where if a developer would forget a project_id in the query condition, rows for all projects would be shown - instead of the single project they are meant to see. For example for "Comments":
comments [id, project_id, message ]
If you forget to filter by project_id you would see all projects. This is caught by tests, sometimes not, but I would rather do a prevention - the dev should see straightaway "WRONG/Empty"!
To get around this, the product manager is insisting on separate tables for comments, like this:
project1_comments [id,message]
project2_comments [id,message]
Here if you forgot the project/table name, if something were to still pass tests and got deployed, you would get nothing or an error.
However the difficulty is then with associated tables. Example "Files" linked to "Comments":
files [ id, comment_id, path ]
3, 1, files/foo/bar
project1_comments
id | message
1 | Hello World
project2_comments
id | message
1 | Bye World
This then turns into a database per project, which seems overkill.
Another possibility, how to add a Behaviour on the Comments model to ensure any find/select query does include the foreign key, eg - project_id?
Many thanks in advance.
In a legacy project we had issues where if a developer would forget a project_id in the query condition
CakePHP generates the join conditions based upon associations you define for the tables. They are automatic when you use contains and it's unlikely a developer would make such a mistake with CakePHP.
To get around this, the product manager is insisting on separate tables for comments, like this:
Don't do it. Seems like a really bad idea to me.
Another possibility, how to add a Behaviour on the Comments model to ensure any find/select query does include the foreign key, eg - project_id?
The easiest solution is to just forbid all direct queries on the Comments table.
class Comments extends Table {
public function find($type = 'all', $options = [])
{
throw new \Cake\Network\Exception\ForbiddenException('Comments can not be used directly');
}
}
Afterwards only Comments read via an association will be allowed (associations always have valid join conditions), but think twice before doing this as I don't see any benefits in such a restriction.
You can't easily restrict direct queries on Comments to only those that contain a product_id in the where clause. The problem is that where clauses are an expression tree, and you'd have to traverse the tree and check all different kinds of expressions. It's a pain.
What I would do is restrict Comments so that product_id has to be passed as an option to the finder.
$records = $Comments->find('all', ['product_id'=>$product_id])->all();
What the above does is pass $product_id as an option to the default findAll method of the table. We can than override that methods and force product_id as a required option for all direct comment queries.
public function findAll(Query $query, array $options)
{
$product_id = Hash::get($options, 'product_id');
if (!$product_id) {
throw new ForbiddenException('product_id is required');
}
return $query->where(['product_id' => $product_id]);
}
I don't see an easy way to do the above via a behavior, because the where clause contains only expressions by the time the behavior is executed.
In my project I have the following tables: Messages, Recipients, Groups and Users. A Message has many Recipients, and a Recipient has one Group and one User.
In my RecipientsTable::beforeFind I have some code to automatically contain Groups and Users for Recipient finds, since I always need to access those associations.
public function beforeFind($event, $query, $options, $primary) {
return $query->contain([
'Groups',
'Users',
]);
}
I don't know if this is a bad design decision but it has worked for me so far.
The problem has come now that I'm trying to filter messages by group, and I tried doing so by using the matching function:
$possible_groups = [1,2,3]; //just an example
$query->matching('Recipients', function($q) use ($possible_groups){
return $q->where(['Recipients.group_id IN' => $possible_groups]);
});
When I execute the query, I get the following error:
Messages is not associated with Groups
Is there any way to keep my beforeFind like that and be able to use matching? Or, is there a better way to automatically load associations without using beforeFind?
TL;DR: A hasMany B, B hasOne C. If a query on table A uses matching on table B and table B's beforeFind uses contain to load C, C ends up contained onto the original query (of A) and the execution fails since A is not associated with C.
Use table classes to load Table associations automatically. Like using hasMany(), hasOne(), belongsTo(), belongsToMany() in necessary table classes.
Official documentation:
https://book.cakephp.org/3.0/en/orm/associations.html
Table files should be in the Table folder inside the Model folder like src\Model\Table.
I have a problem retrieving values of a column from relations in Laravel.
I have a User - Model. This model has relation to a table btw. a model named Userhobbies.
For now we have:
User ::: hasMany >>> Userhobbies
Now with User::all()->load('hobbies') I'm getting right results like
{"id":"1","username":"jdoe","first_name":"Joe","last_name":"Doe","birth":"
1992-04-11","picture_id":"f3dca65323e876026b409b9ba3d49c56","hobbies":
[{"hobby_id":"1","user_id":"1"},{"hobby_id":"2","user_id":"1"},
{"hobby_id":"3","user_id":"1"},{"hobby_id":"4","user_id":"1"}]}
As you can see Userhobbies contains only primary-key relations between hobby - table (Hobby Model) and user - table (User Model).
(Hobby model also has hasMany relation to Userhobbies)
My question now is - how to retrieve all hobby-names (from hobby - table) in my call over (after load('hobbies') ) and is it possible without writting a lot of code?
For better understanding of my idea the result which I want to retrieve:
{"id":"1","username":"jdoe","first_name":"Joe","last_name":"Doe","birth":"
1992-04-11","picture_id":"f3dca65323e876026b409b9ba3d49c56","hobbies":
["golf", "cards", "games", "football"]}
EDIT:
If I try following (I tried with belongsToMany in User and Hobby):
User::with('hobbies')->get()->first()
And I'm getting the whole values from the hobbies - table:
{user-specific data ...
hobbies:[{"id":"1","name":"golf","created_at":"2015-04-07
14:15:02","updated_at":"2015-04-07 14:15:02","pivot":
{"user_id":"1","hobby_id":"1"}},
{"id":"2","name":"cards","created_at":"2015-04-07
14:15:02","updated_at":"2015-04-07 14:15:02","pivot":
{"user_id":"1","hobby_id":"2"}},
{"id":"3","name":"games","created_at":"2015-04-07
14:15:02","updated_at":"2015-04-07 14:15:02","pivot":
{"user_id":"1","hobby_id":"3"}},
{"id":"4","name":"football","created_at":"2015-04-07
14:15:02","updated_at":"2015-04-07 14:15:02","pivot":
{"user_id":"1","hobby_id":"4"}}]}
Same try with ->load('hobbies'). I really don't know how to go on.
To explain it a bit more what I need one could imagine such query as follows:
User::all(['id', 'name'])->load(array('hobbies.id','hobbies.name'))->get();
From my knowledge, I know that it's possible to use a closure to set constraints on the query that performs the load, like so:
User::all()->load(['hobbies' => function($query)
{
$query->select('id', 'name');
}]);
By doing it, when you cast it to array, it will produce a result near to what you want. You can even add 'pivot' to your $hidden property on your Hobby model to hide this information.
I'm working on an app with extensive database relationships and I need to return a specific data object. I'm having trouble with it. Here's what I need it to look like:
AccommodationGroup =>
array(
['name']=>'Group1',
['AccommodationRoom']=> array(
[0]=> array(
['id']=>1,
['RoomIdentifier']=>'Cabin001',
),
[1]=> array(
['id']=>2,
['RoomIdentifier']=>'Cabin002'
)
)
)
AccommodationRoom is related to camp by camp_id. AccommodationGroup is related to AccommodationRoom by accommodation_group_id.
As you can see in the data object example, I need to retrieve all the groups for a particular camp with the rooms of each group as a nested array of the group, and all this restricted to a particular camp.
I've tried to get at it by selecting all the applicable groups using findAllByCampId() which, of course, doesn't work (knew it wouldn't but tried it anyway). Another table, AccommodationRoomsCamp is a transitional table between AccommodationRooms and Camps. One way to do it would be:
$this->AccommodationRoomsCamp->findAllByCampId($id, array('group'=>'AccommodationRoomsCamp.accommodation_group_id'))
but accommodation_group_id is not stored in AccommodationRoomsCamp because it is already being stored in the AccommodationRoom table. I think that I need to do more than one operation, but I'm baffled. Little bit of a newb. Ideas?
Have a little bit of a hard time following how your tables are related - but I think you can solve your problem using recursive finds. Something like this in your controller:
$this->AccomodationRoomsCamp->recursive = 2;
$this->AccomodationRoomsCamp->find('all, array('conditions' => array(bla bla bla bla
See http://book.cakephp.org/view/1063/recursive
I have a n...n structure for two tables, makes and models. So far no problem.
In a third table (products) like:
id
make_id
model_id
...
My problem is creating a view for products of one specifi make inside my ProductsController containing just that's make models:
I thought this could work:
var $uses = array('Make', 'Model');
$this->Make->id = 5; // My Make
$this->Make->find(); // Returns only the make I want with it's Models (HABTM)
$this->Model->find('list'); // Returns ALL models
$this->Make->Model->find('list'); // Returns ALL models
So, If I want to use the list to pass to my view to create radio buttons I will have to do a foreach() in my make array to find all models titles and create a new array and send to the view via $this->set().
$makeArray = $this->Make->find();
foreach ($makeArray['Model'] as $model) {
$modelList[] = $model['title'];
}
$this->set('models', $models)
Is there any easier way to get that list without stressing the make Array. It will be a commom task to develops such scenarios in my application(s).
Thanks in advance for any hint!
Here's my hint: Try getting your query written in regular SQL before trying to reconstruct using the Cake library. In essence you're doing a lot of extra work that the DB can do for you.
Your approach (just for show - not good SQL):
SELECT * FROM makes, models, products WHERE make_id = 5
You're not taking into consideration the relationships (unless Cake auto-magically understands the relationships of the tables)
You're probably looking for something that joins these things together:
SELECT models.title FROM models
INNER JOIN products
ON products.model_id = models.model_id
AND products.make_id = 5
Hopefully this is a nudge in the right direction?
Judging from your comment, what you're asking for is how to get results from a certain model, where the condition is in a HABTM related model. I.e. something you'd usually do with a JOIN statement in raw SQL.
Currently that's one of the few weak points of Cake. There are different strategies to deal with that.
Have the related model B return all ids of possible candidates for Model A, then do a second query on Model A. I.e.:
$this->ModelB->find('first', array('conditions' => array('field' => $condition)));
array(
['ModelB'] => array( ... ),
['ModelA'] => array(
[0] => array(
'id' => 1
)
)
Now you have an array of all ids of ModelA that belong to ModelB that matches your conditions, which you can easily extract using Set::extract(). Basically the equivalent of SELECT model_a.id FROM model_b JOIN model_a WHERE model_b.field = xxx. Next you look for ModelA:
$this->ModelA->find('all', array('conditions' => array('id' => $model_a_ids)));
That will produce SELECT model_a.* FROM model_a WHERE id IN (1, 2, 3), which is a roundabout way of doing the JOIN statement. If you need conditions on more than one related model, repeat until you have all the ids for ModelA, SQL will use the intersection of all ids (WHERE id IN (1, 2, 3) AND id IN (3, 4, 5)).
If you only need one condition on ModelB but want to retrieve ModelA, just search for ModelB. Cake will automatically retrieve related ModelAs for you (see above). You might need to Set::extract() them again, but that might already be sufficient.
You can use the above method and combine it with the Containable behaviour to get more control over the results.
If all else fails or the above methods simply produce too much overhead, you can still write your own raw SQL with $this->Model->query(). If you stick to the Cake SQL standards (naming tables correctly with FROM model_as AS ModelA) Cake will still post-process your results correctly.
Hope this sends you in the right direction.
All your different Make->find() and Model->find() calls are completely independent of each other. Even Make->Model->find() is the same as Model->find(), Cake does not in any way remember or take into account what you have already found in other models. What you're looking for is something like:
$this->Product->find('all', array('conditions' => array('make_id' => 5)));
Check out the Set::extract() method for getting a list of model titles from the results of $this->Make->find()
The solution can be achieved with the use of the with operation in habtm array on the model.
Using with you can define the "middle" table like:
$habtm = " ...
'with' => 'MakeModel',
... ";
And internally, in the Model or Controller, you can issue conditions to the find method.
See: http://www.cricava.com/blogs/index.php?blog=6&title=modelizing_habtm_join_tables_in_cakephp_&more=1&c=1&tb=1&pb=1