In a cakePHP application I am building, a profile can have multiple locations; the tables are called "profiles" and "locations" and in the model classes I have defined a HasMany relationship. Now I want the user to be able to search profiles based on their locations. After reading some questions here and the CakePHP Cookbook, I have decided I need to use SQL joins (in reality more tables are involved, and the result of a search should be based on conditions concerning different tables).
I have written the following function inside my Profile model:
public function findProfiles($long, $lat){
$options['joins'] = array(
array('table' => 'locations',
'alias' => 'Location',
'type' => 'Inner',
'conditions' => array('Location.profile_id = Profile.id'))
);
$options['order'] = array('Location.lng ASC'); //this is just as trial
return $this->find('all',$options);
}
The code works, but I get a copy of a profile for each location it possesses. That is, if a profile possesses 5 positions, I get five instances of that profile (each instance containing all five positions!)
How can I achieve this?
[edit]
eg. let's assume I only have one profile, with two positions. I get:
result[0][Profile]
[Position][0]
[1]
[1][Profile]
[Position][0]
[1]
Where the data in result[0] and result[1] is identical.
The problem happens because of the type of join used. With inner join you'll get this return with your query
profile_id location_id
---------------------------
1 2
1 3
And cake understands that as two records of Profile, so you get repeated Profiles with the same info.
If this were all the extent of your problem, I'd say "go with Containable behaviour and forget joins", but since you said there are more tables involved, maybe the type of join can't be changed. So to get the unique Profile without repetitions, you'll have to GROUP BY the query to get
profile_id location_id
---------------------------
1 2 & 3
with a code similar to this
$options['joins'] = array(
array('table' => 'locations',
'alias' => 'Location',
'type' => 'Inner',
'conditions' => array('Location.profile_id = Profile.id')),
'group' => 'Profile.id'
);
and you'll get rid of repetitions. For future problems like this, is best to first check the actual query that gets send to the DB, check yourself if the result that the DB gives you is what you want, and if not, see what you can do in cake to change it.
Related
I have this database structure:
I need the user to get only those reservations in ReservationsSeachModel that are in the warehouse belonging to the user. Model user_has_warehouse has a function getUserHasWarehouse($user_id, $warehouse_id) to check for relations. I am able to get the relations when viewing single, deleting and updating records in controllers. But I cannot make the SearchModel to return reservations that belong to current user.
I cannot depend on Reservation->user_id, because that value is not mandatory and reservations can be inserted publicly with no user_id. How to make the SearchModel work properly so it only shows reservations belonging to the warehouse where user has the warehouse?
ReservationsSearchModel:
$warehouse_id = User_Has_Warehouse::find()->select(['warehouse_id'])->where(['user_id' => Yii::$app->user->id]);
$freezer_type_id = Freezer_type::find()->select(['id'])->where(['warehouse_id' => $warehouse_id]);
$reservation_id = Reservation_freezer_type::find()->select(['reservation_id'])->where(['freezer_type_id'=>$freezer_type_id]);
...
$query->andFilterWhere([
'id' => $reservation_id,
'customer_id' => $this->customer_id,
'user_id' => $this->user_id,
]);
...
But the following did not work for me. What am I doing wrong?
Managed to solve it.
$h = User_Has_Warehouse::find()->select(['warehouse_id'])->where(['user_id' => Yii::$app->user->id]);
$rt = Freezer_type::find()->select(['id'])->where(['warehouse_id' => $h]);
$res = Reservation_freezer_type::find()->select(['reservation_id'])->where(['freezer_type_id' => $rt]);
// grid filtering conditions
$query->andFilterWhere(['id' => $res]);
Already received a great answer at this post
Laravel Query using GroupBy with distinct traits
But how can I modify it to include more than just one field. The example uses pluck which can only grab one field.
I have tried to do something like this to add multiple fields to the view as such...
$hats = $hatData->groupBy('style')
->map(function ($item){
return ['colors' => $item->color, 'price' => $item->price,'itemNumber'=>$item->itemNumber];
});
In my initial query for "hatData" I can see the fields are all there but yet I get an error saying that 'colors', (etc.) is not available on this collection instance. I can see the collection looks different than what is obtained from pluck, so it looks like when I need more fields and cant use pluck I have to format the map differently but cant see how. Can anyone explain how I can request multiple fields as well as output them on the view rather than just one field as in the original question? Thanks!
When you use groupBy() of Laravel Illuminate\Support\Collection it gives you a deeper nested arrays/objects, so that you need to do more than one map on the result in order to unveil the real models (or arrays).
I will demo this with an example of a nested collection:
$collect = collect([
collect([
'name' => 'abc',
'age' => 1
]),collect([
'name' => 'cde',
'age' => 5
]),collect([
'name' => 'abcde',
'age' => 2
]),collect([
'name' => 'cde',
'age' => 7
]),
]);
$group = $collect->groupBy('name')->values();
$result = $group->map(function($items, $key){
// here we have uncovered the first level of the group
// $key is the group names which is the key to each group
return $items->map(function ($item){
//This second level opens EACH group (or array) in my case:
return $item['age'];
});
});
The summary is that, you need another loop map(), each() over the main grouped collection.
I have a model named Application. And Application is associated to has_many model named Location.
Application has many Location
In my Application query:
$this->Application->find('all', array('conditions' => 'Application.status' => 'accepted'));
I'm finding applications where status is accepted.
Next thing that I would like to achieve is to find Application records where associated Location is empty/null or in other words where count of Location records is 0.
I tried to make join query like this:
$join_query = array(
'table' => 'locations',
'alias' => 'Location',
'type' => 'INNER',
'conditions' => array(
'Location.application_id = Application.id',
'OR' => array(
array('Location.id' => NULL)
)
)
);
But seems like it's just querying Application records that do have associated Location records.
Thanks in advanced if you guys have any idea(s).
You need to use a left join, not an inner join. Inner join will get only those results that have a row in both of the tables you are joining, where you want only results where there is only a row in the left table. Left joins will get all the results in the left table, regardless if there's a row associated with it in the right table. Then add a condition after the join is complete, to only select those joined results where Location.id is null.
$this->Application->find('all',
array(
'conditions' => array('Location.id' => null),
'joins' => array(
array(
'table' => 'locations',
'alias' => 'Location',
'type' => 'LEFT',
'conditions' => array('Location.application_id = Application.id')
),
),
)
);
Your query says "find any application and its location with application_id = id, AND (1 OR where location.id = null)", so that will match any application that has location.
What I'd do is to leave joins and just use containable and counts. With plain sql I'd use a left join and count the Locations, like in this example. But cake doesn't behave well with not named columns, like "COUNT(*) AS num_locations", so I tend to avoid that.
I'd transform your query to a containtable one
$apps = this->Application->find('all', array('contains'=>'Location'));
foreach($apps as $app) {
if (count($app['Location']) <= 0)
//delete record
}
You could also implement a counterCache, and keep in a BD column the number of locations per application, so the query can be a simple find like
$this->Application->find('all', array('conditions'=>array('location_count'=>0)));
Ooooor, you could add a virtual field with "SUM(*) as num_locations" and then use your join with "left outter join" and compare "num_locations = 0" on the conditions.
Those are the options that comes to mind. Personally I'd use the first one if the query will be a one time/not very used one. Probably put it in the Application model like
public function findAppsWithNoLocations() {
$apps = this->Application->find('all', array('contains'=>'Location'));
foreach($apps as $app) {
if (count($app['Location']) <= 0)
//delete record
}
}
But the other two options would be better if the sum of locations per app is going to be a recurrent query you'll search for.
EDIT
And of course Kai's answer options that does what you want xD. This tendency to complicate things will be the end of me... Well, will leave the answer here to show a reference to other convoluted options (specifically counterCache if you'll need to count the relations a lot of times).
i know this is already some time ago.
i could manage it this way:
public function getEmpty($assoc) {
foreach($this->find('all') as $c){
if(empty($c[$assoc])) $return[] = $c;
}
return $return;
}
now i got all entries that have an empty associated data.
in my controller i call the function like this:
$ce = $this->Company->getEmpty('CompaniesUsers');
companies Users is the Empty Associated model i want to check.
I'm not savvy with MySQL or databases generally, so here's a model of my data (table{cols}]) in order to make my question coherent:
Domains{id, name} Note: 'domains' here does not refer to web domains
Subdomains{id, domain_id, name}
Items{id, subdomain_id, name}
SubdomainsItems{id, subdomain_id, item_id} no domain_id column!
My Items Controller has a function, fetchWithin($domains, $subdomains) which, ultimately, should just execute one of two complexish find(). It's the complexish I can't get past.
Programmatically I can achieve this, but I'm quite certain the better way is by clever joins and the like. Alas, currently this is approach:
If $domainsis empty, do only steps 2&3, otherwise:
foreach($domains as $d): get all the rows of Subdomains where Subdomain.domain_id = Domains.id as $subdomains
foreach($subdomains as $s) : go get all the rows of SubdomainsItems where SubdomainsItems.subdomain_id = Subdomains.id as $item_ids
foreach($items_ids as $i): get all the rows of Items where Items.id = SubdomainsItems.items_id
This works, but I think this is obviating the power of a relational database and I'd like to understand how this should be done (ie. according to either Cakephp convention or simply by whatever MySQL statement would achieve this).
Help would be hugely appreciated, I try to learn the more complex aspects of SQL but it just goes right over my head. :S
Understanding the necessary query
With the structure described in the question the kind of query necessary is of the form:
SELECT
*
FROM
items
LEFT JOIN
subdomains ON (
items.subdomain_id = subdomains.id
)
LEFT JOIN
domains ON (
subdomains.domain_id = domains.id
)
WHERE
domains.name = "foo"
AND
subdomains.name IN ('some', 'list', 'of', 'subdomains');
Compared to the logic in the question this joins all three tables together and permits finding all items by domain name, or subdomain name (or any other criteria involving any or all three tables); Generally speaking if you want to find data in a db and use more than one query to get it - there's a more efficient way to do it.
Implementing the find call
There are a number of ways of creating such a query with Cake. The simplest, probably, is to use the join key and just specify the joins explicitly:
function fetchWithin($domains = null, $subdomains = null) {
$params = array(
'joins' => array(
array('table' => 'subdomains',
'alias' => 'Subdomain',
'type' => 'LEFT',
'conditions' => array(
'Subdomain.id = Item.subdomain_id',
)
),
array('table' => 'domains',
'alias' => 'Domain',
'type' => 'LEFT',
'conditions' => array(
'Domain.id = Subdomain.domain_id',
)
)
)
);
if ($domains) { // single value or an array
$params['conditions']['Domain.name'] = $domains;
}
if ($subdomains) { // single value or an array
$params['conditions']['Subdomain.name'] = $subdomains;
}
return $this->find('all', $params);
}
I have three models linked in this manner: Item->Order->Payment
Order hasMany Item
Order hasOne Payment
Now, I am paginating Items and want to add a condition in it to find only items of that order which has payment of a particular id. I hope that makes sense :P
I added the condition as:
array('Payment.id'=>$id)
but it doesn't work. Obviously cause Payment is not associated with Item.
So, how can I go about this?
I am new to cakephp, maybe I am completily wrong but as I understand it you can use other models in your controller with the $uses variable. First make a query on payment model to get your order id, than you can use this id to find the corresponding items.
$uses=array('Item','Order','Payment');
$order_id=$this->Payment->find('first',array('fields'=>'order_id','conditions'=>array('id'=>$payment_id)));
$items=$this->Item->find('all',array('conditions'=>array('order_id'=>$order_id)));
I hope it help.
Why don't you add a condition:
array('Order.payment_id'=>$id)
I think this should work.
If you specify that you want two levels of recursion this should work. Im assuming you have
in Payment.php
//recursion level 1
var $belongsTo = array('Order');
in Order.php
//recursion level 2
var $hasMany = array('Items')
You are right that for paginate to work you must query the model you wish to page and sort the lists by.
in PaymentController.php
//Query two levels deep, so the $payment['Order']['Item'][0-n] will be present
var $paginate = array('recursive' => 2);
Note this method does generate another query for each row to retrieve items.
Make sure the debug level in app/config/core.php is set to 2 to see the database calls.
1) You can use Containable behaviour, in which case you need to put this in your Item model:
var $actsAs = array('Containable');
and this into your Items controller:
$items = $this->Item->find('all'
, array (
'contain' => array('Order' => array('Payment'))
, 'conditions' => array('Payment.id' => $paymentId)
)
)
However I suspect that that will do a left join onto the Payments table (as its a hasMany relationship). So you won't filter Items in any way.
2) If you can't get contains to work then I often use explict joins (read this bakery article by nate on joins) in my find queries. So in your Items controller you'd have:
$items = $this->Item->find('all'
, array (
, 'joins' => array(
array(
'table' => 'payments'
, 'alias' => 'Payment'
, 'type' => 'INNER'
, 'conditions' => array(
'Option.id = Payment.option_id'
)
)
)
, 'conditions' => array('Payment.id' => $paymentId)
)
)
You may also need to specify the join onto the options table.