So I've searched the internet for similar cases, and I just got lost from all contradicting answers, and unrelated scenarios. So I thought to put my case hoping to get some specific answers.
I am new to Laravel, and creating small application. In the application I have to search for offeres and show the result in a blade view. Since the query is complex, and output of the search does not belong to a specific Model, I just wish to keep it as a raw query.
I have placed the query in the controller, but I just don't feel it's the right place. Especially if I need to reuse the query in many places.
Here's the method from the OfferController:
public function search(Request $request)
{
$area = $request->area;
$size = $request->size;
$sql = "SELECT distinct product_name,product_offer.quantity, product_offer.price
FROM product
inner join brand on product.brand_id = brand.brand_id
inner join brand_area on brand_area.brand_id = brand.brand_id
inner join area on area.area_id = brand_area.area_id
inner join product_offer on product_offer.product_id = product.product_id
where area.area_id = :area
and product.size_id = :size ";
$params = array(
'area'=>$area,
'size'=>$size
);
$offers = DB::select( DB::raw($sql), $params);
return view('searchresult')->with($offers);
}
So in short: should I move the query to the model, create DAL class, or keep it here? Baring in mind the project is small scale.
in my opinion, if you are going to reuse it, create a service that will perform that query and gives you back a result, something like SearchService that looks like this:
<?php
class SearchService
{
public static function perform(array $params){
$sql = "SELECT distinct product_name,product_offer.quantity, product_offer.price
FROM product
inner join brand on product.brand_id = brand.brand_id
inner join brand_area on brand_area.brand_id = brand.brand_id
inner join area on area.area_id = brand_area.area_id
inner join product_offer on product_offer.product_id = product.product_id
where area.area_id = :area
and product.size_id = :size ";
return DB::select( DB::raw($sql), $params);
}
}
?>
And by doing so, you can just call
SearchService::perform([...]);
to get the results.
Obviously this is version1.0, you can improve it in a lot of ways, for example making it not static in order to make it testable, and also to allow getter and setter to exists, and a lot of other things, that might be usefull
You have a fair point saying it does not look right to place query in the controller. I would offer you to have a look at the concept of Laravel repository pattern:
https://dev.to/asperbrothers/laravel-repository-pattern-how-to-use-why-it-matters-1g9d
Also, I think you could use Laravel DB for this kind of query without the need to write it as a raw query. I do not think there is a huge need to have a raw query here. Laravel DB table(), select(), where() and other methods should be enough.
Actually, you could potentially write this using models and their relationships, but if the query is quite slow, it is better to use query builder for better efficiency.
EDIT:
also for some specific queries which do not belong to anything I remember seeing custom trait used, could also be a solution.
I'm writing my previous comment as an answer since I think its really what you are looking for:
I suggest you use a Trait to do this. The reason why is that it will keep your code clean and you'll be able to reuse this code later for other uses. A Trait is made for reusable code.
You will create a Trait:
<?php
namespace App\Traits; // Optional, but you could create a namespace for all your Traits if you need others
trait MyCustomTrait
{
public function perform(array $params) {
$sql = "SELECT distinct product_name,product_offer.quantity, product_offer.price
FROM product
inner join brand on product.brand_id = brand.brand_id
inner join brand_area on brand_area.brand_id = brand.brand_id
inner join area on area.area_id = brand_area.area_id
inner join product_offer on product_offer.product_id = product.product_id
where area.area_id = :area
and product.size_id = :size ";
return DB::select( DB::raw($sql), $params);
}
}
To use it, only write: use MyCustomTrait in the scope of your controller and call your function like this: $this->perform([...]).
Here's some links to learn more about traits:
https://www.w3schools.com/php/php_oop_traits.asp
https://www.develodesign.co.uk/news/laravel-traits-what-are-traits-and-how-to-create-a-laravel-trait/.
Hope it helps you!
Related
I'm working with L5 and elequent
My table structure is..
users
id
other field
auctions
id
other field
lots
id
auction_id
other field
lot_user
lot_id
user_id
I want to find auctions for a user.
How can i do this?
$user = User::find($id);
$auctions = $user->auctions();
I have got an idea to do this with eloquent..
$auctions = $user->lots()
->join('auctions','auctions.id','=','lots.id')
->select(['auctions.*'])
->get();
I'm not sure Eloquent is going to be very efficient here, but you could do something like :
In your User(s) class, you need to define a many-many relationship like :
public function lots()
{
return $this->belongsToMany('App\Lot');
}
In your Lot(s) class, you need to define the inverse of a one-to-many relationship like:
public function auctions()
{
return $this->belongsTo('App\Auction')
}
Then, to get Lots for a user, you'd do something like :
$user->lots();
To get auctions, you'd need to loop over lots and call $lot->auctions() for each one, and then filter by id to get the unique auctions.
This is a case where it would probably be easier to use the DB facade to built a query instead of just trying to use Eloquent.
About DB facade.
Raw query will looks like this:
SELECT * FROM auctions AS a
INNER JOIN lots AS l ON (l.auction_id = a.id)
INNER JOIN lot_user AS lu ON (lu.lot_id = l.id AND lu.user_id = $findUserId)
GROUP BY a.id
And using query-builder you can do it like this:
DB::table('auctions')
->join('lots', 'lots.auction_id', '=', 'auctions.id')
->join('lot_user', function ($join) {
$join->on('lot_user.lot_id', '=', 'lots.id')
->where('lot_user.user_id', '=', $findUserId);
})
->groupBy('auctions.id')
->get();
I have been trying to convert a right join query to left join query in order to use it inside laravel query builder. Here is my Sql statement and it result wich works flawlessly
select `weekday`.`name`, `open_time`, `close_time`
from `schedule`
join `restaurants_has_schedule` on `schedule`.`id` = `restaurants_has_schedule`.`schedule_id`
and `restaurants_has_schedule`.`restaurants_id` = 1
right join `weekday` on `weekday`.`id` = `schedule`.`weekday_id`
ORDER BY `weekday`.`id`
|------
|name|open_time|close_time
|------
|Domingo|NULL|NULL
|Lunes|NULL|NULL
|Martes|NULL|NULL
|Miercoles|NULL|NULL
|Jueves|14:11:51|14:11:51
|Vienes|09:11:21|17:00:00
|Sábado|NULL|NULL
but when convert It to left join it stop working, displaying me the same data for every single restaurants_id. This is my left join statement.
select `weekday`.`name`, `open_time`, `close_time`
from `weekday`
left join `schedule` on `weekday`.`id` = `schedule`.`weekday_id`
join `restaurants_has_schedule` on `schedule`.`id` = `restaurants_has_schedule`.`schedule_id`
and `restaurants_has_schedule`.`restaurants_id` = 1
ORDER BY `weekday`.`id`
What am I doing wrong? Is There another alternative? Thak you in advance
Try use Laravels Eloquent ORM, which handles relationships really cool! no need anymore to concat or write sql-queries
See here about: Laravels Eloquent ORM & Schema Builder
Or maybe about orm's in general, you should really give it a try:
Object Role Modeling
Object Role Modeling
Example from Laravel doc:
One Post may have many comments
One to many:
class Post extends Eloquent {
public function comments()
{
return $this->hasMany('Comment');
}
}
Where 'Comment' is the model.
The "reverse" to define:
class Comment extends Eloquent {
public function post()
{
return $this->belongsTo('Post');
}
}
Where 'Post' is the model.
And then as query:
$comments = Post::find(1)->comments; //by Primary Key
Or
$comments = Post::where('somefield', '=', 'somevalue')->comments->get();
....Really cool, for many to many see the docs#Laravels Eloquent ORM
For example, if I run one query in model:
public function list_users() {
$q = "SELECT user_id, username FROM users";
return $q->result_array();
}
And now, to lists posts from that user, I need to refer to it's id within this function:
public function list_posts() {
$q = "SELECT post_id, post_title, post_content FROM posts
WHERE user_id = what??";
return $q->result_array();
}
OK both of these functions are in Model. Now, How to use RESULT from list_users() in list_posts(). Please have in mind that I need to pass ARRAY of IDs and, to use it only for particular id from list_users() which also returns ARRAY
I know I can use joined query, but that's not the point at all, as I have lots of queries that I need to split
why arent you using a JOIN statement, and making two queries into one?
This will reduce db load, decrease query times, and also reduce clutter in your models.
SELECT p.post_id, p.post_title, p.post_content, u.user_id, u.username FROM posts p LEFT JOIN users u ON u.user_id = p.user_id
you can also do this using active records. Which will avoid having to use full blown queries, and more of a CI methodology to SQL.
http://codeigniter.com/user_guide/database/active_record.html
$this->db->select('p.post_id, p.post_title, p.post_content, u.user_id, u.username');
$this->db->from('posts p');
$this->db->join('users u', 'u.user_id = p.user_id');
$q = $this->db->get();
$q->result();
Edit:
You can return the value as an object.. IE: $this->user_id then reference it in the posts function. Ideally you should call the first function in your Controller, return $user_id and then reference that in your next function.. This is definitely not best case though, you should use JOINs as they are less taxing on the db.
//controller
function test(){
$users = $this->exampleModel->list_users();
//manipulate user data if needed
$posts = $this->exampleModel->list_posts($users);
}
Is it possible to make sub-queries in ActiveRecord in Yii?
i have a query like this:
select * from table1
where table1.field1 in (select table2.field2 from table2)
i'm currently using the fallowing code:
object1::model()->findAll(array('condition'=>'t.field1 in (select table2.field2 from table2)'))
[Edit]
i would like to know if there is a manner to construct the sub-query without using SQL, and without using joins.
Is there any solution ?
and thanks in advance.
First find doublets by db fields:
$model=new MyModel('search');
$model->unsetAttributes();
$criteria=new CDbCriteria();
$criteria->select='col1,col2,col3';
$criteria->group = 'col1,col2,col3';
$criteria->having = 'COUNT(col1) > 1 AND COUNT(col2) > 1 AND COUNT(col3) > 1';
Get the subquery:
$subQuery=$model->getCommandBuilder()->createFindCommand($model->getTableSchema(),$criteria)->getText();
Add the subquery condition:
$mainCriteria=new CDbCriteria();
$mainCriteria->condition=' (col1,col2,col3) in ('.$subQuery.') ';
$mainCriteria->order = 'col1,col2,col3';
How to use:
$result = MyModel::model()->findAll($mainCriteria);
Or:
$dataProvider = new CActiveDataProvider('MyModel', array(
'criteria'=>$mainCriteria,
));
Source: http://www.yiiframework.com/wiki/364/using-sub-query-for-doubletts/
No, there is not a way to programmatically construct a subquery using Yii's CDbCriteria and CActiveRecord. It doesn't look like the Query Builder has a way, either.
You can still do subqueries a few different ways, however:
$results = Object1::model()->findAll(array(
'condition'=>'t.field1 in (select table2.field2 from table2)')
);
You can also do a join (which will probably be faster, sub-queries can be slow):
$results = Object1::model()->findAll(array(
'join'=>'JOIN table2 ON t.field1 = table2.field2'
);
You can also do a direct SQL query with findAllBySql:
$results = Object1::model()->findAllBySql('
select * from table1 where table1.field1 in
(select table2.field2 from table2)'
);
You can, however, at least provide a nice AR style interface to these like so:
class MyModel extends CActiveRecord {
public function getResults() {
return Object1::model()->findAll(array(
'condition'=>'t.field1 in (select table2.field2 from table2)')
);
}
}
Called like so:
$model = new MyModel();
$results = $model->results;
One interesting alternative idea would be to create your subquery using the Query Builder's CDbCommand or something, and then just pass the resulting SQL query string into a CDbCritera addInCondition()? Not sure if this will work, but it might:
$sql = Yii::app()->db->createCommand()
->select('*')
->from('tbl_user')
->text;
$criteria->addInCondition('columnName',$sql);
You can always extend the base CDbCriteria class to process and build subqueries somehow as well. Might make a nice extension you could release! :)
I hope that helps!
I know this an old thread but maybe someone (like me) still needs an answer.
There is a small issues related to the previous answers. So, here is my enhancement:
$model=new SomeModel();
$criteria=new CDbCriteria();
$criteria->compare('attribute', $value);
$criteria->addCondition($condition);
// ... etc
$subQuery=$model->getCommandBuilder()->createFindCommand($model->getTableSchema(),$criteria)->getText();
$mainCriteria=new CDbCriteria();
$mainCriteria->addCondition($anotherCondition);
// ... etc
// NOW THIS IS IMPORTANT
$mainCriteria->params = array_merge($criteria->params, $mainCriteria->params);
// Now You can pass the criteria:
$result = OtherModel::model()->findAll($mainCriteria);
I have the following query I want to build using CakePHP. How should I go about this?
SELECT
`Artist`.`id`,
CONCAT_WS(' ', `Person`.`first_name`, `Person`.`last_name`, `Person`.`post_nominal_letters`) AS `name`,
`Portfolio`.`count`
FROM
`people` as `Person`,
`artists` as `Artist`
LEFT OUTER JOIN
(SELECT
`Product`.`artist_id`,
COUNT(DISTINCT `Product`.`id`) AS `count`
FROM
`product_availabilities` AS `ProductAvailability`,
`products` AS `Product`
LEFT OUTER JOIN
`order_details` AS `OrderDetail`
ON
`Product`.`id` = `OrderDetail`.`product_id`
LEFT OUTER JOIN
`orders` AS `Order`
ON
`Order`.`id` = `OrderDetail`.`order_id`
WHERE
`ProductAvailability`.`id` = `Product`.`product_availability_id`
AND
`Product`.`online` = true
AND
(`ProductAvailability`.`name` = 'For sale')
OR
((`ProductAvailability`.`name` = 'Sold') AND (DATEDIFF(now(),`Order`.`order_date`) <= 30))
GROUP BY
`Product`.`artist_id`)
AS
`Portfolio`
ON
`Artist`.`id` = `Portfolio`.`artist_id`
WHERE
`Artist`.`person_id` = `Person`.`id`
AND
`Artist`.`online` = true
GROUP BY
`Artist`.`id`
ORDER BY
`Person`.`last_name`, `Person`.`first_name`;
I think that the model is Artist and It has a belongsTo Relationship with , then you could use the CakePHP ORM on this way and hasMany with Portfolio
first you mus distroy the relation between Artist and Portfolio
$this->Artist->unbindModel(array('hasMany'=>array('Portfolio')));
and then Build the relations
$this->Artist->bindModel(array('hasOne'=>array('Portfolio')));
Finally you must create the other relationsships
$this->Artist->bindModel(array(
'belongsTo'=>array(
'Product'=>array(
'clasName'=>'Product',
'foreignKey'=> false,
'conditions'=>'Product.id = Artist.product_id'
),
'ProductAvaibility'=>array(
'clasName'=>'ProductAvaibility',
'foreignKey'=> false,
'conditions'=>'ProductAvaibility.id = Product.product_avaibility_id'
),
'OrderDetail'=>array(
'clasName'=>'OrderDetail',
'foreignKey'=> false,
'conditions'=>'Product.id = OrderDetail.product_id'
),
'Order'=>array(
'clasName'=>'Order',
'foreignKey'=> false,
'conditions'=>'Order.id = OrderDetail.order_id'
),
)
));
Now, when the relationships are done, you could do your find
$this->Artist->find('all', array(
'conditions'=>array(
'Artist.online'=>true
),
'group'=>array(
'Artist.id'
),
'order'=>array(
'Person.last_name',
'Person.first_name',
)
))
I hope it could be useful for you
You should place it in your model. If you haven't read yet, I suggest you to read about skinny controllers and fat models.
class YourModel extends AppModel {
public function getArtistsAndPortfolioCounts() {
return $this->query("SELECT ... ");
}
}
So in your controller:
class ArtistsControllre extends AppController {
public function yourAction() {
debug($this->YourModel->getArtistsAndPortfolioCounts());
}
}
Do not try to build that query using ORM. It will be a disaster in slow motion.
Instead you should try to do that request as "close to metal" as possible. I am not familiar with CakePHP's API ( as i tend to avoid it like black plague ), but you should be able to create a Model which is not related to ActiveRecord, and most likely have access to anything similar to PDO wrapper. Use that to execute the query and map results to variables.
That said, you might want to examine your query ( you seem to have some compulsive quoting sickness ). One of improvements you could do would be stop using id columns in tables.
This whole setup could really benefit from creation of another column in Artists table : portfolio_size. Which you can update each time it should be changed. Yes , it would de-normalize your table, but it also will make this query trivial and blazing fast, with minor costs elsewhere.
As for names of columns, if table Artists has a id, then keep it in column artist_id , and if Products has a foreign key referring to Artists.artist_id then name it too artist_id. Same thing should have same name all over your database. This would additionally let you use in SQL USING statement. Here is a small example :
SELECT
Users.name
Users.email
FROM Users
LEFT JOIN GroupUsers USING (user_id)
LEFT JOIN Groups USING (group_id)
WHERE Groups.name = 'wheel'
I assume that this query does not need explanation as it is a simple many-to-many relationship between users and groups.
You could use the query method, but that method is treated as a last resort, when find or any of the other convenience methods don't suffice.
But it's also possible to manually specify joins in the Cake find method, see the cookbook. I'm not entirely sure, but I believe nested joins are also supported. CONCAT_WS could just be called, with the relevant fields, in the field property of the find method.