Jak to zrobić w Laravel, subquery gdzie w

Jak mogę zrobić to zapytanie w Laravel: 

SELECT 
    `p`.`id`,
    `p`.`name`, 
    `p`.`img`, 
    `p`.`safe_name`, 
    `p`.`sku`, 
    `p`.`productstatusid` 
FROM `products` p 
WHERE `p`.`id` IN (
    SELECT 
        `product_id` 
    FROM `product_category`
    WHERE `category_id` IN ('223', '15')
)
AND `p`.`active`=1

Mógłbym to zrobić również za pomocą join, ale potrzebuję tego formatu do wydajności.

Author: Donald Duck, 2013-05-29
Source

6 answers

Rozważ ten kod: 

Products::whereIn('id', function($query){
    $query->select('paper_type_id')
    ->from(with(new ProductCategory)->getTable())
    ->whereIn('category_id', ['223', '15'])
    ->where('active', 1);
})->get();
 128
Author: lukaserat,
Warning: date(): Invalid date.timezone value 'Europe/Kyiv', we selected the timezone 'UTC' for now. in /var/www/agent_stack/data/www/doraprojects.net/template/agent.layouts/content.php on line 54
2016-05-24 06:56:45

Spójrz na zaawansowaną dokumentację wheres dla Fluent: http://laravel.com/docs/queries#advanced-wheres

Oto przykład tego, co próbujesz osiągnąć: 

DB::table('users')
    ->whereIn('id', function($query)
    {
        $query->select(DB::raw(1))
              ->from('orders')
              ->whereRaw('orders.user_id = users.id');
    })
    ->get();

To da: 

select * from users where id in (
    select 1 from orders where orders.user_id = users.id
)
 40
Author: drewjoh,
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2014-01-01 07:57:06

Możesz użyć zmiennej używając słowa kluczowego "use ($category_id)" 

$category_id = array('223','15');
Products::whereIn('id', function($query) use ($category_id){
   $query->select('paper_type_id')
     ->from(with(new ProductCategory)->getTable())
     ->whereIn('category_id', $category_id )
     ->where('active', 1);
})->get();
 17
Author: Ramesh,
Warning: date(): Invalid date.timezone value 'Europe/Kyiv', we selected the timezone 'UTC' for now. in /var/www/agent_stack/data/www/doraprojects.net/template/agent.layouts/content.php on line 54
2014-07-29 09:49:32

Następujący kod zadziałał dla mnie: 

$result=DB::table('tablename')
->whereIn('columnName',function ($query) {
                $query->select('columnName2')->from('tableName2')
                ->Where('columnCondition','=','valueRequired');

            })
->get();
 2
Author: Aditya Singh,
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2017-01-19 11:29:41

Laravel 4.2 i więcej, można użyć TRY relationship querying: - 

Products::whereHas('product_category', function($query) {
$query->whereIn('category_id', ['223', '15']);
});

public function product_category() {
return $this->hasMany('product_category', 'product_id');
}
 0
Author: LC Yoong,
Warning: date(): Invalid date.timezone value 'Europe/Kyiv', we selected the timezone 'UTC' for now. in /var/www/agent_stack/data/www/doraprojects.net/template/agent.layouts/content.php on line 54
2017-01-20 20:47:17 
Product::from('products as p')
->join('product_category as pc','p.id','=','pc.product_id')
->select('p.*')
->where('p.active',1)
->whereIn('pc.category_id', ['223', '15'])
->get();
 0
Author: 潘庆强,
Warning: date(): Invalid date.timezone value 'Europe/Kyiv', we selected the timezone 'UTC' for now. in /var/www/agent_stack/data/www/doraprojects.net/template/agent.layouts/content.php on line 54
2017-04-01 07:15:16