How to do this in Laravel, subquery where in

问题

How can I make this query in 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

I could also do this with a join, but I need this format for performance.

回答1:

Consider this code:

Products::whereIn('id', function($query){
    $query->select('paper_type_id')
    ->from(with(new ProductCategory)->getTable())
    ->whereIn('category_id', ['223', '15'])
    ->where('active', 1);
})->get();

回答2:

Have a look at the advanced wheres documentation for Fluent: http://laravel.com/docs/queries#advanced-wheres

Here's an example of what you're trying to achieve:

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

This will produce:

select * from users where id in (
    select 1 from orders where orders.user_id = users.id
)

回答3:

You can use variable by using keyword "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();

回答4:

The following code worked for me:

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

            })
->get();

回答5:

You can use Eloquent in different queries and make things easier to understand and mantain:

$productCategory = ProductCategory::whereIn('category_id', ['223', '15'])
                   ->select('product_id'); //don't need ->get() or ->first()

and then we put all together:

Products::whereIn('id', $productCategory)
          ->where('active', 1)
          ->select('id', 'name', 'img', 'safe_name', 'sku', 'productstatusid')
          ->get();//runs all queries at once

This will generate the same query that you wrote in your question.

回答6:

Laravel 4.2 and beyond, may use 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');
}

回答7:

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();

回答8:

using a variable

$array_IN=Dev_Table::where('id',1)->select('tabl2_id')->get();
$sel_table2=Dev_Table2::WhereIn('id',$array_IN)->get();

回答9:

Please try this online tool sql2builder

DB::table('products')
    ->whereIn('products.id',function($query) {
                            DB::table('product_category')
                            ->whereIn('category_id',['223','15'])
                            ->select('product_id');
                        })
    ->where('products.active',1)
    ->select('products.id','products.name','products.img','products.safe_name','products.sku','products.productstatusid')
    ->get();

来源：https://stackoverflow.com/questions/16815551/how-to-do-this-in-laravel-subquery-where-in