如何在Laravel,子查询中进行此操作

Mar*_*rke 102 php subquery where-in laravel laravel-3

如何在Laravel中进行此查询:

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
Run Code Online (Sandbox Code Playgroud)

我也可以通过连接来做到这一点,但我需要这种格式来提高性能.

luk*_*rat 172

考虑以下代码:

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

  • @lukaserat有问题的查询是对产品表应用AND`p` .active` = 1,而你的查询将它应用于ProductCategory的表....对吗?还是有什么我想念的? (3认同)

dre*_*joh 47

看看Fluent的高级文档:http://laravel.com/docs/queries#advanced-wheres

以下是您要实现的目标示例:

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

这将产生:

select * from users where id in (
    select 1 from orders where orders.user_id = users.id
)
Run Code Online (Sandbox Code Playgroud)


Ram*_*esh 19

您可以使用关键字"use($ category_id)"来使用变量

$category_id = array('223','15');
Products::whereIn('id', function($query) use ($category_id){
   $query->select('paper_type_id')
     ->from(with(new ProductCategory)->getTable())
     ->whereIn('category_id', $category_id )
     ->where('active', 1);
})->get();
Run Code Online (Sandbox Code Playgroud)


Dar*_*uct 15

这是我从多个答案中收集的Laravel 8.x方法:

  • 使用查询生成器,不要直接编写 SQL。
  • 使用模型并从那里确定一切。不要使用硬编码的表名称或任何名称(列等)。
Product::select(['id', 'name', 'img', 'safe_name', 'sku', 'productstatusid'])
    ->whereIn('id', ProductCategory::select(['product_id'])
        ->whereIn('category_id', ['223', '15'])
    )
    ->where('active', 1)
    ->get();
Run Code Online (Sandbox Code Playgroud)


Mad*_*ota 9

The script is tested in Laravel 5.x and 6.x. The static closure can improve performance in some cases.

Product::select(['id', 'name', 'img', 'safe_name', 'sku', 'productstatusid'])
            ->whereIn('id', static function ($query) {
                $query->select(['product_id'])
                    ->from((new ProductCategory)->getTable())
                    ->whereIn('category_id', [15, 223]);
            })
            ->where('active', 1)
            ->get();
Run Code Online (Sandbox Code Playgroud)

generates the SQL

SELECT `id`, `name`, `img`, `safe_name`, `sku`, `productstatusid` FROM `products` 
WHERE `id` IN (SELECT `product_id` FROM `product_category` WHERE 
`category_id` IN (?, ?)) AND `active` = ?
Run Code Online (Sandbox Code Playgroud)


小智 8

你可以在不同的查询中使用 Eloquent,让事情更容易理解和维护:

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

然后我们把所有东西放在一起:

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

这将生成您在问题中编写的相同查询。


Adi*_*ngh 5

以下代码对我有用:

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

            })
->get();
Run Code Online (Sandbox Code Playgroud)