一尘不染

Laravel雄辩的LEFT JOIN WHERE NULL

mysql

我正在尝试使用Eloquent在数据库种子期间执行以下查询:

SELECT
    *
FROM
    customers
LEFT JOIN
    orders
    ON customers.id = orders.customer_id
WHERE
    orders.customer_id IS NULL

这是我在Eloquent中的实现:

$c = Customer::leftJoin('orders', function($join) {
      $join->on('customers.id', '=', 'orders.customer_id');
    })
    ->whereNull('orders.customer_id')
    ->first();

尽管第一个查询始终返回完整结果,但Eloquent等价项始终为表的emailphone字段以外的所有内容返回空元素customers。由于CustomersOrders模型都是工匠生成的骨架,因此我无所适从。

例如:

class Customer extends \Eloquent {

    // Add your validation rules here
    public static $rules = [
        // 'title' => 'required'
    ];

    // Don't forget to fill this array
    protected $fillable = [];

}

这是当我对种子(最初由Faker生成)的第一个口才查询dd()时输出的数组:

protected $original =>
  array(25) {
    'id' =>
    NULL
    'first_name' =>
    NULL
    'last_name' =>
    NULL
    'email' =>
    string(24) "luther.braun@example.org"
    'phone' =>
    string(17) "642.150.9176x5684"
    'address1' =>
    NULL
    'address2' =>
    NULL
    'city' =>
    NULL
    'state' =>
    NULL
    'county' =>
    NULL
    'district' =>
    NULL
    'postal_code' =>
    NULL
    'country' =>
    NULL
    'notes' =>
    NULL
    'created_at' =>
    NULL
    'updated_at' =>
    NULL
    'customer_id' =>
    NULL
    'total' =>
    NULL
}

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2020-05-17

共1个答案

一尘不染

可以通过指定特定表中所需的特定列名来解决,如下所示:

$c = Customer::leftJoin('orders', function($join) {
      $join->on('customers.id', '=', 'orders.customer_id');
    })
    ->whereNull('orders.customer_id')
    ->first([
        'customers.id',
        'customers.first_name',
        'customers.last_name',
        'customers.email',
        'customers.phone',
        'customers.address1',
        'customers.address2',
        'customers.city',
        'customers.state',
        'customers.county',
        'customers.district',
        'customers.postal_code',
        'customers.country'
    ]);
2020-05-17