仅当关系结果存在时才获取对象 [laravel]

Question

仅当关系查询计数大于 0 时才获取数据时出现问题。

这是我的客户关系模型

class Customer extends Model
{
protected $table = 'customers';
public function contracts()
{
return $this->hasMany('AppContract');
}

这是我的合同模型

class Contract extends Model
{
public function customer()
{
return $this->belongsTo('AppCustomer');
}
}

最后，我只需要他们签约的客户

$customers = Customer::with(['contracts' => function($query)
{
$query->where('data_end','>=','2017-07-01')
->where('data_end','<=','2017-07-31') 
->where('typ','=','U') ;
}
])->paginate(10);

但是我有所有的客户。它看起来像这样：

"Customer 1"
"Customer 2"
"Customer 3"
*"Contract 1"
*"Contract 2"
*"Contract 3"
"Customer 4"
*"Contract 1"
*"Contract 2"
*"Contract 3"  
"Customer 4"  

在此示例中，我不需要客户 1、2 和 5。我怎样才能用急切的加载和末端有关系的对象来做到这一点。

在此处输入图像描述

发生这种情况，我不需要屏幕截图上有 X 的客户 - 我的意思是，我不需要从哪里查询 0 份合同的客户

--来自 dd(( 的对象--

此查询结束 2 个客户，第一个有 2 个合同，第二个有 0 个合同

LengthAwarePaginator {#217 ▼
#total: 75000
#lastPage: 37500
#items: Collection {#213 ▼
#items: array:2 [▼
0 => Customer {#220 ▼
#table: "customers"
#connection: null
#primaryKey: "id"
#keyType: "int"
+incrementing: true
#with: []
#perPage: 15
+exists: true
+wasRecentlyCreated: false
#attributes: array:5 [▼
"id" => 1
"customer_number" => "46071600600"
"name" => "Nikodem Zalewski"
"customer_contact" => "507614445"
"customer_type" => "P"
]
#original: array:5 [▶]
#casts: []
#dates: []
#dateFormat: null
#appends: []
#events: []
#observables: []
#relations: array:1 [▼
"contracts" => Collection {#224 ▼
#items: array:2 [▼
0 => Contract {#227 ▼
#connection: null
#table: null
#primaryKey: "id"
#keyType: "int"
+incrementing: true
#with: []
#perPage: 15
+exists: true
+wasRecentlyCreated: false
#attributes: array:10 [▶]
#original: array:10 [▶]
#casts: []
#dates: []
#dateFormat: null
#appends: []
#events: []
#observables: []
#relations: []
#touches: []
+timestamps: true
#hidden: []
#visible: []
#fillable: []
#guarded: array:1 [▶]
}
1 => Contract {#229 ▶}
]
}
]
#touches: []
+timestamps: true
#hidden: []
#visible: []
#fillable: []
#guarded: array:1 [▶]
}
1 => Customer {#221 ▼
#table: "customers"
#connection: null
#primaryKey: "id"
#keyType: "int"
+incrementing: true
#with: []
#perPage: 15
+exists: true
+wasRecentlyCreated: false
#attributes: array:5 [▶]
#original: array:5 [▼
"id" => 2
"customer_number" => "81050371854"
"name" => "Adam Wróbel"
"customer_contact" => "560047958"
"customer_type" => "P"
]
#casts: []
#dates: []
#dateFormat: null
#appends: []
#events: []
#observables: []
#relations: array:1 [▼
"contracts" => Collection {#222 ▼
#items: []
}
]
#touches: []
+timestamps: true
#hidden: []
#visible: []
#fillable: []
#guarded: array:1 [▶]
}
]
}
#perPage: 2
#currentPage: 1
#path: "*"
#query: []
#fragment: null
#pageName: "page"
}

Answer 1

你的代码几乎就在那里了。

您没有指定您正在使用哪个版本的 Laravel，但您应该查看 wherehas(至少从 5.0 开始就存在于 Laravel 中(

https://laravel.com/docs/5.0/eloquent#querying-relations (v5.0(

访问模型的记录时，您可能希望根据关系的存在来限制结果。例如，您希望拉取至少具有一条评论的所有博客文章。为此，您可以使用 has 方法：

如果你需要更多的功能，你可以使用 wherehas 和 orWhereHas 方法来对你的 has 查询放置 "where" 条件：

请尝试以下代码。

$customers = Customer::with('contracts')->whereHas('contracts', function($query)
{
$query->where('data_end','>=','2017-07-01')
->where('data_end','<=','2017-07-31') 
->where('typ','=','U') ;
}
)->paginate(10);

这将使用客户加载合同对象，但仅返回具有与查询匹配的合同的客户。