Php 拉威尔在两个对象之间求并
我有以下内容来获取登录用户拥有的所有组织Php 拉威尔在两个对象之间求并,php,laravel,object,eloquent,Php,Laravel,Object,Eloquent,我有以下内容来获取登录用户拥有的所有组织 $user_owned_orgs = Organization::where('user_id', Auth::user()->id)->get(); 并获取登录用户所属的所有组织 $user_orgs = Organization::whereIn('id', function($query){ $query->select('organization_id') ->from(with(new Organiza
$user_owned_orgs = Organization::where('user_id', Auth::user()->id)->get();
并获取登录用户所属的所有组织
$user_orgs = Organization::whereIn('id', function($query){
$query->select('organization_id')
->from(with(new OrganizationUser)->getTable())
->where('user_id', Auth::user()->id);
})->get();
考虑到我想要一个两者并集的对象,我测试运行
$obj_merged = (object) array_merge((array) $user_owned_orgs, (array) $user_orgs);
那么我只有一个组织
{#1362 ▼
#items: array:1 [▼
0 => App\Organization {#1390 ▶}
]
}
如果我跑
$user_owner_or_member = array_unique(array_merge($user_owned_orgs->toArray(),$user_orgs->toArray()), SORT_REGULAR);
然后我得到了
array:2 [▼
0 => array:7 [▼
"id" => 1
"name" => "organization1"
"slug" => "organization1"
"is_visible" => 0
"user_id" => 1
"created_at" => "2021-02-12T10:14:23.000000Z"
"updated_at" => "2021-02-12T10:14:23.000000Z"
]
1 => array:7 [▼
"id" => 5
"name" => "organization12"
"slug" => "organization12"
"is_visible" => 1
"user_id" => 1
"created_at" => "2021-03-01T08:34:25.000000Z"
"updated_at" => "2021-03-01T08:34:25.000000Z"
]
]
虽然结果在显示两条记录的意义上是正确的,但它是一个数组。这意味着需要修改为处理对象而创建的视图(这很好,但并不理想)。以下面的方式将其更改为对象也不是一个选项
(object) array_unique(array_merge($user_owned_orgs->toArray(),$user_orgs->toArray()), SORT_REGULAR);
换句话说,我要说的(这可能是最简单的选项)是如何更改查询以获取登录用户拥有的所有组织(
organization.user\u id=Auth::user()->id
)和/或同一用户是(organization\u users.organization\u id=organization.id和organization\u users.user\u id=Auth::user)的成员()->id
)?您在Eloquent中使用Eloquent和更好的解决方案来合并两个结果如下:
$obj_merged = $user_owned_orgs->merge($user_orgs);
现在您有了一个具有所有有用属性的雄辩对象