我有一个表user_childrens,其中包含id_parent和id_user。 我正在尝试用此列出父母的所有孩子:
代码:
//relation in model via belongsTo
$idparent = auth('api')->user()->id;
$list = UserChildren::where('id_parent',$idparent)
->with('child:id,name,email')
->get();
return $list->toJson();
返回值为:
[
{
"id": 1,
"id_parent": 1,
"id_user": 1,
"created_at": null,
"updated_at": null,
"child": {
"id": 1,
"name": "Mr. Davin Conroy Sr.",
"email": "prempel@example.com"
}
},
{
"id": 4,
"id_parent": 1,
"id_user": 2,
"created_at": null,
"updated_at": null,
"child": {
"id": 2,
"name": "Krystel Lehner",
"email": "cernser@example.net"
}
}
]
但是它是API,所以我只想要像这样的子列:
[
{
"id": 1,
"name": "Mr. Davin Conroy Sr.",
"email": "prempel@example.com"
},
{..}
]
UserChildren模型:
public function child() {
return $this->belongsTo('App\User','id_user','id');
}
我知道我可以通过集合上的.map()来执行此操作,但是此查询上可能已经有其他解决方案了
答案 0 :(得分:0)
您可以使用此代码
$idparent = auth('api')->user()->id;
$childs = User::whereHas('user_childrens', function ($query) use ($idparent) {
$query->where('id_parent', $idparent);
})->get(['id', 'name', 'email']);
dd($childs->toJson());
用户模型定义user_childrens
关系。
public function user_childrens()
{
return $this->hasMany('App\UserChildren','id_user','id');
}
另请参阅文档https://laravel.com/docs/5.5/eloquent-relationships#querying-relationship-existence