调用字符串上的成员函数 addEagerConstraints() - Laravel 8
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【中文标题】调用字符串上的成员函数 addEagerConstraints() - Laravel 8【英文标题】:Call to a member function addEagerConstraints() on string - Laravel 8 【发布时间】:2021-12-10 03:32:28 【问题描述】:我想通过 type_name 属性获得响应,而不在表中添加新字段
"status": 200,
"message": "OK",
"data":
"id": 23,
"uuid": "9b1d33f9-0e44-4161-9936-ec41309697a5",
"sender_id": null,
"receiver_id": 2,
"type": 0,
"coin": 200,
"balance": 27000,
"description": "Topup 200 coin",
"type_name": "Topup"
因此我尝试在 CoinTransaction 模型中创建一个名为 typeName() 的方法,希望可以像这样通过 with() 方法调用该方法:
$transaction = CoinTransaction::create([
'receiver_id' => auth()->user()->id,
'coin' => $request->coin,
'balance' => $predefineCoin->balance ?? 1000,
'type' => 0,
'description' => $request->description
]);
$transaction = CoinTransaction::with(['typeName'])->find($transaction->id);
但它返回一个错误:
Error: Call to a member function addEagerConstraints() on string...
在我的 CoinTransaction 模型中
class CoinTransaction extends Model
use HasFactory;
protected $guarded = ['id'];
public function sender()
return $this->belongsTo(User::class, 'sender_id');
public function receiver()
return $this->belongsTo(User::class, 'receiver_id');
public function typeName()
$typeName = null;
switch($this->type)
case 0: $typeName = 'Topup'; break;
default: $typeName = 'Unknown';
return $typeName;
【问题讨论】:
【参考方案1】:typeName
不是relationship method
所以你有typeName()
如下所示
$coin=CoinTransaction::find($transaction->id);
dd($coin->typeName());
或
要将type_name
属性添加到现有响应,您可以使用Mutators
参考:https://laravel.com/docs/8.x/eloquent-mutators#accessors-and-mutators
所以在CoinTransaction
中添加下面的属性和方法
protected $appends = ['type_name'];
和方法
public function getTypeNameAttribute()
$typeName = null;
switch($this->type)
case 0: $typeName = 'Topup'; break;
default: $typeName = 'Unknown';
return $typeName;
所以在控制器中
$transaction = CoinTransaction::find($transaction->id);
【讨论】:
感谢先生,效果很好以上是关于调用字符串上的成员函数 addEagerConstraints() - Laravel 8的主要内容,如果未能解决你的问题,请参考以下文章
调用字符串上的成员函数 addEagerConstraints() - Laravel 8