在string上调用成员函数addEagerConstraints()

Question

我想用type_name获得响应属性，无需在表

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，希望可以通过with()方法调用该方法，如下所示:

$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);

但是它返回一个错误:

Error: Call to a member function addEagerConstraints() on string...

在我的CoinTransaction模型

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;
}
}

Answer 1

typeName不是relationship method所以下面是typeName()

$coin=CoinTransaction::find($transaction->id); 
dd($coin->typeName());

或

要将type_name属性添加到现有响应中，可以使用Mutators

裁判:https://laravel.com/docs/8.x/eloquent-mutators器和写值器

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;    
}

so in controller

$transaction = CoinTransaction::find($transaction->id);