为什么在whereHas为假的情况下加载模型？

Question

我有一个疑问：

$result = PortingItem::whereHas('porting', function ($query) {
        $query->whereIn('status', [
            Porting::STATUS_REQUESTED,
            Porting::STATUS_ACCEPTED,
            Porting::STATUS_DELAYED,
        ]);
    })->where(function ($query) use ($numbers) {
        $query->whereBetween('phone_number_start', [$numbers[0], $numbers[1]])
            ->orWhereBetween('phone_number_end', [$numbers[0], $numbers[1]]);
    })->orWhere([
        ['phone_number_start', '<=', $numbers[0]],
        ['phone_number_end', '>=', $numbers[1]],
    ])->get();

当移植关系为空时，PortingItem模型仍然返回查询结果。我不明白为什么会这样。

这是我的Porting模型关系

public function items()
{
    return $this->hasMany(PortingItem::class);
}

这是我的PortingItem模型关系：

public function porting()
{
    return $this->belongsTo(Porting::class);
}

Answer 1

，您应该始终对orWhere调用进行分组，以避免在应用全局作用域时发生意外行为。

$result = PortingItem::whereHas('porting', function ($query) {
        $query->whereIn('status', [
            Porting::STATUS_REQUESTED,
            Porting::STATUS_ACCEPTED,
            Porting::STATUS_DELAYED,
        ]);
    })->where(function ($query) use ($numbers) {
        $query->where(function ($query) use ($numbers) {
            $query->whereBetween('phone_number_start', [$numbers[0], $numbers[1]])
                ->orWhereBetween('phone_number_end', [$numbers[0], $numbers[1]]);
        })->orWhere(function ($query) use ($numbers) {
            $query->where('phone_number_start', '<=', $numbers[0])
                ->where('phone_number_end', '>=', $numbers[1]);
        });
    })->get();

https://laravel.com/docs/8.x/queries#logical-grouping