Eloquent：如何在不删除任何内容的情况下同步一对多关系（FK 可以为空）？

Question

分行可能有零(*) 或只有一个假期计划。假期计划可以“链接”（附加）到零个或多个分支。

(*) 这很重要：它是Foreign Key，可能是null

型号

这是我的Branch 模特：

class Branch extends Model
{
    protected $table = 'branch';
    protected $fillable = ['name', 'holiday_plan_id'];
    public $timestamps = false;

    public function holidayPlan()
    {
        return $this->hasOne('App\HolidayPlan', 'id', 'holiday_plan_id');
    }
}

---

这是我的HolidayPlan 模特：

class HolidayPlan extends Model
{
    protected $table = 'holiday_plan';
    protected $fillable = ['name'];
    public $timestamps = false;

    public function branches()
    {
        return $this->hasMany('App\Branch', 'holiday_plan_id', 'id');
    }
}

---

查看

现在我想将假期计划链接/取消链接到不同的分支机构

$branches = Branch::all();
$plan = HolidayPlan::find(1);

---

@foreach($branches as $branch)
<label>
    <input type="checkbox" name="branch[]"
        value="{{ $branch->id }}"{{ ($branch->holiday_plan_id == $plan->id) ? ' checked' : '' }}>
    {{ $branch->name }}
</label>
@endforeach

---

控制器

public function linkBranches($planId, Request $request) // post
{
    $plan = HolidayPlan::where('id', $planId)->first();
    $branches = Branch::findMany($request->input('branch'));
    //
    $this->solution1($plan, $branches); // see next
    //
    return redirect()->back();
}

---

可能的解决方案（无效）

private function solution1(HolidayPlan $plan, $branches)
{
    $plan->branches()->sync($branches); // Not working because this function is for Many-to-Many relations
}

---

private function solution2(HolidayPlan $plan, $branches)
{
    $plan->branches()->delete(); // This will delete my branches, which is not what I want to do, of course
    $plan->branches()->saveMany($branches);
}

---

private function solution3(HolidayPlan $plan, $branches)
{
    foreach($plan->branches as $branch) {
        $branch->holiday_plan_id = null;
        $branch->save();
    }
    $plan->branches()->saveMany($branches); // Not working; I don't know why: strange behavior. Maybe because of locks; I'm not sure. One time it detaches all branches, and only next time it attaches them again.
}

---

终于

实际上，原始 SQL 应该是：

UPDATE branch SET holiday_plan_id = NULL WHERE holiday_plan_id = 1;
UPDATE branch SET holiday_plan_id = 1 WHERE branch.id IN (2, 5, 7)/* for example */;

但是如何使用 Eloquent 函数和关系（没有原始 sql）来做到这一点？

---

编辑

我已经编辑了我的Branch 模型关系，但是solution3 仍然不起作用（当我尝试添加一个已经存在的holiday_plan_id 时，它没有添加它= 它变为空）：

public function holidayPlan()
{
    return $this->belongsTo('App\HolidayPlan', 'holiday_plan_id');
}

Answer 1

临时解决方案：

public function solution4(HolidayPlan $plan, $inputBranches)
{
    $oldLinks = $plan->branches;
    $plan->branches()->saveMany($inputBranches);
    foreach ($oldLinks as $oldLink) {
        if (! $inputBranches->contains($oldLink)) {
            $oldLink->holiday_plan_id = null;
            $oldLink->save();
        }
    }
}