Laravel从一种形式插入多条记录 [英] Laravel Insert multiple records from one form

问题描述

我正在执行一项向项目添加其他内容的任务，我有一个选择字段来添加item_id并将其保存到数据库中的Extras表中.

I'm working on a task to add extra to items, I have a select field to add item_id and save it to extras table in the database.

我试图添加1个以上的项目以节省时间，而不是将相同的额外项目添加到其他项目中.

I'm trying to add more than 1 items to save my time instead of adding same extra to another items.

我将仅分享一项的当前代码工作:

I'll share the current code work for one item ONLY :

查看:

 {!! Form::select('food_id', $food, null, ['class' => 'select2 form-control']) !!}

我尝试使其插入多个ID:

my try do make it insert multiple ids:

{!! Form::select('food_id[]', $food, null, ['class' => 'select2 form-control', 'multiple'=>'multiple']) !!}

控制器:

public function create()
{
    $this->foodRepository->pushCriteria(new FoodsOfUserCriteria(auth()->id()));
    $food = $this->foodRepository->groupedByRestaurants();
    $extraGroup = $this->extraGroupRepository->pluck('name', 'id');

    $hasCustomField = in_array($this->extraRepository->model(), setting('custom_field_models', []));
    if ($hasCustomField) {
        $customFields = $this->customFieldRepository->findByField('custom_field_model', $this->extraRepository->model());
        $html = generateCustomField($customFields);
    }
    return view('extras.create')->with("customFields", isset($html) ? $html : false)->with("food", $food)->with("extraGroup", $extraGroup);
}

显示错误:

ErrorException(E_NOTICE)数组到字符串的转换

ErrorException (E_NOTICE) Array to string conversion

错误突出显示:

    foreach ($segments as $segment) {
        $result .= (array_shift($replace) ?? $search).$segment;
    }

'food_id'=>整数"

'food_id' => 'integer'

如果我 print_r($ request-> all()); 它向我显示此结果:

if I print_r($request->all()); it show me this result :

Array ( [_token] => p1oO45NAnBfqvMHm [name] => 23 => [price] => 9 [food_id] => Array ( [0] => 33 [1] => 34 ) [extra_group_id] => 4 ) 

希望我能解释清楚，等待帮助:)谢谢

I hope I explain it clearly, waiting for a help :) Thanks

推荐答案

感谢所有试图帮助我的人员，我想分享为实现这一目标而做出的解决方案:

Thank you for all personnel trying to help me, I would like to share the solution I made to make it done :

     $input = $request->all();
     $food_ids = $input['food_id'];
     foreach($food_ids as $food_id) {
       $input['food_id'] = $food_id;
       ...
       ...

       $extra = $this->extraRepository->create($input);
     }

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