问题描述

我想更新一对多关系.例如，我有一个名为 Product

I want to update a one to many relationship. For example I have a model called Product

class Product extends Model
{
    protected $primaryKey = 'product_id';
    public $timestamps = FALSE;

    public function Size(){
        return $this->hasMany('AppSize');
    }

}

和一个名为 Size

class Size extends Model
{
    public $timestamps = FALSE;
    protected $fillable = ['product_id','size'];

    public function Product(){
        return $this->belongsTo('AppProduct');
    }

这是我的控制器:

public function update(Request $request){
    for($i = 1; $i <= $sizeCounter; $i++){
        $selectedSize = "size_$i";

        if($request->$selectedSize){
            $array = ['size'=> $request->$selectedSize];
        }
    }

    $Size = Size::where('product_id' , $request->id)->update($array);
}

但是它正在做什么用选定的产品 id 更新所有尺寸记录到最后一个输入者尺寸.我想用选定的不同尺寸而不是最后选择的尺寸更新所有尺寸

But what it is doing updating all the records of size with the selectedd product id to the last enterder size. I want to update all the sizes with the slected diffrent sizes not the last selected sizes

如何更新特定产品的尺寸.像多对多关系的同步方法一样，有什么方法可以更新记录.

How to update the sizes of a particular product. Like many to many relations' sync method is there any way to update the records.

推荐答案

你在循环中覆盖你的$array.在 for 循环之后，$array 可能是最后一个大小.

You override your $array in the loop. After the for loop, $array has possibly the last size.

总的来说，您的查询是正确的，但它与执行的位置有关.它应该如下所示:

In general, your query is correct, but it's about the place of execute. It should be like below:

if (isset($request->$selectedSize)) { // or is_int() ?
    $Size = Size::where('product_id', $request->id)->update(['size'=> $request->$selectedSize]);
}

这篇关于更新 Laravel 5.3 中的一对多关系的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！