問題描述

我想更新一對多關系.例如，我有一個名為 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');
    }

}

和一個名為 Size

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

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

這是我的控制器:

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選定的產品 id 更新所有尺寸記錄到最后一個輸入者尺寸.我想用選定的不同尺寸而不是最后選擇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

如何更新特定產品的尺寸.像多對多關系的同步方法一樣，有什么方法可以更新記錄.

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

推薦答案

你在循環中覆蓋你的$array.在 for 循環之后，$array 可能是最后一個大小.

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

總的來說，您的查詢是正確的，但它與執行的位置有關.它應該如下所示:

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]);
}

這篇關于更新 Laravel 5.3 中的一對多關系的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！