在laravel 5.3中更新一对多关系 [英] Updating One to many relation in laravel 5.3
问题描述
我想更新一对多的关系.例如,我有一个名为产品
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('App\Size');
}
}
和一个名为 Size
class Size extends Model
{
public $timestamps = FALSE;
protected $fillable = ['product_id','size'];
public function Product(){
return $this->belongsTo('App\Product');
}
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);
}
但是它正在执行的操作是将具有选定产品ID的所有尺寸记录更新为最后输入的尺寸.我想用选定的不同尺寸而不是最后选择的尺寸来更新所有尺寸
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.就像多对多关系的同步方法一样,有任何方法可以更新记录.
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
您在循环中覆盖您的$array
.在for循环之后,$array
可能具有最后一个大小.
You override your $array
in the loop. After the for loop, $array
has possibly the last size.
通常,您的查询是正确的,但这与执行的位置有关.它应该如下所示:
In general, your query is correct, but it's about the place of execute. It should be like below:
if (isset($request->$selectedSize)) { // or is_int() ?
$Size = Size::where('product_id', $request->id)->update(['size'=> $request->$selectedSize]);
}
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