Question

拜托，我的控制器中有一个功能。我想查询股票表以选择product_id列表，在select之后，它应该转到产品表以向用户显示产品的NAME，而不是{{1} }。

product_id

我的PHP刀片有

public function getStockList()
{
    $stock = DB::table("stocks")->pluck("product_id","product_id")->all();
    $products = DB::table("products")->pluck("name")->where('id', $stock);
    return view('shop.shop')->with(['stocks' => $stocks, 
    'stock' => $stock, 'products' => $products]);
}

基本上，我不希望向用户显示{!! Form::select ( 'stock', ['' => 'Select'] +$stock,'', array( 'class'=>'form-control', 'id'=>'country', 'style'=>'width:350px;' ) )!!} {!! Form::select ( 'product_id', [''=>'Select Supplier'] + $products, null, ['class'=>'form-control'] )!!}，而是显示产品名称。

Answer 1

试试这个

public function getStockList()
{
    $stock = DB::table("stocks")->pluck("product_id");
    $products = DB::table("products")->whereIn('id', $stock)->pluck("name","id");
    return view('shop.shop')->with([
        'stocks'   => $stocks,
        'stock'    => $stock,
        'products' => $products
    ]);
}

使用以下

替换产品列表选择框

{!! Form::select('product_id',$products,null, ['class'=>'form-control','placeholder'=>'Select Products'])!!}

Answer 2

你需要数组传递形式

/* $product_array = ['1' => 'product 1',  '2' => 'product 2', '3' => 'product 3']; */

所以在你的控制器中

$product_array = []
  foreach($products as $product){
    $product_array[$product->id] = $product->name;
  }

return view('shop.shop')->with(['stocks' => $stocks, 
'stock' => $stock, 'products' => $products , 'product_array'=>$product_array]);

在你看来

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

希望这会有所帮助。

使用Laravel Eloquent中另一个表的ID显示表的名称

2 个答案: