laravel 类型错误:函数App\Http\Controllers\gst_billing\WOSBillingController::index()的参数太少,传递了0,需要1

x3naxklr  于 2023-04-07  发布在  其他
关注(0)|答案(1)|浏览(96)

我必须显示索引页面与相应的id,当我dd()在索引方法中查询我得到空
控制器中的索引方法

public function index($id){
    
     $bill = DB::table('proforma_invoice as b')
                ->leftjoin('states as ss', 'ss.id', '=', 'b.shipping_state_id')
                ->leftjoin('countries as sco', 'sco.id', '=', 'ss.country_id')
                ->leftjoin('states as s', 's.id', '=', 'b.billing_state_id')
                ->leftjoin('countries as co', 'co.id', '=', 's.country_id')
                ->join('customer as c', 'c.id', '=', 'b.customer_id')
                ->select('b.billing_name', 'b.customer_id', 'b.*', 's.state_name', 'co.country_name', 'ss.state_name as shipping_state', 'sco.country_name as shipping_country')
                ->where('b.id', $id)
                ->first();
                
                dd($bill);
   return view('gst_billing.wos_billing.index')->with('bill',$bill);
    
    }

从表中获取到索引页的具有相应id的按钮:

<a href="{{ route('gst_billing.wos_billing.index',['id' => $b->id]) }}" title="Send to Invoice"><i class="fas fa-file-invoice"></i></a>

路线:

Route::get('/wos_billing/{id}',['as' => 'gst_billing.wos_billing.index', 'uses' => 'WOSBillingController@index'])->name('gst_billing.wos_billing.index');

有人能看出我哪里做错了吗

jecbmhm3

jecbmhm31#

用这个

<a href="{{ route('gst_billing.wos_billing.index', $b->id) }}" title="Send to Invoice"><i class="fas fa-file-invoice"></i></a>

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