我有一個evaluation模型有很多trials. 我可以通過我的試驗創建一個試驗和查詢，并將它們輸出到我的evaluation.show視圖中。我的問題是當我想創建指向我的trial.edit或trial.show路線的鏈接時，我不斷收到錯誤訊息Missing required parameter for [Route: trial.show] [URI: evaluation/{evaluation}/trial/{trial}] [Missing parameter: trial]。我知道我遺漏了一些明顯的東西，但我已經耗盡了我的大腦輸出。
網頁.php
...
Route::get('/evaluation/{evaluation}/trial/create', App\Http\Livewire\Trial\Create::class)->name('trial.create');
Route::get('/evaluation/{evaluation}/trial/{trial}/edit', App\Http\Livewire\Trial\Edit::class)->name('trial.edit');
Route::get('/evaluation/{evaluation}/trial/{trial}', App\Http\Livewire\Trial\Show::class)->name('trial.show');
...
livewire/評估/show.blade.php
...
@foreach($trials as $trial)
<a href="{{route('trial.show', $trial->id)}}" class="block hover:bg-gray-50">
@endforeach
...
Livewire/評估/Show.php
...
public function mount(Evaluation $evaluation, Trial $trial) {
$this->evaluation = $evaluation;
$this->trial = $trial;
}
public function render()
{
$trials = Trial::where('evaluation_id', $this->evaluation->id)->get();
return view('livewire.evaluation.show', compact('trials'));
}
uj5u.com熱心網友回復：
你的路線有兩個引數（{evaluation}和{trial}），你只傳入一個{{route('trial.show', $trial->id)}}。
添加評估 id（假設$evaluation->id持有該 id）：
<a href="{{route('trial.show', [$evaluation->id, $trial->id])}}" class="block hover:bg-gray-50">
轉載請註明出處，本文鏈接：https://www.uj5u.com/qita/452473.html