laravel 根据id在页面上显示数据

Question

我有一个页面显示 links，其中包含在数据库中检索到的企业名称，如下所示：

控制器：

public function viewBusiness() {
// Return our "website" object
$business = Business::all();
// Pass the contents of the "html" property to the view
return view('viewBusiness', ['business' => $business]);
}

查看：

@extends('master') @section('title', 'Live Oldham') @section('content')
@section('content')
            @foreach ($business as $businesses)
        <a target="_blank" href="{{ url('business/' . $businesses->name) }}"> {{($businesses->name) }}
        </a> @endforeach       
@endsection

路线：

Route::get('business/list', 'BusinessController@viewBusiness')->name('viewBusiness');

然后我添加了一个功能，用户点击 link，它会被带到一个显示该特定业务的所有数据的页面，但是它显示所有数据，但不是所有业务。

控制器：

function displayBusiness() {
    $business = Business::all();
    $address = Address::all();
    return view('displayBusiness', ['business' => $business, 'address' => $address]);
}

查看：

@foreach ($business as $businesses)
    <p>{{$businesses->name}}</p>
    <p>{{$businesses->email}}</p>
@endforeach
@foreach ($address as $addresses)
    <p>{{$addresses->firstline_address}}</p>
    <p>{{$addresses->secondline_address}}</p>
    <p>{{$addresses->town}}</p>
    <p>{{$addresses->city}}</p>
    <p>{{$addresses->postcode}}</p>
    <p>{{$addresses->telephone}}</p>
@endforeach

路线：

Route::get('business/{name}', 'BusinessController@displayBusiness')->name('displayBusiness');

现在的问题是，如何修改此代码以便仅显示匹配 bussiness->name 或 business->id 的企业。（我想当用户点击一个名字时，名字就被取走了。

另外一个问题是如何限制url使得if localhost/business/{name} is not equal to any business->name in the database returns错误？目前，无论您输入什么，它都会显示该页面。

谢谢！

Answer 1

路由参数在控制器函数中作为参数可用。现在您可以使用此函数构建查询。如果您的查询没有 return 任何结果，您可以将用户发送回业务概览。

function displayBusiness($name) {
    $business = Business::where('name', $name)->orWhere('id', $name)->first();
    if ($business === null)
    {
        // No business with this name or id found.
        // Redirect to businesses list page.
    }

    $address = Address::all();
    return view('displayBusiness', ['business' => $business, 'address' => $address]);
}

Answer 2

我不知道我是否理解了这个问题，但这可能是解决方案的开始...

第一个视图：

@extends('master') @section('title', 'Live Oldham')
@section('content')
            @foreach ($business as $businesses)
        <a target="_blank" href="{{ url('business/' . $businesses->id) }}"> {{($businesses->name) }}
        </a> @endforeach       
@endsection

第二个控制器：

function displayBusiness($id) {
    $business = Business::find($id);
    $address = Address::find($id);
    return view('displayBusiness', compact('business', 'address'));
}

第二个视图：

    <p>{{$business->name}}</p>
    <p>{{$business->email}}</p>
    <p>{{$address->firstline_address}}</p>
    <p>{{$address->secondline_address}}</p>
    <p>{{$address->town}}</p>
    <p>{{$address->city}}</p>
    <p>{{$address->postcode}}</p>
    <p>{{$address->telephone}}</p>

第二条路线：

Route::get('business/{id}', 'BusinessController@displayBusiness')->name('displayBusiness');

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