轉換笨查詢Laravel查詢

Question

我在笨

$html = array(); 
     $sqltourtypes = 'SELECT * FROM tourtypes ORDER BY nTourTypeID ASC'; 
     $sqltours = 'SELECT * FROM tours WHERE nTourTypeID = ? ORDER BY _kpnID ASC'; 

     $tourtypes = $this->db->query($sqltourtypes)->result(); 
     for($i = 0; $i < count($tourtypes); $i++){ 
      $html[] = '<li><a href="#">'.$tourtypes[$i]->_kftDescription.'</a>'; 
      $tours = $this->db->query($sqltours,array($tourtypes[$i]->nTourTypeID))->result(); 
      if(count($tours)>0){ 
       $html[] = '<ul>'; 
       for($ia = 0; $ia < count($tours); $ia++){ 
        $html[] = '<li>'.$tours[$ia]->tDescription.'</li>'; 
       } 
       $html[] ='</ul></li>'; 
      }else { 
       $html[] = '</li>'; 
      } 
     } 
     return implode('',$html); 

下面的代碼我最近不得不改用Laravel框架。我無法讓我的查詢在Laravel中工作。基本上我有兩個表，tourtypes和旅遊。 _kftDescription用於列出ul標籤下的巡視類型，tDescription用於將特定組下的巡迴名稱列爲li標籤。

嘗試轉換查詢時，我總是收到錯誤。任何人都可以建議如何從CodeIgniter實現我的代碼？當nTourTypeID是「1」時，它們屬於遊覽類型「遊輪」。希望是有道理的。

更新：我的應用程序\ HTTP \控制器\ BookingsController.php文件看起來像這樣

namespace App\Http\Controllers; 

use App\Models\Bookings; 
use Illuminate\Http\Request; 
use Illuminate\Pagination\LengthAwarePaginator as Paginator; 

use Illuminate\Support\Facades\DB; 
use App\Http\Controllers\Controller; 
use Validator, Input, Redirect ; 
class BookingsController extends Controller { 
    public function index() 
    { 
    $tourTypes = collect(DB::table('tourtypes')->orderBy('nTourTypeID')->get()) 
     ->map(function ($item) { 
      $item->tours = DB::table('tours')->where('nTourTypeID', $item->nTourTypeID)->get(); 

      return $item; 
     }); 

     return view('bookings', compact('tourTypes')); 

    } 

和預訂的路線是這樣的（我的路線是預訂我不有路線導覽）：

Route::get('bookings','BookingsController@getIndex'); 

最後\ resources \ views \ bookingings \ index.blade.php文件看起來像是th是：

@extends('layouts.app') 

@section('content') 
{{--*/ usort($tableGrid, "SiteHelpers::_sort") /*--}} 
@if(count($tourTypes)) 

    <ul> 
     @foreach($tourTypes as $tourType) 
      <li> 
       <a href="#">{{ $tourType->_kftDescription }}</a> 

       @if(count($tourType->tours)) 
        <ul> 
         @foreach($tourType->tours as $tour) 
          <li>{{ $tour->tDescription }}</li> 
         @endforeach 
        </ul> 
       @endif 
      </li> 
     @endforeach 
    </ul> 

@endif 

我仍然得到錯誤

未定義的變量：tourTypes（查看： d：\ XAMPP \ htdocs中\預訂\資源\意見\預訂\ index.blade.php）

當我寫

$tourTypes = DB::table('tourtypes')->orderBy('nTourTypeID', 'ASC')->get(); 
print_r($tourTypes); 

打印

照亮\支持\集合對象（[項目：保護] =>數組（ [0] => stdClass的對象（[nTourTypeID] => 1 [_kftDescription] => 遊輪[_kftColourID] => 003399 ）1 => stdClass Object（ [nTourTypeID] => 2 [_kftDescription] => 4WD [_kftColourID] =>）[2] => stdClass對象（[nTourTypeID] => 3 [_kftDescription] =>珍珠農場 [ _kftColourID] => 00ccff）

因此，該查詢正在工作，但無法打印ul和li標籤，其值爲using;

@if(count($tourTypes)) 
    <ul> 
     @foreach($tourTypes as $tourType) 
      <li> 
       <a href="#">{{ $tourType->_kftDescription }}</a> 
       @if(count($tourType->tours)) 
        <ul> 
         @foreach($tourType->tours as $tour) 
          <li>{{ $tour->tDescription }}</li> 
         @endforeach 
        </ul> 
       @endif 
      </li> 
     @endforeach 
    </ul> 
@endif 

Answer 1

請注意，這個答案只是給你一個你可以在Laravel做什麼的例子。

說你的路線網址爲/tours，你可以這樣做：

Route::get('tours', function() { 

    $tourTypes = collect(DB::table('tourtypes')->orderBy('nTourTypeID')->get()) 
     ->map(function ($item) { 
      $item->tours = DB::table('tours')->where('nTourTypeID', $item->nTourTypeID)->get(); 

      return $item; 
     }); 

    return view('tours', compact('tourTypes')); 
}); 

然後創建一個文件resources/views/tours.blade.php並添加以下內容：

@if(count($tourTypes)) 

    <ul> 
     @foreach($tourTypes as $tourType) 
      <li> 
       <a href="#">{{ $tourType->_kftDescription }}</a> 

       @if(count($tourType->tours)) 
        <ul> 
         @foreach($tourType->tours as $tour) 
          <li>{{ $tour->tDescription }}</li> 
         @endforeach 
        </ul> 
       @endif 
      </li> 
     @endforeach 
    </ul> 

@endif 

上面的例子只會輸出UL認證。本教程會幫助你多一點： https://laracasts.com/series/laravel-5-from-scratch/episodes/5

---

此外，作爲@PaulSpiegel在評論中提到它會更有利於你使用Eloquent廣告它減少了你的路由/控制器代碼，並與幫助急切的加載。

要做到這一點，你可以創建以下文件：

應用程序/ Tour.php

<?php 

namespace App; 

use Illuminate\Database\Eloquent\Model; 

class Tour extends Model 
{ 
    protected $primaryKey = 'kpnID'; 

    public function Tourtypes() 
    { 
     return $this->belongsTo(Tourtype::class, 'nTourTypeID', 'nTourTypeID'); 
    } 
} 

應用程序/ Tourtype.php

<?php 

namespace App; 

use Illuminate\Database\Eloquent\Model; 

class Tourtype extends Model 
{ 
    /** 
    * The primary key for the model. 
    * 
    * @var string 
    */ 
    protected $primaryKey = 'nTourTypeID'; 

    /** 
    * Tours Relationship 
    * 
    * @return \Illuminate\Database\Eloquent\Relations\HasMany 
    */ 
    public function tours() 
    { 
     return $this->hasMany(Tour::class, 'nTourTypeID', 'nTourTypeID'); 
    } 
} 

在我假設上述旅遊的主要關鍵是kpnID。如果不是，那就改變它。

那麼你的路由看起來是這樣的：

Route::get('tours', function() { 

    $tourTypes = \App\Tourtype::with('tours')->get(); 

    return view('tours', compact('tourTypes')); 
}); 

https://laravel.com/docs/5.1/eloquent

https://laravel.com/docs/5.1/eloquent-relationships#one-to-many

https://laravel.com/docs/5.1/blade#defining-a-layout

希望這有助於！