將mysql查詢邏輯轉換為Laravel查詢生成器

Question

我正在嘗試將mysql查詢邏輯轉換為Laravel查詢生成器。 II不知道如何將其轉換為laravel查詢。

我的查詢邏輯是

SELECT id,name,
case 
    when visibility_status = '1' 
    then 'Visible' 
    when visibility_status = '0' 
    then 'Invisible'
    end as visibility_status FROM `flowers`

通常我使用查詢生成器編寫一個選擇查詢，但是無法實現上述邏輯

$result = DB::table('flowers')
        ->select('flowers.id as id', 'flowers.name as name',
'flowers.visibility_status as visibility_status');

Answer 1

嘗試這個

$users = DB::table('flowers')
->select(["id", "name",
      DB::raw("
       case 
          when visibility_status = '1' 
          then 'Visible' 
          when visibility_status = '0' 
          then 'Invisible'
          end as visibility_status
    ")])->get();

這是它的參考http://laravel.com/docs/4.2/queries#raw-expressions