将mysql查询逻辑转换为Laravel查询生成器 [英] Convert mysql query logic to Laravel query builder

问题描述

我正在尝试将mysql查询逻辑转换为Laravel查询生成器.我不知道如何将其转换为laravel查询.

I am trying to convert my mysql query logic to Laravel query builder. I I have no Idea how to convert it as laravel query.

我的查询逻辑是

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

通常，我使用查询生成器编写一个选择查询，但无法实现上述逻辑

generally I write a select query using query builder but cant implement above logic

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

推荐答案

尝试一下

$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();

这是它的参考 http://laravel.com/docs/4.2/queries#raw-expressions

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