将 mysql 查询逻辑转换为 Laravel 查询构建器

Question

我正在尝试将我的 mysql 查询逻辑转换为 Laravel 查询构建器。我不知道如何将其转换为 laravel 查询。

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`

通常我会使用查询生成器编写 select 查询，但无法实现上述逻辑

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

Answer 1

试试这个

$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();

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

将 mysql 查询逻辑转换为 Laravel 查询构建器

Convert mysql query logic to Laravel query builder

php

mysql

query-builder

laravel

laravel-4