MySQL

Question

我想运行一个MySQL这样的查询-

 SELECT country_ID*2/id*3.159 as my_id
 FROM `state`
 WHERE my_id>2;

当我运行它时，我收到这样的错误-

1054 - Unknown column 'my_id' in 'where clause'

在我新创建的虚拟列 my_id 中搜索是否有任何替代解决方案？

实际上，我正在尝试在 Laravel Query Builder 中进行这样的搜索-

  DB::table(    'project')->select( 'project.id as id',
                                    'project.completion_date as completion_date',
                                     DB::raw('FORMAT(project.total_cost_to_dispose - project.actual_cost_dispose, 2) as disposal_savings')
                                   )
                            ->where(disposal_savings>100);

我可以吗？

如果不是，那么Laravel或MySQL的解决方案是什么？

Answer 1

您不能在 WHERE 中引用别名，请改用：

SELECT country_ID*2/id*3.159 as my_id 
FROM `state`
WHERE (country_ID*2/id*3.159)>2;

或使用子查询：

SELECT t.*
FROM
(
   SELECT country_ID*2/id*3.159 as my_id 
   FROM `state`
) as t
WHERE t.my_id>2

Simplified logical query processing，SELECT 几乎是最后一个，所以 WHERE 不知道 my_id 别名：

^{图片来源：https://social.technet.microsoft.com/wiki/contents/articles/20724.all-at-once-operations-in-t-sql.aspx}

Answer 2

在 where 子句中再次使用完整条件：

  DB::table(    'project')->select( 'project.id as id',
                                    'project.completion_date as completion_date',
                                     DB::raw('FORMAT(project.total_cost_to_dispose - project.actual_cost_dispose, 2) as disposal_savings')
                                   )
                            ->where FORMAT(project.total_cost_to_dispose - project.actual_cost_dispose, 2) > 100;

MySQL - 搜索自定义列

MySQL - Search into a custom Column

sql

query-builder

laravel

laravel-5