Laravel，数据表，低效查询答案

【问题标题】：Laravel, Datatables, inefficient queriesLaravel，数据表，低效查询
【发布时间】：2018-04-25 10:58:30
【问题描述】：

一些开发人员使用 Laravel 和 Datatables 制作了 PHP 软件。当我看到应用程序生成的数据库查询时，我发现一个效率很低。示例（涉及的列列表从一种软件功能更改为另一种）：

 select count(*) as aggregate from (select '1' as [row_count] from [assignments]
  left join [jobs] on [assignments].[id] = [jobs].[name] 
  left join [categories] on [categories].[id] = [jobs].[vol_cat_type]
   where ([assignments].[id] LIKE '%XYZ%' or [assignments].[person_id] LIKE '%XYZ%' or 
    (select count(1) from [people] where [assignments].[person_id] = [people].[person_id] and [people].[name_title] LIKE '%XYZ%') >= 1 or 
    (select count(1) from [people] where [assignments].[person_id] = [people].[person_id] and [people].[first_name] LIKE '%XYZ%') >= 1 or 
    [assignments].[title] LIKE '%XYZ%' or 
    (select count(1) from [people] where [assignments].[person_id] = [people].[person_id] and [people].[gender] LIKE '%XYZ%') >= 1 or 
    (select count(1) from [states_master] where [states_master].[identity] = [assignments].[current_state] and [states_master].[name] LIKE '%XYZ%') >= 1 or
        [assignments].[updated_datetime] LIKE '%XYZ%' or
        (select count(1) from [people] where [assignments].[person_id] = [people].[person_id] and [people].[nationality] LIKE '%XYZ%') >= 1 or 
    (select count(1) from [people] where [assignments].[person_id] = [people].[person_id] and [people].[place_of_birth] LIKE '%XYZ%') >= 1 or 
    (select count(1) from [people] where [assignments].[person_id] = [people].[person_id] and [people].[civil_status] LIKE '%XYZ%') >= 1 or 
    (select count(1) from [people] where [assignments].[person_id] = [people].[person_id] and [people].[email] LIKE '%XYZ%') >= 1 or 
    (select count(1) from [people] where [assignments].[person_id] = [people].[person_id] and [people].[date_of_birth] LIKE '%XYZ%') >= 1 or 
    (select count(1) from [people] where [assignments].[person_id] = [people].[person_id] and [people].[nationality2] LIKE '%XYZ%') >= 1
      or (select count(1) from [people] where [assignments].[person_id] = [people].[person_id] and [people].[skype] LIKE '%XYZ%') >= 1))
       count_row_table

涉及的表很大，而且这些查询不仅执行需要 20-30 秒，而且会影响服务器的总体性能。我不喜欢的是许多 LIKE 应用于所有列，尤其是那些应用于非字符串的列（如最后修改日期）。我的建议是：

1) 对所有非字符串使用 = 而不是 LIKE：数字、单字符代码（如性别 M/F、布尔值）

2) 配置Microsoft SQL Server中包含的全文搜索，写CONTAINS(column,'XYZ')而不是LIKE

开发者的反对意见是 Laravel 和 Datatables 会产生这些查询，我们对此无能为力。

是否可以在不更改库代码的情况下配置 Laravel 和 Datatables，为每一列指定是否使用 LIKE、= 或 CONTAINS？

【问题讨论】：

标签： php sql-server laravel datatables

【解决方案1】：

Eloquent 通常用于创建更简单的模型查询。

对于更复杂的 SQL 查询，请使用查询生成器。

这是文档中有关如何使用它的示例。

$users = DB::table('users')
                     ->select(DB::raw('count(*) as user_count, status'))
                     ->where('status', '<>', 1)
                     ->groupBy('status')
                     ->get();

如果您愿意，也可以使用原始 sql。

$results = DB::select('select * from users where id = :id', ['id' => 1]);

参考资料：

https://laravel.com/docs/5.6/queries

https://laravel.com/docs/5.6/database#running-queries

【讨论】：