如何在 Laravel 中将 sql 转换为查询生成器答案

【问题标题】：how to convert the sql to QueryBuilder in Laravel如何在 Laravel 中将 sql 转换为查询生成器
【发布时间】：2021-02-23 12:15:40
【问题描述】：

$old_records = DB::connection("mysql_old_s")->table('leads')
        ->select(DB::raw('count(*) as  source_count, source'))
        ->groupBy('source')
        ->get();
dd($old_records); // works right

这个查询工作正常。我想再添加一列，例如 id 或 created_at 来选择

我不能这样做：

$old_records = DB::connection("mysql_old_s")->table('leads')
        ->select('id',DB::raw('count(*) as source_count, source'))
        ->groupBy('source')
        ->get();
dd($old_records); // it gives an error

它给了我一个错误：

SQLSTATE[42000]: Syntax error or access violation: 1055 Expression #1 of SELECT list is not in GROUP 
BY clause and contains nonaggregated column 'crmclini_crmv2.leads.id' which is not functionally 
dependent on columns in GROUP BY
clause; this is incompatible with sql_mode=only_full_group_by (SQL: select `id`, count(*) as 
source_count, source , id from `leads` group by `source`)

【问题讨论】：

标签： laravel eloquent laravel-query-builder laravel-8

【解决方案1】：

您应该在 group by 子句中包含您选择的列以按它们分组：

$old_records = DB::connection("mysql_old_s")->table('leads')
        ->select('id',DB::raw('count(*) as source_count, source'))
        ->groupBy(['id','source'])
        ->get();

第二个选项（女巫我不推荐），你可以在config/database.php中禁用only_full_group_by选项：

'连接'=> [ ...

'mysql' => [
...
    'strict' => false,
    ...

],
]

除非您有特殊情况，否则按您选择的列进行查询分组

【讨论】：