运行原始SQL查询

Question

我有一个MYSQL查询，我需要在Laravel 5.6查询构建器上运行。我的查询是

SELECT paper_id,user_id,COUNT(payments.user_id),users.district
FROM payments
LEFT JOIN users ON payments.user_id = users.id
WHERE payments.paper_id=3
GROUP BY users.district HAVING COUNT(payments.user_id)>=0;

我尝试使用此代码

在Laravel DB::Raw上运行此功能

$data=DB::Raw('SELECT paper_id,user_id,COUNT(payments.user_id),users.district
                    FROM payments
                    LEFT JOIN users ON payments.user_id = users.id
                    WHERE payments.paper_id='.$paper_id.'
                    GROUP BY users.district HAVING 
                    COUNT(payments.user_id)>=0 
                    ');

以及通过此代码

$data2=DB::table('payments')
                    ->leftJoin('users','payments.user_id','users.id')
                    ->select('paper_id','user_id','users.district',
                      DB::Raw('COUNT(payments.user_id)'))
                    ->where('payments.paper_id',$paper_id)
                    ->groupBy('users.district')
                    ->select(DB::Raw('HAVING COUNT(payments.user_id)>=0'))
                    ->get();

我遇到$data（第一次查询）查询时出错

[{}]

我得到一个空的回复

我遇到`$ data2（第二次查询）查询错误

SQLSTATE [42000]：语法错误或访问冲突：1064您有错误在你的SQL语法中;检查与MariaDB服务器对应的手册用于在'HAVING COUNT（payments.user_id）＆gt; = 0附近使用的正确语法的版本付款在第1行留下加入用户付款（SQL：选择HAVING 来自付款的COUNT（payments.user_id）＆gt; = 0加入用户 payments.user_id = users.id其中payments.paper_id = 1 group by users.district）

但我得到SQL语法错误有人可以告诉我如何将SQL转换为Laravel查询构建器格式。

Answer 1

试试这个：

使用Raw查询

时

   $data = DB::select('SELECT paper_id,user_id,COUNT(payments.user_id),users.district
            FROM payments
            LEFT JOIN users ON payments.user_id = users.id
            WHERE payments.paper_id='.$paper_id.'
            GROUP BY users.district HAVING 
            COUNT(payments.user_id)>=0 
            ');

OR

        $data2= DB::table('payments')->select([
            'payments.paper_id','payments.user_id','users.district',
            DB::Raw('COUNT(payments.user_id)')
            ])
            ->leftJoin('users','payments.user_id','users.id')
            ->where('payments.paper_id',$paper_id)
            ->havingRaw("COUNT(payments.user_id)>=0")
            ->groupBy('users.district')
            ->get();

我认为select函数应该只在查询中使用一次，不需要添加多个时间。在添加任何questin总是搜索您的问题之前。希望它能起作用..

为您的错误

SQLSTATE [42000]：语法错误或访问冲突：1055

在此文件中 vendor\laravel\framework\src\Illuminate\Database\Connectors\Connector.php 在laravel的这个位置替换此文件laravel5.4 PDO::ATTR_EMULATE_PREPARES => true,

Answer 2

运行原始SQL查询

$data = DB::select('SELECT paper_id,user_id,COUNT(payments.user_id),users.district
                FROM payments
                LEFT JOIN users ON payments.user_id = users.id
                WHERE payments.paper_id= ?
                GROUP BY users.district HAVING 
                COUNT(payments.user_id)>=0', [$paper_id]);

参数绑定可防止SQL注入。

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

groupBy / have

->select(DB::Raw('HAVING COUNT(payments.user_id)>=0'))
=>
->having(DB::Raw('COUNT(payments.user_id)'), '>=', 0)

https://laravel.com/docs/5.6/queries#ordering-grouping-limit-and-offset

在Laravel`DB :: Raw`上运行mysql查询

2 个答案:

运行原始SQL查询

groupBy / have