Laravel 分页不适用于 group by 子句

Laravel pagination not working with group by clause

提问by Anam

It seems Laravel pagination deos not working properly with group by clause. For example:

Laravel 分页 deos 似乎无法与 group by 子句一起正常工作。例如：

            $users = Subject::select(DB::raw('subjects.*, count(user_subjects.id) as total_users'))
            ->join('user_subjects', 'user_subjects.subject_id', '=', 'subjects.id')
            ->whereNull('user_subjects.deleted_at')
            ->groupBy('subjects.id')
            ->orderBy('subjects.updated_at', 'desc')
            ->paginate(25);

Produced

出品

            select subjects.*, count(user_subjects.id) as total_users 
            from `subjects` inner join `user_subjects` on `user_subjects`.`subject_id` = `subjects`.`id` 
            where `subjects`.`deleted_at` is null and `user_subjects`.`deleted_at` is null 
            group by `subjects`.`id` 
            order by `subjects`.`updated_at` desc

note that, there is no limitclause on the query.

请注意，limit查询中没有子句。

Working fine if no group by clause in the query:

如果查询中没有 group by 子句，则工作正常：

$users = Subject::select(DB::raw('subjects.*, count(user_subjects.id) as total_users'))
            ->join('user_subjects', 'user_subjects.subject_id', '=', 'subjects.id')
            ->whereNull('user_subjects.deleted_at')
            ->orderBy('subjects.updated_at', 'desc')
            ->paginate(25);

produced the following query:

产生了以下查询：

select subjects.*, count(user_subjects.id) as total_users from `subjects` 
            inner join `user_subjects` on `user_subjects`.`subject_id` = `subjects`.`id`
            where `subjects`.`deleted_at` is null and `user_subjects`.`deleted_at` is null 
            order by `subjects`.`updated_at` desc 
            limit 25 offset 0

does anyone has any idea how can i fix this?

有谁知道我该如何解决这个问题？

回答by Mahendran Kannan

Check the documentation https://laravel.com/docs/5.2/pagination

检查文档 https://laravel.com/docs/5.2/pagination

Currently, pagination operations that use a groupBystatement cannot be executed efficiently by Laravel. If you need to use a groupBywith a paginated result set, it is recommended that you query the database and create a paginator manually.

目前，groupByLaravel 无法有效执行使用语句的分页操作。如果您需要使用groupBy带有分页结果集的 a，建议您查询数据库并手动创建分页器。

回答by naT erraT

This works for me in laravel 5.2

这在 Laravel 5.2 中对我有用

Select(\DB::RAW("assignment_descendant_child.assignment_descendant_child_id, assignment_descendant_child.assignment_descendant_child_name, COUNT(assignment_descendant.assignment_descendant_id) as xNum"))
            ->leftJoin(
                'assignment_descendant',
                'assignment_descendant.assignment_descendant_child_id',
                '=',
                'assignment_descendant_child.assignment_descendant_child_id'
            )
            ->orderBy('assignment_descendant_child_name')
            ->groupBy('assignment_descendant_child.assignment_descendant_child_id')
            ->paginate(\Config::get('constants.paginate_org_index'))

回答by user8838201

I think this works if you want to group by and paginate.

如果你想分组和分页，我认为这有效。

$code = DB::table('sources')
->select(DB::raw('sources.id_code,sources.title,avg(point) point'))
->join('rating','sources.id_code','rating.id_code')
->groupBy('sources.id_code')
->groupBy('sources.title')
->groupBy('sources.language')
->groupBy('sources.visited')
->groupBy('sources.')
->paginate(5);

回答by Atiqur

1. create a database viewnamedvw_anything. MySql query will be like

   create view vw_anything as select subjects.*, count(user_subjects.id) as total_users from subjects inner join user_subjects on user_subjects.subject_id = subjects.id wheresubjects.deleted_at is null and user_subjects.deleted_at is null group bysubjects.id;

2. Now create a new model named UserSubModelfor this view, protected $table = 'vw_anything';

3. Now your paginate query will be like UserSubModel::orderBy('subjects.updated_at', 'desc')->paginate(25);

1. 创建一个view名为vw_anything. MySql 查询会像

   create view vw_anything as select subjects.*, count(user_subjects.id) as total_users from subjects inner join user_subjects on user_subjects.subject_id = subjects.id wheresubjects. deleted_at is null and user_subjects. deleted_at is null group bysubjects. id;

2. 现在为此视图创建一个名为UserSubModel的新模型，protected $table = 'vw_anything';

3. 现在你的分页查询会像 UserSubModel::orderBy('subjects.updated_at', 'desc')->paginate(25);

.

.

To answer this questioin Laravel Pagination group by year and month only

仅按年和月回答Laravel 分页组中的此问题

View query will be :

查看查询将是：

create view vw_anything as select gallery.*, DATE_FORMAT(created_at, "%Y-%m") as tanggal,count(created_at) as jumlah from gallery group by tanggal;

Let you model is VwModel then your paginate query will be

让你的模型是 VwModel 那么你的分页查询将是

VwModel::where('type','Foto')->orderBy('tanggal','desc')->paginate(2);

回答by Rafik Farhad

I know it is an old question, by I am sharing my solution for future reference.

我知道这是一个老问题，我正在分享我的解决方案以供将来参考。

I managed to write a function based on this linkwhich does the heavy job of determining the pagination of a complex query. Just pass the 'QueryBuilder' and it will return the paginated object/collection.

我设法根据此链接编写了一个函数，该函数完成了确定复杂查询分页的繁重工作。只需传递“QueryBuilder”，它就会返回分页的对象/集合。

Additionally, this procedure can track and maintain the other parameters except for page=.

此外，此过程可以跟踪和维护除 之外的其他参数page=。

public function mergeQueryPaginate(\Illuminate\Database\Eloquent\Builder $query): \Illuminate\Pagination\LengthAwarePaginator
    {
        $raw_query = $query;
        $totalCount = $raw_query->get()->count();

        $page = request('page', 1);
        $skip = $perPage * ($page - 1);
        $raw_query = $raw_query->take($perPage)->skip($skip);

        $parameters = request()->getQueryString();
        $parameters = preg_replace('/&page(=[^&]*)?|^page(=[^&]*)?&?/', '', $parameters);
        $path = url(request()->getPathInfo() . '?' . $parameters);

        $rows = $raw_query->get();

        $paginator = new LengthAwarePaginator($rows, $totalCount, $perPage, $page);
        $paginator = $paginator->withPath($path);
        return $paginator;
    }