过滤分页的口才集合 [英] filtering a paginated eloquent collection

问题描述

我试图过滤分页的口才收集，但是每当使用任何一种收集方法时，我都会失去分页。

I am trying to filter a paginated eloquent collection, but whenever I use any of the collection methods, I lose the pagination.

$models = User::orderBy('first_name','asc')->paginate(20);

$models = $models->each(function($model) use ($filters) {
    if(!is_null($filters['type'])) {
        if($model->type == $filters['type'])
            return $model;
    }

    if(!is_null($filters['state_id'])) {
        if($model->profile->state_id == $filters['state_id'])
            return $model;
    }

    if(!is_null($filters['city_id'])) {
        if($model->profile->city_id == $filters['city_id'])
            return $model;
    }
});

return $models;

我正在使用Laravel 4.2，是否有任何方法可以保持分页？

I am working with Laravel 4.2, is there any way to persist the pagination?

推荐答案

根据具体情况扩展mininoz的答案：

Expanding on mininoz's answer with your specific case:

//Start with creating your object, which will be used to query the database

$queryUser = User::query();

//Add sorting

$queryUser->orderBy('first_name','asc');

//Add Conditions

if(!is_null($filters['type'])) {
    $queryUser->where('type','=',$filters['type']);
}

if(!is_null($filters['state_id'])) {
    $queryUser->whereHas('profile',function($q) use ($filters){
        return $q->where('state_id','=',$filters['state_id']);
    });
}

if(!is_null($filters['city_id'])) {
    $queryUser->whereHas('profile',function($q) use ($filters){
        return $q->where('city_id','=',$filters['city_id']);
    });
}

//Fetch list of results

$result = $queryUser->paginate(20);

通过对SQL查询应用适当的条件，您可以限制返回到您的PHP脚本，从而加快了处理过程。

By applying the proper conditions to your SQL query, you are limiting the amount of information that comes back to your PHP script, and hence speeding up the process.

来源： http://laravel.com/docs/4.2/eloquent#querying-relations

这篇关于过滤分页的口才集合的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！