Laravel Eloquent 用于向用户列出过滤器名称和 link

Question

我正在使用来自 url 的过滤器来过滤学校，我需要向用户显示这些过滤器的链接。好吧，实际上我只需要向用户展示使用后的过滤器而不是 return 空结果。我尝试使用范围和关系，但出现此错误：

SQLSTATE[42S22]: Column not found: 1054 Unknown column 'specialties.school_id' in 'where clause' (SQL: select `id`, `name` from `specialties` where ((select count(*) from `schools` where `specialties`.`school_id` = `schools`.`id` and `specialties` like %""%) >= 1))

在我的例子中，我需要按专业、地区、直辖市、城市等过滤学校。这是我的示例学校模型：

<?php

class School extends Eloquent {
    protected $table = 'schools';

    public function scopeWhereSpecialties($query, $specialties)
    {
        if( !is_array($specialties) )
        {
            $specialties = [$specialties];
        }

        return $query->where(function($q) use ($specialties)
        {
            foreach($specialties as $specialty){
                $q->where('specialties', 'like', '%"'.$specialty.'"%');
            }
        });
    }

    // I just delete other scopes to shorten the code

    public function listSchoolsEndUser($filters)
    {
        $schools_data = School::query();

        foreach($filters as $filter => $value)
        {
            call_user_func(array($schools_data, 'where' . studly_case($filter)), $value);
        }

        return $schools_data->paginate(12);
    }

    public function listFilters($filters)
    {
        $specialties_filters = Specialty::select('id', 'name')->whereFilterAvailable($filters)->get()->toArray();

        return $specialties_filters;
    }
}

还有我的示例专业模型：

<?php

class Specialty extends Eloquent {
    protected $table = 'specialties';

    public function scopeWhereFilterAvailable($query, $filters)
    {
        $specialty = $this->id;
        return $query->where(function($q) use ($specialty, $filters)
        {
            $q->whereHas('school', function($q) use ($specialty, $filters) {
                $q->where('specialties', 'like', '%"'.$specialty.'"%');
            });
        });
    }

    public function school(){
        return $this->belongsTo('School');
    }
}

Table学校结构类似于：

____________________________________
| id | name    | specialties       |
|____|_________|___________________|
| 1  | example | ["1","2","3","4"] |
|____|_________|___________________|

而table专业结构类似于：

________________
| id | name    |
|____|_________|
| 1  | example |
|____|_________|

Answer 1

问题在于，使用 select('id', 'name') 时，您将查询限制为仅那些属性，因此 school_id 不可用。您应该包括您（和您的关系）需要的所有列：

$specialties_filters = Specialty::select('id', 'name', 'school_id')->whereFilterAvailable($filters)->get()->toArray();

也许 lists() 也是一个选项。它构造一个数组，其中一个属性作为键，另一个作为值：

$specialties_filters = Specialty::whereFilterAvailable($filters)->lists('name', 'id');

Laravel Eloquent 用于向用户列出过滤器名称和 link

Laravel Eloquent for listing filters name and link to user

php

sql

laravel

eloquent