我在Laravel 5.1中有一个全局范围设置,运行良好。但是在我的一些页面上,我使用mysqljoins使用雄辩的构建器。这将导致不明确的错误:

Column 'cust_id' in where clause is ambiguous

我不知道如何避免这个问题。我知道我可以使用子查询,但是否没有其他解决方案?
这是我的作用域文件:
<?php

namespace App\Scopes;

use Illuminate\Database\Eloquent\Model;
use Illuminate\Database\Eloquent\Builder;
use Illuminate\Database\Eloquent\ScopeInterface;
use App\Customers;
use DB, Session;

class MultiTenantScope implements ScopeInterface
{
    /**
     * Create a new filter instance.
     *
     * @param  UsersRoles $roles
     * @return void
     */
    public function __construct()
    {
        $this->custId = Session::get('cust_id');
    }

    /**
     * Apply scope on the query.
     *
     * @param Builder $builder
     * @param Model $model
     * @return void
     */
    public function apply(Builder $builder, Model $model)
    {
        if($this->custId)
        {
            $builder->whereCustId($this->custId);
        }
        else
        {
            $model = $builder->getModel();
            $builder->whereNull($model->getKeyName());
        }
    }

    /**
     * Remove scope from the query.
     *
     * @param Builder $builder
     * @param Model $model
     * @return void
     */
    public function remove(Builder $builder, Model $model)
    {
        $query = $builder->getQuery();
        $query->wheres = collect($query->wheres)->reject(function ($where)
        {
            return ($where['column'] == 'cust_id');
        })->values()->all();
    }
}

最佳答案

为了消除字段的歧义,应添加表名作为前缀:

public function apply(Builder $builder, Model $model)
{
    if($this->custId)
    {
        $fullColumnName = $model->getTable() . ".cust_id";
        $builder->where($fullColumnName, $this->custId);
    }
    else
    {
        $model = $builder->getModel();
        $builder->whereNull($model->getKeyName());
    }
}

07-26 09:37