Laravel 全局作用域和连接

Question

我在 Laravel 5.1 中有一个全局范围设置，运行良好。然而，在我的一些页面上，我使用 Eloquent 构建器使用 MySQL joins。这会导致一个模棱两可的错误：

Column 'cust_id' in where clause is ambiguous

我不确定如何避免这个问题。我知道我可以使用子查询来代替，但是没有其他解决方案吗？

这是我的范围文件：

<?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();
    }  
}

Answer 1

为了消除字段的歧义，您应该添加表名作为前缀：

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()); 
    }
}