php - Laravel SQLSTATE[42000] : Syntax error or access violation

Question

我是 Laravel 的新手，我一直与我的关系人保持联系，如下所示

类别

    id  name            slug
-----------------------------------------------------------------
    3   Location        location
    4   Outfits         outfits
    5   Other           other

子类别

  id    category_id     name                slug
-----------------------------------------------------------------------------
     12     3           Club                club
     13     3           Home / Hotel        home-hotel
     14     3           Outdoor             outdoor
     15     3           Studio              studio
     16     4           Bikini / Swimwear   bikini-swimwear
     17     4           Dress               dress
     19     4           Jeans               jeans
     35     5           Dancing             dancing

类别模型

<?php

class Category extends Eloquent {

    public $timestamps = false;

    public function subcategory()
    {
        return  $this->belongsToMany('subcategory', "sub_categories");
    }
} 

我收到以下错误

SQLSTATE[42000]: Syntax error or access violation: 1066 Not unique table/alias: 'sub_categories' (SQL: select `sub_categories`.*, `sub_categories`.`category_id` as `pivot_category_id`, `sub_categories`.`subcategory_id` as `pivot_subcategory_id` from `sub_categories` inner join `sub_categories` on `sub_categories`.`id` = `sub_categories`.`subcategory_id` where `sub_categories`.`category_id` = ?) (Bindings: array ( 0 => 1, )) 

有人可以指出我做错了什么吗？

Answer 1

您应该为同一个表指定不同的别名

改变这个

 inner join `sub_categories`

至

   inner join `sub_categories` as sc   
                                   ^^--//-this alias use it instead of sub_categories

在您的查询中它将是

    select `sub_categories`.*, sc.`category_id` as `pivot_category_id`, sc.`subcategory_id` as `pivot_subcategory_id` from `sub_categories` inner join `sub_categories` sc on sc.`id` = `sub_categories`.`subcategory_id` where `sub_categories`.`category_id` = ?) (Bindings: array ( 0 => 1, )

示例:

 select * from table1 as t1
 inner join table1 as t2
on t1.id = t2.id

错误说明: 表/别名不唯一

您正在将同一个表连接到不同的表，而无需使用别名。