查詢和子查詢在laravel中加入

Question

我在laravel上嘗試這個，但我得到這樣的錯誤

SQLSTATE[42601]: Syntax error: 7 ERROR: subquery in FROM must have an alias
 LINE 3: (select * from subcategory a inner join s...
HINT: For example, FROM (SELECT ...) [AS] foo. (SQL: select * from 
mastercategory c
inner join 
 (select * from subcategory a inner join subling b on 
a.idsubcategory=b.idsubcategory_subcategory) on c.idcategory=a.idsubcategory_subcategory 

我試圖加入mastercatergory與（子類別和subling）的連接子查詢

   $users = DB::select(DB::raw(" select *  from mastercategory c
            inner join  (select * from subcategory a inner join subling b on 
            a.idsubcategory=b.idsubcategory_subcategory) on 
            c.idcategory=a.idsubcategory_subcategory "))->get();

我怎么能做到這一點我不想提前使用模型

這是我的更新

   $users  =   DB::raw(" select a.*,b.*  from mastercategory a
                  inner join  
               (select b.*,c.* from subcategory b inner join subling c  on  
           b.idsubcategory=c.idsubcategory_subcategory) on 
          a.idcategory=b.idcategory_mastercategory"); 

為什么我得到空值

Answer 1

 $users = DB::select("select *  from mastercategory c
            inner join  (select * from subcategory a inner join subling b on 
            a.idsubcategory=b.idsubcategory_subcategory) as d on 
            c.idcategory=d.idsubcategory_subcategory");

當您使用DB::select ，執行該語句並返回結果。 你不需要使用get和DB::raw （ 請點擊這里 ）。

您收到此錯誤是因為您沒有為內部子查詢提供別名。

Answer 2

您必須為每個子查詢命名。 例如，使用此：

select * from mastercategory c
inner join 
  (select * from subcategory a 
   inner join subling b on a.idsubcategory=b.idsubcategory_subcategory) as sub 
on c.idcategory=sub.idsubcategory_subcategory

我添加了粗體部分：a.idsubcategory = b.idsubcategory_subcategory）作為sub