两次加入同一张桌子

Question

使用Codeigniter活动记录类我正在模型中加入我的表：

  $subquery =
            $this->db
            ->from('houseconstructiontypelist')
            ->where('HouseFk', $id)
            ->where('discount_id', '0')
            ->get_compiled_select();

  $query = 
            $this->constructiontype_model
            ->select('*, constructiontype.PkID AS ConstructionTypeID')
            ->join('('.$subquery.') t2', 'constructiontype.PkID = t2.ConstructionTypeFk', 'left outer')
            ->find_all();

  return $query;

并且因为第二个表中没有与'discount_id' = 0匹配的行，我得到了我的结果：

array (size=5)
  0 => 
    object(stdClass)[50]
      public 'PkID' => null
      public 'Name' => string 'Standard' (length=8)
      public 'Descr' => string 'Standard' (length=8)
      public 'HouseFk' => null
      public 'ConstructionTypeFk' => null
      public 'Price' => null
      public 'Discount' => null
      public 'Subvention' => null
      public 'discount_id' => null
      public 'ConstructionTypeID' => string '1' (length=1)
  1 => 
    object(stdClass)[51]
      public 'PkID' => null
      public 'Name' => string 'Prestige' (length=8)
      public 'Descr' => string 'Prestige' (length=8)
      public 'HouseFk' => null
      public 'ConstructionTypeFk' => null
      public 'Price' => null
      public 'Discount' => null
      public 'Subvention' => null
      public 'discount_id' => null
      public 'ConstructionTypeID' => string '2' (length=1)

我得到了第一个表（constructiontype）的所有结果，无论第二个表（houseconstructiontypelist）中是否有任何匹配结果，一切正常。而且总会有一个！在第二个表中匹配结果，请注意子查询的->where('discount_id', '0')部分。

现在如果我还要加入第二个表->where_not_in('discount_id', '0')的其他结果怎么办？但我也想保留第一个查询给我的结果！我想得到这个：

array (size=5)
  0 => 
    object(stdClass)[50]
      public 'PkID' => null
      public 'Name' => string 'Standard' (length=8)
      public 'Descr' => string 'Standard' (length=8)
      public 'HouseFk' => null
      public 'ConstructionTypeFk' => null
      public 'Price' => null
      public 'Discount' => null
      public 'Subvention' => null
      public 'discount_id' => null
      public 'ConstructionTypeID' => string '1' (length=1)
  1 => 
    object(stdClass)[51]
      public 'PkID' => string '610' (length=3)
      public 'Name' => string 'Standard' (length=8)
      public 'Descr' => string 'Standard' (length=8)
      public 'HouseFk' => string '135' (length=3)
      public 'ConstructionTypeFk' => string '1' (length=1)
      public 'Price' => string '4564' (length=4)
      public 'Discount' => string '0' (length=1)
      public 'Subvention' => string '0' (length=1)
      public 'discount_id' => string '3' (length=1)
      public 'ConstructionTypeID' => string '1' (length=1)
  2 => 
    object(stdClass)[51]
      public 'PkID' => null
      public 'Name' => string 'Prestige' (length=8)
      public 'Descr' => string 'Prestige' (length=8)
      public 'HouseFk' => null
      public 'ConstructionTypeFk' => null
      public 'Price' => null
      public 'Discount' => null
      public 'Subvention' => null
      public 'discount_id' => null
      public 'ConstructionTypeID' => string '2' (length=1)

数组（1）的第二个元素现在填充了值，因为第二个表中存在匹配，其中第一个元素（0）具有空值，因为就像我说的那样仍然没有匹配！但是我需要返回那些空值。

我尝试使用第二个子查询（使用->where_not_in('discount_id', '0')）和另一个sql连接部分：

    $subquery =
            $this->db
            ->from('houseconstructiontypelist')
            ->where('HouseFk', $id)
            ->where('discount_id', '0')
            ->get_compiled_select();

    $subquery2 =
            $this->db
            ->from('houseconstructiontypelist')
            ->where('HouseFk', $id)
            ->where_not_in('discount_id', '0')
            ->get_compiled_select();

    $query = 
            $this->constructiontype_model
            ->select('*, constructiontype.PkID AS ConstructionTypeID')
            ->join('('.$subquery.') t2', 'constructiontype.PkID = t2.ConstructionTypeFk', 'left outer')
            ->join('('.$subquery2.') t3', 'constructiontype.PkID = t3.ConstructionTypeFk', 'left outer')
            ->find_all();

但是这只给了我带有填充值的结果，因为它找到了匹配并且它完全忽略了第一次使用子查询->where('discount_id', '0')的连接调用：

  0 => 
    object(stdClass)[51]
      public 'PkID' => string '610' (length=3)
      public 'Name' => string 'Standard' (length=8)
      public 'Descr' => string 'Standard' (length=8)
      public 'HouseFk' => string '135' (length=3)
      public 'ConstructionTypeFk' => string '1' (length=1)
      public 'Price' => string '4564' (length=4)
      public 'Discount' => string '0' (length=1)
      public 'Subvention' => string '0' (length=1)
      public 'discount_id' => string '3' (length=1)
      public 'ConstructionTypeID' => string '1' (length=1)
  1 => 
    object(stdClass)[51]
      public 'PkID' => null
      public 'Name' => string 'Prestige' (length=8)
      public 'Descr' => string 'Prestige' (length=8)
      public 'HouseFk' => null
      public 'ConstructionTypeFk' => null
      public 'Price' => null
      public 'Discount' => null
      public 'Subvention' => null
      public 'discount_id' => null
      public 'ConstructionTypeID' => string '2' (length=1)