INNER JOIN别名sintax

我有这个INNER JOIN别名sintax

选择
p.id， p.name， p.code， clA.code为claCode， sclA.code为sclaCode， clB.code作为clbCode， sclB.code为sclbCode， clC.code为clcCode， sclC.code作为sclcCode

FROM 
      products p 
       INNER JOIN catleathers clA 
        ON p.catleather_id_a = clA.id 
       INNER JOIN subcatleathers sclA 
        ON p.subcatleather_id_a = sclA.id 
       INNER JOIN catleathers clB 
        ON p.catleather_id_b = clB.id 
       INNER JOIN subcatleathers sclB 
        ON p.subcatleather_id_b = sclB.id 
       INNER JOIN catleathers clC 
        ON p.catleather_id_c = clC.id 
       INNER JOIN subcatleathers sclC 
        ON p.subcatleather_id_c = sclC.id

但我需要这样的事

function getdatatableajaxcost() 
    { 

    $this->load->library('datatables'); 
    $this->datatables 

    ->select(" 



      products.id as productid, 
      products.code as code, 
      products.name as name, 

      clA.code as claCode, 
      sclA.code as sclaCode, 
      clB.code as clbCode, 
      sclB.code as sclbCode, 
      clC.code as clcCode, 
      sclC.code as sclcCode, 

      products.details", FALSE) 

      ->from('products') 
      ->join('catleathers clA', 'products.catleather_id_a = clA.id', 'INNER') 
      ->join('subcatleathers sclA', 'products.subcatleather_id_a = sclA.id', 'INNER') 
      ->join('catleathers clB', 'products.catleather_id_a = clB.id', 'INNER') 
      ->join('subcatleathers sclB', 'products.subcatleather_id_a = sclB.id', 'INNER') 
      ->join('catleathers clC', 'products.catleather_id_a = clC.id', 'INNER') 
      ->join('subcatleathers sclC', 'products.subcatleather_id_a = sclC.id', 'INNER') 
      ->group_by("products.id"); 

      $this->datatables->unset_column('productid'); 
      $this->datatables->unset_column('image'); 

      echo $this->datatables->generate();

请一定帮忙！

来源

2014-10-01 gatuso

很难理解你在问什么。 – Rahul 2014-10-01 21:42:41

您是否试图将查询移植到其他语言？ – Degustaf 2014-10-01 21:45:54

不是真的.. 我有一个数据表函数，我需要在里面添加更多的collumns。请看看上面编辑的。这在笨 – gatuso 2014-10-01 21:59:38

解决！谢谢大家！

->select(" 

      products.id as productid, 
      products.image as image, 
      products.code as code, 
      products.name as name, 
      categories.name as cname, 
      subcategories.name as subcategory, 
      products.cost, 
      products.price, 

      clA.code as claCode, 
      sclA.code as sclaCode, 
      clB.code as clbCode, 
      sclB.code as sclbCode, 
      clC.code as clcCode, 
      sclC.code as sclcCode, 

      catSoleA.code as catSoleACode, 
      subcatSoleA.code as subcatSoleACode, 

      catSoleB.code as catSoleBCode, 
      subcatSoleB.code as subcatSoleBCode, 

      catSoleC.code as catSoleCCode, 
      subcatSoleC.code as subcatSoleCCode, 

      products.details", FALSE) 

      ->from('products') 
      ->join('categories', 'products.category_id=categories.id', 'left') 
      ->join('subcategories', 'products.subcategory_id=subcategories.id', 'left') 

      ->join('catleathers clA', 'products.catleather_id_a = clA.id', 'INNER') 
      ->join('subcatleathers sclA', 'products.subcatleather_id_a = sclA.id', 'INNER') 

      ->join('catleathers clB', 'products.catleather_id_b = clB.id', 'INNER') 
      ->join('subcatleathers sclB', 'products.subcatleather_id_b = sclB.id', 'INNER') 

      ->join('catleathers clC', 'products.catleather_id_c = clC.id', 'INNER') 
      ->join('subcatleathers sclC', 'products.subcatleather_id_c = sclC.id', 'INNER') 

      ->join('catsoles catSoleA', 'products.catsole_id_a = catSoleA.id', 'INNER') 
      ->join('subcatsoles subcatSoleA', 'products.subcatsole_id_a = subcatSoleA.id', 'INNER') 

      ->join('catsoles catSoleB', 'products.catsole_id_b = catSoleB.id', 'INNER') 
      ->join('subcatsoles subcatSoleB', 'products.subcatsole_id_b = subcatSoleB.id', 'INNER') 

      ->join('catsoles catSoleC', 'products.catsole_id_c = catSoleC.id', 'INNER') 
      ->join('subcatsoles subcatSoleC', 'products.subcatsole_id_c = subcatSoleC.id', 'INNER') 


      ->group_by("products.id");

来源

2014-10-02 16:40:00 gatuso

INNER JOIN别名sintax

回答

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