选择数据库时使用AS的codeigniter

我在用户指南中找不到任何关于加入的信息，以下是我正在尝试做的：

SELECT 
c.id
,c.name AS companyName
,con.id AS conID
,l.id AS lid

FROM
     company AS c
LEFT JOIN
     contacts AS con ON
     c.primary_contact = con.id
LEFT JOIN
     locations AS l ON
    c.id = l.cid

任何人都有一个快速的解决方案，或者给我指一下指南中我在AS语句中找不到的部分

SELECT 
    c.id
   ,c.name AS companyName
   ,con.id AS conID
   ,l.id AS lid

FROM
   company c
LEFT JOIN
   contacts con ON
   c.primary_contact = con.id
LEFT JOIN
   locations l ON
   c.id = l.cid

您不必在AS中提及联接表名称，如果在表名称之后给出名称，它将自动为您要联接的表别名

试词

$this->db->select('c.id
                  ,c.name AS companyName
                  ,con.id AS conID
                  ,l.id AS lid'); 
$this->db->from('company c');        
$this->db->join('contacts con','c.primary_contact = con.id','left');
$this->db->join('locations l','c.id = l.cid','left');       

---

我对此不确定，但您可能想试试：

$this->db->select('company.id','name','contacts.id','locations.id');
$this->db->from('company');
$this->db->join('contacts','company.primary_contact = contacts.id','left');
$this->db->join('locations','contacts.id = locations.id','left');

---

我不知道我会按照你将如何添加到模型页面。下面是我的模型页面的样子：`$this->db->select'c.id'$这->数据库->来自“公司为c”$这->数据库->加入'contacts'，'contacts.id=company.primary_contact'$query=$this->db->get；返回$query->result\u数组`很抱歉，我的评论中的代码乱七八糟，显然你不能在评论中添加代码，这看起来很愚蠢。

$this->db->select('t1.*, t2.*')
         ->join('table2 AS t2', 't1.column = t2.column', 'left');

return $this->db->get('table1 AS t1')->result();