codeignitor选择DATABSE

Question

我无法找到用户指南做JOIN任何信息在使用AS，这里是我尝试做的事：

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 

人有一个快速的解决方案，或我指向在引导部分，我似乎无法找到与AS语句

Answer 1

尝试这样

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 

你不必提也加入表名与AS，如果你在表名后给其命名会AUTOMATICA lly别名你的桌子，你加入。

尝试词像

$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');  

Answer 2

林不知道这个，但你可以试试这个：

$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'); 

Answer 3

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

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