我想用codeigniter做一个内部联接,我已经尝试了很多但没有成功。
内连接的Mysql代码,
SELECT * FROM shop INNER JOIN city ON city.city_id = shop.city_id WHERE city.city_name = 'Bangalore'
上面的SQL查询在phpmyadmin中运行得很好。在codeigniter中转换此代码时,它无效。
Codeigniter代码是,
$this->db->select('*');
$this->db->from('shop');
$this->db->join('city', 'city.city_id = shop.city_id');
$query = $this->db->where('city', array('city.city_name' => 'Bangalore'));
if($query->num_rows() > 0) {
return $result = $query->result_array();
}
else {
return 0;
}
哪里出错了。我是codeigniter的初学者。
答案 0 :(得分:2)
您的查询应运行良好:
$this->db->select('*');
$this->db->from('shop');
$this->db->join('city', 'city.city_id = shop.city_id');
//$this->db->where('city', array('city.city_name' => 'Bangalore'));
$this->db->where('city.city_name', 'Bangalore');
$query = $this->db->get();
答案 1 :(得分:1)
试试这个:
$query = $this->db->select('*')
->from('shop')
->join('city', 'city.city_id = shop.city_id', 'inner')
->where('city.city_name', 'Bangalore');
->get();
答案 2 :(得分:1)
您错过了来自db的<{1}}数据,
将此更改为
get()
此
$query = $this->db->where('city', array('city.city_name' => 'Bangalore'));
您无需$this->db->where('city', array('city.city_name' => 'Bangalore'));
$query = $this->db->get();// add this line.
即可使用from()
并将表名传递给它。
get()