我有两个表。voting_ip和country。我想从country表检索结果，其中voting_ip表的open_id_fk (外键)等于country表的open_id (主键)。如何编写sql查询来组合这些查询并返回结果。我在我的codeigniter模型中使用以下代码来检索voting_ip表中open_id_fk的出现次数。
public function mostvotedans()
{
$this->db->select('open_id_fk, COUNT(*) as total');
$this->db->group_by('open_id_fk');
$this->db->order_by('total', 'desc');
$query = $this->db->get('voting_ip', 5);
return $query;
$query = $this->db->query("SELECT * FROM country WHERE open_id_fk=open_id;");
return $query;
}
修改如下:
public function mostvotedans()
{
$this->db->select('c.open_id,COUNT(ip.open_id_fk) as total')->from('voting_ip ip');
$this->db->join('country c','c.open_id=ip.open_id_fk');
$this->db->group_by('ip.open_id_fk');
$this->db->order_by('total', 'desc');
$this->db->limit(5);
$query = $this->db->get();
return $query;
}
使用join语句:
$query = $this->db->select('v.open_id_fk, COUNT(v.*) AS total, c.country_name')
->join('country AS c', 'c.open_id = v.open_id_fk')
->from('voting_ip AS v')
->group_by('v.open_id_fk')
->order_by('total', 'DESC')
->limit(5);
应该可以，我放了'country_name'因为我不知道你的表