将mysql查询转换为codeigniter

Question

如何将此查询转换为codeigniter框架：

SELECT sname FROM alumni WHERE email IN (SELECT * FROM alumni_user WHERE uname = '')

Answer 1

试试这个..

$this->db->select('*');

$this->db->from('alumni_user');

$this->db->where('uname','');

$result = $this->db->get();

Answer 2

试试这个：

$q = $this->db->query("SELECT sname FROM alumni WHERE email IN (SELECT * FROM alumni_user WHERE uname = '')");

if($q->num_rows() > 0)
{
    $data = $q->result_array();
    print_r($data);
}
else
{
    echo '';
}