我正在尝试将此查询转换为代码点火器查询。但我不知道从哪里开始。我是代码点火器的新手。请帮忙。
public function get_loginsecuritydetails($security_date,$apt_id) {
$sql = "SELECT s.* , IF( DATE( s.security_date ) = DATE( '$security_date' ) , 1, 0 ) AS loginstatus,a.agency_companyname FROM security s JOIN agency a ON(a.agency_id=s.security_cat) WHERE apartment_id='$apt_id'; ";
$res=mysqli_query($sql) or die(mysqli_error());
return $res;
}
您可以在代码点火器查询生成器中转换它,如下所示:
$this->db->select("s.* , IF( DATE( s.security_date ) = DATE( '$security_date' ) , 1, 0 ) AS loginstatus,a.agency_companyname");
$this->db->join("agency as a","a.agency_id=s.security_cat");
$this->db->where('apartment_id',$apt_id);
$query = $this->db->get('security as s');
$result = $query->result();
return $result;
Try like this -
$this->db->select("s.* , IF( DATE( s.security_date ) = DATE( '$security_date' ) , 1, 0 ) AS loginstatus,a.agency_companyname");
$this->db->join("agency as a","a.agency_id=s.security_cat");
$this->db->where('apartment_id',$apt_id);
$querys = $this->db->get('security as s');
$result = $querys->result();
你应该从阅读手册开始,https://www.codeigniter.com/userguide3/database/index.html
话虽如此,最简单的解决方案是简单地运行查询,如下所示:
$this->db->query('YOUR QUERY HERE');
显示"您的查询在此处"的部分应替换为实际查询,例如:
$this->db->query("SELECT s.* , IF( DATE( s.security_date ) = DATE( '$security_date' ) , 1, 0 ) AS loginstatus,a.agency_companyname FROM security s JOIN agency a ON(a.agency_id=s.security_cat) WHERE apartment_id='$apt_id'");
现在我很确定你想要清理它并开始在你的查询中使用参数而不是PHP变量。
因此,您应该研究查询生成器类,https://www.codeigniter.com/userguide3/database/query_builder.html