Reputation: 53
I'm trying to convert this query into code igniter query. But i don't know where to start. I'm new to code igniter.Please help.
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;
}
Upvotes: 0
Views: 61
Reputation: 1059
You should start at reading the manual, https://www.codeigniter.com/userguide3/database/index.html
Having said that the easiest solution is to simply run the query like:
$this->db->query('YOUR QUERY HERE');
The part that says 'YOUR QUERY HERE' should be replaced with the actual query like:
$this->db->query("SELECT s.* , IF( DATE( s.security_date ) = DATE( '$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'");
Now I'm pretty sure you'd want to clean this up and start using paramaters inside your query instead of PHP variables.
So you should look into the query builder class, https://www.codeigniter.com/userguide3/database/query_builder.html
Upvotes: 0
Reputation: 130
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();
Upvotes: 1
Reputation: 16446
You can convert it in codeigniter query builder as below:
$this->db->select("s.* , IF( DATE( s.security_date ) = DATE( '$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;
Upvotes: 1