get column with the same value on the database using codeigniter - php

Question

Here are my resources first
As you can see on my database I have a Referal_Code and Referral_Link
Now what I am trying to here is I want to get all those Referral_Link that has the same value on the Referal_Code
I know I am no where near to what I am trying to do but I am trying my best really to get this done.
Here what I have tried so far
On my model
public function get_same_referral_code()
{
$this->db->select('referal_code');
$this->db->where('referal_code', 'PoVsw0Epne');
$query = $this->db->get('investor');
if ($query->num_rows() > 0) {
foreach ($query->result() as $row) {
$data[] = $row;
}
return $data;
}
}
and on my controller
public function get_all()
{
$data = array();
$data['title'] = "Rewards";
$data["data"] = $this->investor->get_all_investor();
$data["get_all_flevel"] = $this->investor->get_all_first_level();
$data["referrals"] = $this->investor->get_same_referral_code();
$this->template->load('default_layout','contents','rewards',$data);
}
and I am calling it on my view like this
<?php
echo ($referrals[0][id]);
echo ($referrals[0][firstname]);
echo ($referrals[0][referal_code]);
echo ($referrals[0][referral_link]);
?>
What my expected result should be like this
Could someone help me .
EDIT
I've tried again this way to get what I want .
First I tried to get all the referal_code
public function get_same_referral_code()
{
$this->db->select('referal_code');
$query = $this->db->get('investor');
if ($query->num_rows() > 0) {
foreach ($query->result() as $row) {
$data[] = $row;
}
return $data;
}
}
and now I will going to use this method get_same_referral_code() to check if
referral_link has the same value on the referal_code
public function get_same_referral_link()
{
$this->db->select('referral_link');
$this->db->where('referral_link','get_same_referral_code();');
$query = $this->db->get('investor');
if($query->num_rows() > 0)
{
foreach($query->result() as $row)
{
$data[] = $row;
}
return $data;
}
}
but the problem is that I got this error now
Message: Invalid argument supplied for foreach()

Answer 1

i think sub query will help you
$SQL= "SELECT * FROM tableName WHERE referal_code EXISTS (SELECT DISTINCT Referral_Link FROM tableName)";
$query = $this->db->query($SQL);
return $query->result_array();
or
$SQL= "SELECT * FROM investor WHERE referral_link IN (SELECT DISTINCT referal_code FROM investor)";
$query = $this->db->query($SQL);
return $query->result_array();