mysql query to check a field exist in an array in codeigniter - php

Question

I post an array of skills from the form and i need to fetch list of candidates list related to the elements in array in codeigniter.
Controller
$data = array(
"mode" => $this->input->post("mode"),
"designation" => $this->input->post("designation"),
"notice_period" => $this->input->post("notice_period"),
"skill" => $this->input->post("skills"),
"cities" => $this->input->post("city")
);
$this->load->model("search_model");
$this->load->model("candidate_model");
$data["candidate"] = $this->search_model->fetch_candidate($data);
$data["designation"] = $this->candidate_model->fetch_designation();
$this->load->view('candidate_profiles', $data);
model
function fetch_candidate($data)
{
$mode = $data['mode'];
$designation = $data['designation'];
$notice_period = $data['notice_period'];
$skill = $data['skill'];
$skills = join(",",$skill);
$city = $data['cities'];
$query = $this->db->query("SELECT DISTINCT(U.user_id), U.firstname, U.lastname, U.username, U.usertype, U.email, U.phone, U.address, U.profile_image, C.name AS country,C.id AS countryid, S.name AS state,S.id AS stateid, A.housename, A.street, A.area, A.po, A.city, CJ.designation, CJ.resume, CJ.biography, CJ.hiring_mode, CJ.notice_period, CJ.current_CTC, CJ.expected_CTC, D.designation_id AS desig, D.designation AS desig_name
FROM users AS U
LEFT JOIN address AS A ON U.user_id=A.cand_id
LEFT JOIN cand_job_details AS CJ ON CJ.cand_id=U.user_id
LEFT JOIN designations AS D ON D.designation_id=CJ.designation
LEFT JOIN countries AS C ON C.id=A.country
LEFT JOIN states AS S ON S.id=A.state
LEFT JOIN user_skills AS US ON U.user_id=US.user_id
LEFT JOIN user_interested_cities AS UC ON U.user_id=UC.user_id
WHERE D.designation LIKE '$designation' && (CJ.hiring_mode='2' || CJ.hiring_mode='$mode') && CJ
.notice_period<='$notice_period' && UC.city='$city' && US.skill IN(".implode(",", $skills).")
");
return $query;
}
How to implement this on codeigniter?
Thanks in advance

Answer 1

You need try to add ' around each argument. After it should looks like IN ('PHP','AngularJS'), because it is a string value and datatype is varchar as well.
$skills = "'";
foreach($skill as $item) {
$skills .= $item."',";
}
$skills = trim($skills,",")

Answer 2

Codeigniter comes with a built in Querybuilder - i strongly suggest to use it, because it makes your life much easier, and protects you against SQL injections - which is clearly a problem in your case.
Try the following
function fetch_candidate($data)
{
$this->db
->select('SELECT DISTINCT(U.user_id), U.firstname, U.lastname, U.username, U.usertype, U.email, U.phone, U.address, U.profile_image, C.name AS country,C.id AS countryid, S.name AS state,S.id AS stateid, A.housename, A.street, A.area, A.po, A.city, CJ.designation, CJ.resume, CJ.biography, CJ.hiring_mode, CJ.notice_period, CJ.current_CTC, CJ.expected_CTC, D.designation_id AS desig, D.designation AS desig_name')
->from('users U')
->join('address A', 'U.user_id=A.cand_id', 'left')
->join('cand_job_details CJ', 'CJ.cand_id=U.user_id', 'left')
->join('designations D', 'D.designation_id=CJ.designation', 'left')
->join('countries C', 'C.id=A.country', 'left')
->join('states S', 'S.id=A.state', 'left')
->join('user_skills US', 'U.user_id=US.user_id', 'left')
->join('user_interested_cities UC', 'U.user_id=UC.user_id', 'left');
if (isset($data['designation']) && !empty($data['designation']))
{
$this->db->like('D.designation', $data['designation']);
}
if (isset($data['mode']) && !empty($data['mode']))
{
$this->db
->group_start()
->where('CJ.hiring_mode', 2)
->or_where('CJ.hiring_mode', $data['mode'])
->group_end();
}
if (isset($data['notice_period']) && !empty($data['notice_period']))
{
$this->db->where('CJ.notice_period <=', $data['notice_period']);
}
if (isset($data['city']) && !empty($data['city']))
{
$this->db->where('UC.city', $data['city']);
}
if (isset($data['skills']) && is_array($data['skills']))
{
$this->db->where_in('US.skill', $data['skills']);
}
return $this->db->get();
}