CODEWITHSUNDEEP
codeignitercodeigniter-query-buildercodeigniter-database
ChocoMartin
ChocoMartin

Reputation: 111

How to convert this SQL Query on Code Igniter Model

Good Day Masters, Can anyone help me how to convert this SQL Query into Code Igniter format (model).

SELECT firstName,  FLOOR(DATEDIFF(CURRENT_DATE, birthDate)/365.25) as age  FROM residents_tbl WHERE  FLOOR(DATEDIFF(CURRENT_DATE, birthDate)/365.25) >= 18

I don't know how to write it on WHERE clause.

    $query = $this->db->select('*');
    $query = $this->db->from('residents_tbl');
    **$query = $this->db->where('isHead', '1');**
    $query = $this->db->order_by('lastName', 'ASC');
    $query = $this->db->get('', 15, $this->uri->segment(3));
    if ($query->num_rows() > 0) {
        return $query->result();
    }

TIA.

Upvotes: 0

Views: 43

Answers (1)

gazdagergo
gazdagergo

Reputation: 6691

This is a simplified version with chaining. I just changed the type of 1 from string to number which might caused the problem.

$query = $this->db
         ->where('isHead', 1)
         ->get('residents_tbl')
         ->order_by('lastName', 'ASC');

Upvotes: 1

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