CODEWITHSUNDEEP
phpmysqlsqlcodeigniter
Nisham Mahsin
Nisham Mahsin

Reputation: 1399

How to give a query result to where condition in CodeIgniter?

i want to check inside where condition that are results of another query.

in English+sql this where condition is like this where(table.field1=$data[field1] , this $data[field1] should be only from some another query results from another results.)

. i have tried this 

if( isset($data['field1']) && $data['field1'] != '' ) {
                $this->db->where('table.field1',$data['field1'] IN (SELECT id FROM table2' WHERE id=2') );
            }

but this causes error

Upvotes: 1

Views: 1335

Answers (2)

Bira
Bira

Reputation: 5506

public function get_category_with_all_sub() {
    $cquery = $this->db->get ( 'category' );
    $cresults = $cquery->result ();

    $return = array ();
    foreach ( $cresults as $c ) {
        $category_name = $c->name;
        $this->db->select ( '*' );
        $this->db->from ( 'sub_category' );
        $this->db->where ( 'category_id', $c->id );

        $squery = $this->db->get ();

        $sresults = $squery->result ();

        foreach ( $sresults as $s ) {
            $return [$category_name] [] = array (
                    'id' => $s->id,
                    'name' => $s->name,
                    'desc' => $s->description 
            );
        }
    }

    return $return;
}

Please build your model function like this, $c->id is coming from another results.

Upvotes: 1

Mhisham
Mhisham

Reputation: 564

try to do it by that way

if( isset($data['field1']) && $data['field1'] != '' ) { 
   $this->db->where('table.field1',$data['field1']);
   $this->db->where('table.field1 in ','(SELECT id FROM table2 WHERE id=2)',false);
}

Upvotes: 1

Related Questions