CODEWITHSUNDEEP
phpmysqlsqlcodeigniter
gadss
gadss

Reputation: 22489

CodeIgniter where and like sql query statement

I am trying to do this sql query on codeigniter

SELECT* FROM person WHERE type = 'staff' AND description LIKE 'university%'

I am not sure if it is correct...

$this->db->get_where('person' , array('type'=>'staff'))
         ->like('description','university%')
         ->result_array();

does anyone have an idea about my case? thanks in advance ...

Upvotes: 7

Views: 20775

Answers (3)

M Khalid Junaid
M Khalid Junaid

Reputation: 64466

Using Active Record you can do so 

$query = $this->db->select('*')
             ->from('person')
             ->where('type','staff')
             ->like('description','university','after')
             ->get();

$result = $query->result_array();

Make sure you pass after as a third parameter in like() function so active record will add the wild card i.e % after university so it will look like LIKE 'university%'

Upvotes: 9

Patrick
Patrick

Reputation: 3367

I never used chaining(even though i know it's possible), but breaking down your question should be easy ; 

$this->db->from('person');
$this->db->where('type','staff');
$this->db->where('description', 'university%');
$query = $this->db->get();
$result = $query->result_array();
return $result;

Upvotes: 0

Ryan
Ryan

Reputation: 14649

The docs has this explained very well, but their API is very simple. To produce the query you need the relevant code which looks like this:

$this->db->select('*')
     ->from('person');
     ->where('type', 'staff')
     ->like('description','university');

$query  = $this->db->get();
$result = $query->result_array();

Upvotes: 0

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