Codeigniter: Weird behaviour of $this->db->like()

Question

I\'ve written a simple query for searching a keyword in the database.

$keyword = \"keyword sample\"; 
$keyword = str_replace(\" \", \"%\", $keyword);   

$th         


        

Answer 1

Use the escape_like_str() method.

The escape_like_str() method should be used when strings are to be used in LIKE conditions so that LIKE wildcards %, _ in the string are also properly escaped. It cannot automatically add the ESCAPE ! condition for you, and so you’ll have to manually do that.

Hope it helps.

$keyword = "keyword sample"; 
$sql = "SELECT id FROM table WHERE column LIKE '%" .
$this->db->escape_like_str($keyword)."%' ESCAPE '!'";

Source:- https://www.codeigniter.com/userguide3/database/queries.html

Answer 2

You just try it like this:

$this->db->select('*')
$this->db->from('table');
$this->db->like('column', $keyword);
return $this->db->get()->result_array();

If you want to control where the wildcard (%) is placed, you can use an optional third argument. Your options are 'before', 'after' and 'both' (which is the default).

example:

$this->db->select('*')
$this->db->from('table');
$this->db->like('column', $keyword, 'before');
return $this->db->get()->result_array();

If you do not want to use the wildcard (%) you can pass to the optional third argument the option 'none'.

example:

$this->db->select('*')
$this->db->from('table');
$this->db->like('column', $keyword, 'none');
return $this->db->get()->result_array();

BUT, for your example you must need to search like "%keyword sample%", or like "%keyword%" OR "%simple%";

For example:

$this->db->like('column', 'keyword simple');
// Produces: WHERE column LIKE '%keyword simple%' 

OR

$this->db->like('column', 'keyword');
$this->db->or_like('column', 'simple');
// Produces: WHERE column LIKE '%keyword%' OR column LIKE '%simple%'  

More, you can read CodeIgniter User Guide

Answer 3

If you dig a bit into CodeIgniter's internals, you'll notice that the $this->db->like() function escapes special characters it contains - including, of course, %.

I don't think like() will help you much with your particular needs. Your best bet, I guess, would be to bypass the problem and use a where function containing your LIKE clause:

$this->db->select('*')->from('table')->where("column LIKE '%$keyword%'")->get()->result_array();