如何在带有绑定参数的代码iginiter query（）中使用like子句？

Question

How to use like clause in code iginiter query() WITH BIND PARAMS?

Eg: When I use

$query = 'SELECT mycol FROM mytable WHERE name LIKE %?';
$name = 'foo';
$db->this->query($query,array($name));
//the clause generated
//SELECT mycol FROM mytable WHERE id LIKE '%'foo'%' 
//I expected this
//SELECT mycol FROM mytable WHERE id LIKE '%foo'

I don't to put param values inside query and use like below: $query = 'SELECT mycol FROM mytable WHERE name LIKE '%foo';

Also I can not use $this->db->like() function as my query consists of: INSERT IGNORE and INSERT INTO table SELECT col FROM table2;

Please suggests? Thanks,

Answer 1

just use CodeIgniter Query Builder

$name = 'foo';

$query = $this
    ->db
    ->select('mycol')
    ->like('name', $name)
    ->get('mytable');

Answer 2

$query = 'SELECT mycol FROM mytable WHERE name LIKE ?';
$name = '%foo';
$this->db->query($query,array($name));

Answer 3

codeigniter will replace ? with 'params' value. if you write this

 $query = 'SELECT mycol FROM mytable WHERE name LIKE %?';
 $name = 'foo';
 //$db->this->query($query,array($name)); //you wrote this line wrong.
 //it should be like this
  $this->db->query($query,array($name));

it will produce

SELECT mycol FROM mytable WHERE name LIKE %'foo' //inverse comma after % ,actually before and after foo.

So your right way will be

 $query = 'SELECT mycol FROM mytable WHERE name LIKE ?';
 $name = '%foo';    
 $this->db->query($query,array($name));

It will produce

SELECT mycol FROM mytable WHERE name LIKE '%foo'

NOTE
You wrote this which is wrong

$db->this->query($query,array($name));

Right way

 $this->db->query($query,array($name));