On Tue, Jul 14, 2009 at 7:33 PM, Mark Wright<[email protected]> wrote:
> I want to insert a variable into a LIKE query using zend_db_statement
> but I'm not sure how to do this correctly. This doesn't work:
>
> $selectObj->where("name LIKE '%?%'", $var );
>
> It gives you "name LIKE '%'some text'%'". I can do this:
>
> $selectObj->where("name LIKE ?", '%' . $var . '%' );
>
> What I really want would be something like this:
>
> $selectObj->where("name LIKE '%" . $adapter->escape($var) . "%'");
That would never work. That's like doing:
$foo->where("name = 'hello ?', 'world'); // 'hell 'world''
The % are part of the value. Can't you do the following?
$foo->where("name LIKE ?', '%' . $var . '%');
Till
