php - compare date with database (birthdate) -

so have row called birthday in database value of let's 2001-04-25 (date type). , want see if it's user birthday. current month , day:

$monthday_today = date('m-d'); 

and query:

$query = "select * users date_format(birthday, '%m-%d')=".$monthday_today; 

but doesn't works, thoughts?

you need quote $monthday_today value.

you query (if output it) this:

select * users date_format(birthday, '%m-%d') !=  04-2014 

but not correct sql query, query should this:

select * users date_format(birthday, '%m-%d') !=  '04-2014' 

the simple solution be:

$query = "select * users date_format(birthday, '%m-%d')!="."'".$monthday_today."'"; 

---

the solution use parameterized prepared statements either positional or named parameters:

$query = "select * users date_format(birthday, '%m-%d') != ?"; 

or

$query = "select * users date_format(birthday, '%m-%d') != :monthday_today"; 

you don't need quote parameterized prepared statements because infer (deduce) if quoting necessary or not type of variable.