Re: [PHP] Date again

2002-12-06 Thread Stephen
Not sure but I think now() doesn't return the date formate you'd want assuming your date section is classified as a date. Here's what I would do: $date = date(Y-m-d); $sql = SELECT * FROM table WHERE date='.$date'; $result = mysql_query($sql, $connection); Hope it works! - Original Message

Re: [PHP] Date again

2002-12-06 Thread Jason Wong
On Saturday 07 December 2002 11:39, Miguel BrĂ¡s wrote: Ok guys, having a problem here. I did my table with 6 fields, they are: id position name timein timeout date I have a form to insert the info on the fields, and at the date field, I have a drop down menu that returns me the next 7