Styles2304 Posted August 25, 2007 Share Posted August 25, 2007 Ok, if anyone read my previous post they'll know a little bit about what I'm talking about. I got the calendar to do everything I need it to except for one thing. The date field in mysql requires the date be 2007-08-07. For some reason, the code has no problem recognizing 08 as 8 but it doesn't recognize 07 as 7. The calendar itself spits out 7 as opposed to 07 so when I compare the two, it doesn't recognize it as part of the array. Does anyone know how to remedy this? Here is the code that actually does the checking: if ($EventDate[$i]) { echo "<td width='24' height='16' class='event'><a href='showevent.php?day=$i' class='calendar'>$i</a></td>"; So like I said it's checking to see if, in this particular case, $EventDate[7] exists but actually . . . it's $EventDate[07] . . . does that make sense? I would think the easiest way would be to convert 07 to 7 someway . . . but I don't know how. Quote Link to comment Share on other sites More sharing options...
AndyB Posted August 25, 2007 Share Posted August 25, 2007 07 is a string. You need to convert it to a number, intval() will do that. Quote Link to comment Share on other sites More sharing options...
Styles2304 Posted August 25, 2007 Author Share Posted August 25, 2007 Thank you . . . it is complete! lol. Seriously, thanks. Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.