strtotime not doing what I need

crash3909

Hi,

I have:

$mysql_date = date('Y-m-d',strtotime($_GET['Party_Date']));

I want it to eventually log to my dbase as 2004-02-24 instead of February 24, 2004. The user enters in the form the date for their event and I need it to convert it to yyyy-mm-dd. But when it logs to the db, it logs today's date instead of the date entered by the user. What change(s) can I make to get this to work?

Thanks in advance.

TomD

If you echo "$mysql_date "; and it shows what you want then there may be a problem in your query and not in this conversion.

crash3909

I echoed "$mysql_date ";

and it gave the date that the form was filled out and not the date the user enters. Any ideas for changes I can make would be helpful.

Thanks

bowchow79

If it's using today's date, my best guess is that the date() function feels it's only receiving one valid argument (the format string) and thus assumes the date argument to be today's date (default behavior). This means strtotime isn't understanding the dates your users are typing in. Try experimenting some sort of form with just a textbox and a corresponding script that calls strtotime on what's inside the textbox. Then print out both the resulting timestamp and what date("Y-m-d",$timestamp) print out. Hopefully that will solve your problem. I have a feeling strtotime gets confused on something like "February 24, 2004" because of the comma, or perhaps user typos.

I always use drop down boxes or some sort of neat little calendar to let users choose dates - that way, I have the dates input in a consistant format.

crash3909

Thanks for your reply. I have been using dropdown boxes for the Month, Date and Year. I then have another small code to turn it into the $Party_Date variable. I have tried:

$Party_Date= "$Year $Month $Date";

$Party_Date= "$Year$Month$Date";

$Party_Date= "$Year-$Month-$Date";

and I have hidden code so I can pass it on to the next page that produces:

<input TYPE=HIDDEN NAME="mysql_date" VALUE="2003-12-10"> -- (wrong date)

from:

$mysql_date = date('Y-m-d',strtotime($_GET['Party_Date']));

Once I see the hidden code produce the date entered by the user, I'll have won.

<input TYPE=HIDDEN NAME="mysql_date" VALUE="2004-02-24"> -- (correct date)

Thanks...