Preparing A Captured Date for Arithmetical Operations

PHPRookie

I'm building a web site that uses html, php and mysql.

I've used an html text input box that successfully captures a date, $userentereddate, that the site visitor inputs in the month/day/fourdigityear (e.g. 01/31/2002) format.

In my php code I've used a $todaysdate = date("m/d/Y") statement to successfully capture today's date and put it into a format that makes sense to site visitors.

What must I do to $userentereddate so that I can compare it to the date the system has generated for me.

I'd like to compare $userentereddate to $todaysdate to make sure that the user entered date is after today's date. (e.g. if ($userentereddate > $todaysdate) {do whatever I need it to do}

Help!

Thanks in advance.

planetsim

read up on mktime();

that will validate the date for you.. or you could just not let the person inputting the info have to insert the date.. why dont you just yse a var liek you are doing now and not validate what the visitor or user is doing...

It will make it a lot easier on yourself

dalecosp

Check out strtotime(), perhaps. Might be more elegant....

$now=strtotime(date("m/d/Y"));
$userenterdate=strtotime($userenterdate);

if ($now <= $userenterdate) { work work work......

HTH, dalecosp🙂

PHPRookie

planetsim - thanks for the mktime () function tip.

dalecosp - the strtotime() function did the job for me. It was not one of the functions in my "MySQL and PHP for Dummies" book but I was able to read up on it in the "Web Database for Applications -- PHP and MySQL" O'Reilly book.

For the record, both books are excellent -- one for its introductory value and one for its comprehensive reference value.

Thanks to both of you for your help.🆒