nike_guy_man — Sat, 31 Aug 2002 15:39:34 +0000

date is 1 field... in the form of date("m\/j\/Y")
I could make it 3 fields... but that would totally complicate things

Mark Hensler — Sat, 31 Aug 2002 07:56:38 +0000

Even easier to through the whole thing at the SQL server...

<?php
// assuming one 'date' field
$query = \"SELECT (TO_DAYS(SYSDATE()) - TO_DAYS(date)) as age FROM dates\";
$result = mysql_query($query);
$row = mysql_fetch_array($result);
echo mysql_error();
$age = $row[age];
echo \"Age is $age\";


// not tested
// assuming 'day', 'month', 'year' fields
$query = \"SELECT (TO_DAYS(SYSDATE()) - TO_DAYS(CONCAT(year,'-',month,'-',day))) as age FROM dates\";
$result = mysql_query($query);
$row = mysql_fetch_array($result);
echo mysql_error();
$age = $row[age];
echo \"Age is $age\";
?>

nike_guy_man — Sat, 31 Aug 2002 05:10:32 +0000

Oh ya know what... that works perfectly with a bit of tweeking
(Just in case someone wants a working version)

<?php
$query = \"SELECT * FROM dates WHERE (date) LIKE ('1') \";
$result = mysql_query($query);
$row = mysql_fetch_array($result);
echo mysql_error();
$day = $row[day]; 
$month = $row[month];
$year = $row[year];

//remember you need bday as day month year 
$bdayunix = mktime (\"\",\"\",\"\",$month,$day,$year); //get unix ts for bday 
$nowunix = time(); //get unix ts for today 
$ageunix = $nowunix - $bdayunix; //work out the difference 
$age = floor($ageunix / (24 * 60 * 60)); //convert from seconds to years 

echo \"Age is $age\"; 
 
?>

Of course, it'd be different values for $row[] but you get the point... that shows up as 242 days... the number of days from January 1st 2002
Thanks a billion busy

nike_guy_man — Sat, 31 Aug 2002 04:55:28 +0000

mktime() is my latest probability... I had just seen that on php.net when I got this reply...
I'll try it out, but if anyone else has better suggestions, I'd love to hear them
Thanks

Busy — Sat, 31 Aug 2002 04:47:36 +0000

here's a script out of "PHP and MYSQL development" book I found

<?php
 
//set date for calculation
$day = 18;
$month = 9;
$year = 1972;

//remember you need bday as day month year
$bdayunix = mktime (\"\",\"\",\"\",$month,$day,$year); //get unix ts for bday
$nowunix = time(); //get unix ts for today
$ageunix = $nowunix - $bdayunix; //work out the difference
$age = floor($ageunix / (365 * 24 * 60 * 60)); //convert from seconds to years

echo \"Age is $age\";
?>

this is for a birthday script to find out how old the person is. Could be easily changed for your needs, I believe
further info on calendar dates at
date and time
http://php.net/manual/ref.datetime.php
http://mysql.com/documentation/mysql/commented/manual.php?section=Date_and_time_functions

calendar functions
http://php.net/manual/ref.calendar.php

nike_guy_man — Sat, 31 Aug 2002 03:57:39 +0000

What do you mean??? How could it count it?
I am running cron programs with this, I could set it up to add 1 day to a table column 'daysince' but that would take a lot of modifying, and I'd prefer to keep it as I have it, if possible

Busy — Sat, 31 Aug 2002 03:55:30 +0000

how about doing it by the hour (or min) then converting that back to days