Php Time help

Okay im trying to create a script which fetches the date, adds three days, then updates this new date and time in the mysql db.

I've read the php manual on time, and I'm still confused. Im useless.


So far I have this:

$mysql_host    = "localhost";
$mysql_username = "username";
$mysql_password = "password";
$mysql_database = "database";
$id   = $_REQUEST['id'];


mysql_query("UPDATE db.table SET time=$thetime WHERE id='"$id"'") or die(mysql_error());
mysql_close();

PHP:

I need the time/date format like this:

2010-10-04 17:45:21


Please help, thanks

 

Figured it out now... took me awhile but I came up with this, please advise if anyone else can think of better method.

<?php

$year = date("Y:m:d");
$time = date("G:i:s");
$future = date("Y:m:d", strtotime("+3 day"));

print "$year $time - - $future $time ";

?>

PHP:

 

Okay, I am having a problem with script... please help...

Error:

Parse error: syntax error, unexpected T_VARIABLE

Im guessing in the mysql query, please help.

<?php

include('include/config.php');
include('include/function.php');

$future = date("Y:m:d G:i:s", strtotime("+3 day"));

$mysql_host    = "localhost";
$mysql_username = "user";
$mysql_password = "pass";
$mysql_database = "db";

$UID = $_REQUEST['UID'];

mysql_query("UPDATE subscriber SET expired_time='"$future"' WHERE UID='"$UID"'") or die(mysql_error());
mysql_close();


?>

PHP:

 

i thing you have not connected to mysql. did u? atleast can you tell what error says?

 

Try this

mysql_query("UPDATE subscriber SET expired_time='".$future."' WHERE UID='".$UID."'") or die(mysql_error());

 

this is correct
mysql_query("UPDATE subscriber SET expired_time='$future' WHERE UID='$UID'") or die(mysql_error());

 

You can probably just do it a lot easier and faster by doing it directly in SQL. Have a look at the adddate function: http://dev.mysql.com/doc/refman/5.1/en/date-and-time-functions.html#function_adddate

 

How about:

mysql_query("UPDATE `db`.`table` SET `time` = ADDDATE(NOW(), 3) WHERE `id` = '$id'") or die(mysql_error());

Code (markup):

 

thanks for the tips

 

yes it is correct.....