PHP TIME() and birthday

[keyboard]

Is there a way to use time() on a website without relying on the user having the correct time set on their pc.


Also for a registration form, if I had three options Date Month Year for them to fill in their birthday how coan I work out how old they are and when it's their birthday, update the sql database with their new age?

[djr33]

time() works on the server, and on the server only. It doesn't matter what time their OS claims it is. This is good, but it also can be a problem if they are in a different timezone. You can use some complicated methods to determine a timezone offset potentially if needed.


Birthdays are easy if you use timestamps. It's just some math to convert DD MM YYYY to a timestamp, then you can do any math you'd like, usually just by adding or subtracting a certain value, such as 60*60*24*365 (1 year). But there are other ways to manipulate timestamps using default functions. Look into this and you'll find your answers.

[keyboard]

Could anyone please give me a good website which has information about timestamps?

[JShor]

Birthdays are easy if you use timestamps. It's just some math to convert DD MM YYYY to a timestamp, then you can do any math you'd like, usually just by adding or subtracting a certain value, such as 60*60*24*365 (1 year).

You're forgetting about leap years. You'd need a conditional to determine how many leap year years have passed, add them together, multiply those days by 60*60*24 and subtract that value from the original equation.

You could attempt to convert a timestamp in DD MM YYYY format using strtotime(), which would give you a universal UNIX timestamp. You could also use strtotime() to subtract one year like strtotime("-1 year", <timestamp>);.

[JShor]

I just realized how unhelpful my reply was to the original post.

Like Daniel said, time() is server-side only. It all depends on the server's time-keeping and the timezone in which the server is located.

Converting a birthday in MM DD YYYY is pretty simple. Let's say the user, through drop-down menus, have entered the following data:
Month: 11
Day: 17
Year: 1992

Well, then add the strings together and convert it using strtotime(). It will convert it to a UNIX timestamp. Then you can just do timestamp / (60*60*24*365). Note that it doesn't count leap year days, so it will probably be a few days off unless the user is 4 years old. That requires a separate algorithm.

For example:

<?php

$month = 11;
$day = 17;
$year = 1992;

$time = strtotime("$month $day $year");

$age = $time / (60*60*24*365);

echo "You are $age years old!";

?>

Would return "You are 18 years old!".

[djr33]

You're right about the correction. I just skipped that to keep things basic. As I said, there are functions (such as strtotime()) that will help with doing more complicated mathematical operations and automating some of it.


Here's plenty of information (including a few tutorials). Time isn't easy to start with, but it's not very complicated once you understand it. Remember, the time is just stored as a number of seconds since 1970. So that is an easy way to deal with "time", since you're just counting seconds.
http://www.google.com/search?q=timestamps+in+php

[keyboard]

I just realized how unhelpful my reply was to the original post.

Like Daniel said, time() is server-side only. It all depends on the server's time-keeping and the timezone in which the server is located.

Converting a birthday in MM DD YYYY is pretty simple. Let's say the user, through drop-down menus, have entered the following data:
Month: 11
Day: 17
Year: 1992

Well, then add the strings together and convert it using strtotime(). It will convert it to a UNIX timestamp. Then you can just do timestamp / (60*60*24*365). Note that it doesn't count leap year days, so it will probably be a few days off unless the user is 4 years old. That requires a separate algorithm.

For example:

<?php

$month = 11;
$day = 17;
$year = 1992;

$time = strtotime("$month $day $year");

$age = $time / (60*60*24*365);

echo "You are $age years old!";

?>

Would return "You are 18 years old!".

<?php

$month = 11;
$day = 17;
$year = 1992;

$time = strtotime("$month $day $year");

$age = $time / (60*60*24*365);

echo "You are $age years old!";

?>

I tried that and it just came up with

You are 0 years old!

Any help?

[keyboard]

Also for a dropdown select so they can enter their age like this

<option value="2010">2010</option>
<option value="2009">2009</option>
<option value="2008">2008</option>
<option value="2007">2007</option>
<option value="2006">2006</option>
<option value="2005">2005</option>
<option value="2004">2004</option>
<option value="2003">2003</option>
<option value="2002">2002</option>
<option value="2001">2001</option>
Thats for the year column but how do I let them select a month and then based on that display how many days e.g

August - and it goes up to 31 days e.g

[JShor]

Whoops, I forgot the slashes. I also forgot the floor(). Try this:

<?php

$month = 11;
$day = 17;
$year = 1992;

$date = "$month/$day/$year";
$time = strtotime($date);

$age = floor($time / (60*60*24*365));

echo "You are $age years old!";

?>

[JShor]

As for the drop-down menus, it would just be easier to populate the HTML using loops.

<?php

$month = array('January', 'February', 'March', 'April', 'May', 'June', 'July', 'August', 'September', 'October', 'November', 'December');

echo " <select name=\"month\">";
foreach($month as $m) {
    echo "<option value=\"$m\">$m</option>";
}
echo "</select>";

$d = 0;

echo " <select name=\"day\">";
    while($d < 31) {
        $d++;
        echo "<option value=\"$d\">$d</option>";
    }
echo "</select>";

$y = date("Y");

echo " <select name=\"year\">";
    while($y >= date("Y", strtotime("-100 year"))) {
        echo "<option value=\"$y\">$y</option>";
        $y--;
    }
echo "</select>";


?>