Date Formatting - yes again :)

jamessmb

Hello,

This has been asked a multitude of times, and I've tried several suggestions I have found.

Why does this date format differently when using the date() function?

<?php echo $row_getAds['startDate']; ?>
shows: 2003-08-01 (how it is in the MySQL DB

and

<?php echo date("m/d/Y",$row_getAds['startDate'])?>
shows: 12/31/1969

Thanks,
James

tekky

quite possibly due to the simple fact that date() takes 1 parameter with an optional second param...

date("format string", [unixtimestamp]);
and format string is single characters that stand for certain things

like date("j M Y"); would print 1 Aug 2003 (day w/o leading zero, 3 letter month, 4 digit year)

to use date() you need to use mktime() to make a timestamp to feed date()

Lirpa

date() tries to calculate the wanted parameters from the second function argument, where it expects a timestamp, which is of type INT.
But the variable $row_getAds['startDate']; is a formatted string. This means its value is "2003-08-01" and that´s not an an integer. So date() doesnt know how to handle this and sets the value of its timestamp to 0 (or perhaps -1, i am not sure).

Try :
echo date("m/d/Y", "Watch me, I am a string");

This will show the same date : 12/31/1969

And why is the result : 12/31/1969 ?

From the Manual:
"The timestamp is a long integer containing the number of seconds between the Unix Epoch (January 1 1970) and the time specified"

Try :
echo date("H.i.s - m/d/Y", 0);
"H.i.s" means Hours, minutes, seconds, the result shows the "beginning of time" for your pc

echo date("H.i.s - m/d/Y", 3600);
now it´s one hour later (60 mins * 60 Secs = 3600)

echo date("H.i.s - m/d/Y", 86400);
the next day (60 mins 60 Secs = 3600 24 = 86400)

echo date("H.i.s - m/d/Y", 31536000);
and one year is gone (60 mins 60 Secs = 3600 24 * 365 = 31536000)

Greetings, Lirpa

fahrvergnuugen

as was already pointed out... when you give date something it doesnt understand [a non unix timestamp] it turns it into 0. 0 is 12/31/1969 [the unix epoch].

try this

echo date("m.d.Y",strtotime($row_getAds['startDate']));

strtotime is a very very useful function 🙂

you might also try using DATE_FORMAT if you are using MySQL.

$query = "select DATE_FORMAT(startDate,'%m/%d/%Y') from ....";

jamessmb

All,

Thanks for your help. strtotime was the key that worked.

James

fahrvergnuugen

dont totally discount DATE_FORMAT. Its atually much faster than running date("m/d/Y",strtotime($date)) and provides you with the same end result.

jamessmb

I tried that late last night and kept getting errors thrown at me. I'll look at it again, after the coffee kicks in. :o

New question.

I need to calculate the difference in days between the date in the DB and the current date. For example

Database date : 8/15/2003
Current date: 8/1/2003
Remaining days: 14

None of the functions in my PHP function manual seem to pertain to what I want to do. Seems like I need a date comparison function.

James

fahrvergnuugen

Again, very easy to do.

$days = (strtotime($database_date) - time()) / (60 * 60 * 24);

echo $days;

or more simply:

$days = (strtotime($database_date) - time()) / 86400;

time() is the number of seconds from the UNIX epoch or a unix timestamp for the current time. strtotime turns the given date into a unix timestamp. you subtract the two to get the difference in seconds between the two dates. then just divide it out to turn it into days (60 seconds in a minute, 60 minutes in an hour, 24 hours in a day)