date() formatting help

Anon

When trying to format a date, pulled from a mysql database into an array, using the date() function I am getting weird results.

Here is the code I am using:

$s_date = date("m/d/y",$myrow['start_date']);

The result I am getting is 12/31/69 for every single date!!! Help...

Thanks.

Anon

have you got:
$s_date = date("m/d/y",$myrow['start_date']);

in 'for' or 'while' loop? becuase you said all the dates are same in the rows. Looks like you are not looping the results that are given by the mySQL, and so you are grabbing only 1 result, and then looping your HTML code or whatever you use to display the result.

check it out!

Anon

thanks, but its in a loop... it works fine without the formatting and besides there is no date even in the database that is 12/31/69... strange.

Anon

ok, then i cant comment any further unless i see the code it self.

anyway, check this article out:

http://www.phpbuilder.com/columns/akent20000610.php3

its about handling the date in PHP and mySQL.

regards,
Daarius...

Anon

thanks for the link to the article but it just says to do it like i am trying... i thought it could be my use of printf but even when i echo it i get 12/31/69. could it be the mysql date column having "-" between the date fields?

here is my code:

$result = mysql_query("SELECT * FROM events ORDER BY start_date",$db);

if ($myrow = mysql_fetch_array($result)) {

// display list if there are records to display

echo "<center><table border=0 cellspacing=15 cellpading=15>\n";
echo "<tr bgcolor=#808080><td>Event Name</td><td>Start Date</td><td>End Date</td><td>Location City</td></tr>\n";

	do {

		$s_date = strftime("%D",$myrow['start_date']);
		$e_date = strftime("%D",$myrow['end_date']);

		printf("<tr><td><a href=\"%s?event_id=%s\">%s</a></td><td>%s</td><td>%s</td><td>%s</td></tr>\n", $PHP_SELF, $myrow["event_id"], $myrow["event_name"], $s_date, $e_date, $myrow["location_city"]);

	} while ($myrow = mysql_fetch_array($result));

Anon

I think I have the answer.

MySQL stores dates in ISO format YYYY-MM-DD. The date() function is looking for a UNIX timestamp. You need to convert the ISO date into a UNIX timestamp before feeding it into date()

e.g. (untested code)

$date_array = explode("-", $iso_date);
$timestamp = mktime(0, 0, 0, $date_array["1"], $date_array["2"], $date_array["0"]);
$formatted_date = date("m/d/y" $timestamp);

or -

$date_array = explode("-", $iso_date);
$formatted_date = $date_array["1"].$date_array["2"].$date_array["0"];

/Mike

Anon

Thanks, I was just doing the:

$date_array = explode("-", $iso_date);
$timestamp = mktime(0, 0, 0, $date_array["1"], $date_array["2"], $date_array["0"]);
$formatted_date = date("m/d/y" $timestamp);

code when i got your email. it works. thanks for everyones help.

Anon

i have read the thread and believe it may be the answer to my problems as well.

is there a site that i may view the result?

thanks ed

Anon

you can use the following code:

this will format m/d/yyyy into yyyy-m-d for insertion into the database.

$array_date = explode("/" ,$date);
$timestamp = mktime(0,0,0,$array_date[0],$array_date[1],$array_date[2]);
$date_insert = date("Y-m-d", $timestamp);

this will format yyyy-m-d into m/d/yyyy for dsiplay.

$array_date = explode("-" ,$myrow['date']);
$timestamp = mktime(0,0,0,$array_date[1],$array_date[2],$array_date[0]);
$date_display = date("m/d/Y", $timestamp);

Anon

What if you have a DATETIME field. Could you help me do the same for that?

Thanks.