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[SOLVED] Explode day, month, year from mysql date


Alexhoward
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Hi guys,

 

back again so soon, sorry.

 

but i'm getting an error code from this piece for script that i don't understand: Undefined offset: 2

 

and : Undefined offset: 1

 

am i not doing this correctly...?

 

<?php
include("whatever.php");

mysql_connect($host, $db, $pass)
or die ("Could not connect to mysql because ".mysql_error());

mysql_select_db($db)
or die ("Could not select database because ".mysql_error());

$mysite_username = $HTTP_COOKIE_VARS["sitename"]; 

$year=mysql_query("SELECT dob FROM $table WHERE username = '$mysite_username'");
while($year1 = mysql_fetch_array($year))
$year1['dob']; 

list ($y, $sm, $d) = explode ('-', $year1);

echo number_format($d);

?>

 

Thanks for all your help,

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$year = mysql_query("SELECT dob FROM $table WHERE username = '$mysite_username'");
$year1 = mysql_fetch_array($year);
$dob = $year1['dob']; // this line is revised
list ($y, $sm, $d) = explode ('-', $dob); // slight change
echo number_format($d);

 

I assume it's a single result, so no while needed.

 

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Hi Andy,

 

Thanks for the reply

 

I got it going like this, which i assume works, but now it's only single digits,

 

e.g: 0 instead of 00

 

is there a way to make number_format double digits..?

 

Thanks

 

<?php
$year=mysql_query("SELECT dob FROM $table WHERE username = '$mysite_username'");
while($year1 = mysql_fetch_array($year))
echo $year1['dob']; 

echo number_format(date($year1,"dd"));

?>

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Here's the answer.

 

include("whatever.php");

mysql_connect($host, $db, $pass)
or die ("Could not connect to mysql because ".mysql_error());

mysql_select_db($db)
or die ("Could not select database because ".mysql_error());

$mysite_username = $HTTP_COOKIE_VARS["sitename"]; 

$year=mysql_query("SELECT dob FROM `profile` WHERE username = '$mysite_username'");
while($year1 = mysql_fetch_array($year))
$test = $year1['dob'];

list($a, $v, $c) = explode('-',$test);

echo $c

 

yipee!

 

i solved one :D

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