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Date Format YYDDD (julian?)

Posted on 2004-08-04
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Last Modified: 2012-05-05
Hello I am looking to get a day format of YYDDD i think it is a julian day.  

when i do

echo date("yz");
 
it gives me 04216, but with leap year the actual date is 04217.

Does anyone know how I can get the actual julian day?
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Question by:jmsloan
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9 Comments
 
LVL 33

Assisted Solution

by:snoyes_jw
snoyes_jw earned 30 total points
ID: 11717859
z gives the day of the year starting with 0.  If you want it starting with 1, just add one.
echo date("yz") + 1;
0
 
LVL 3

Author Comment

by:jmsloan
ID: 11717958
what is going to give me the format i need  YYDDD

what if it is the 2 day of the year I need it to say

04002
0
 
LVL 33

Expert Comment

by:snoyes_jw
ID: 11718008
as I said, date("yz") + 1;
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LVL 27

Accepted Solution

by:
Diablo84 earned 20 total points
ID: 11718022
try

$var = date("y");
$var2 = date("z");
$var2 = str_pad($input, 3, "0", STR_PAD_LEFT);
$out = $var.$var2;

echo $out;
0
 
LVL 27

Expert Comment

by:Diablo84
ID: 11718026
sorry, should be

$var = date("y");
$var2 = date("z");
$var2 = str_pad($var2, 3, "0", STR_PAD_LEFT);
$out = $var.$var2;

echo $out;
0
 
LVL 33

Expert Comment

by:snoyes_jw
ID: 11718029
ah, but then we lose leading zeros.  I'll find it here in a second...
0
 
LVL 33

Expert Comment

by:snoyes_jw
ID: 11718038
There, what Diablo84 said, but you'll still need to increment it, otherwise the first day of the year will come out as 04000.
0
 
LVL 27

Expert Comment

by:Diablo84
ID: 11718059
true, so...

$var = date("y");
$var2 = date("z") +1;
$var2 = str_pad($var2, 3, "0", STR_PAD_LEFT);
$out = $var.$var2;

echo $out;
0
 
LVL 3

Author Comment

by:jmsloan
ID: 11718668
Sorry that should have been the other way around
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