01-03-2007
parse variable
I have a variable (it is a date actually -> 2007-01-03) which
would be passed in as parameter, what I want is to parse in and put
year, month, and day in separate variables, I have tried the following
but doesn't work
echo $dt | awk -F- '{print $1 $2 $3}' | read y m d
Thanks in advance
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LEARN ABOUT PHP
gregoriantojd
GREGORIANTOJD(3) 1 GREGORIANTOJD(3)
gregoriantojd - Converts a Gregorian date to Julian Day Count
SYNOPSIS
int gregoriantojd (int $month, int $day, int $year)
DESCRIPTION
Valid Range for Gregorian Calendar 4714 B.C. to 9999 A.D.
Although this function can handle dates all the way back to 4714 B.C., such use may not be meaningful. The Gregorian calendar was not
instituted until October 15, 1582 (or October 5, 1582 in the Julian calendar). Some countries did not accept it until much later. For exam-
ple, Britain converted in 1752, The USSR in 1918 and Greece in 1923. Most European countries used the Julian calendar prior to the Grego-
rian.
PARAMETERS
o $month
- The month as a number from 1 (for January) to 12 (for December)
o $day
- The day as a number from 1 to 31
o $year
- The year as a number between -4714 and 9999
RETURN VALUES
The julian day for the given gregorian date as an integer.
EXAMPLES
Example #1
Calendar functions
<?php
$jd = GregorianToJD(10, 11, 1970);
echo "$jd
";
$gregorian = JDToGregorian($jd);
echo "$gregorian
";
?>
SEE ALSO
jdtogregorian(3), cal_to_jd(3).
PHP Documentation Group GREGORIANTOJD(3)