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Top Forums UNIX for Dummies Questions & Answers Any simple method to check date format (YYDDD)? Post 302154880 by vino on Wednesday 2nd of January 2008 09:19:20 AM
Old 01-02-2008
Quote:
Originally Posted by ganapati
Sorry Vino but thanks,

My question is for the format of the date (YYDDD). i.e,
  1. YYDDD field should not be blank
  2. YY should be only from 00-99, not any other charecters and come first before DDD.
  3. DDD should be from 001-365 for normal years and 001-366 for leap years. This should come after YY

Iam looking for simpler code to accomplish this.

Thanks again,
Ganapati
Ah, I misread your question. Smilie
You will have to split the input into fields of 2 and 3. Compare the first against the limits you have in mind. And likewise with the other field. Looking at your requirements, why dont you use the information you asked for in the other thread ? https://www.unix.com/unix-dummies-que...er-number.html
 

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DATETIME.SUB(3) 							 1							   DATETIME.SUB(3)

DateTime::sub - Subtracts an amount of days, months, years, hours, minutes and seconds from a DateTime object

       Object oriented style

SYNOPSIS
public DateTime DateTime::sub (DateInterval $interval) DESCRIPTION
Procedural style DateTime date_sub (DateTime $object, DateInterval $interval) Subtracts the specified DateInterval object from the specified DateTime object. PARAMETERS
o $object -Procedural style only: A DateTime object returned by date_create(3). The function modifies this object. o $interval - A DateInterval object RETURN VALUES
Returns the DateTime object for method chaining or FALSE on failure. EXAMPLES
Example #1 DateTime.sub(3) example Object oriented style <?php $date = new DateTime('2000-01-20'); $date->sub(new DateInterval('P10D')); echo $date->format('Y-m-d') . " "; ?> Procedural style <?php $date = date_create('2000-01-20'); date_sub($date, date_interval_create_from_date_string('10 days')); echo date_format($date, 'Y-m-d'); ?> The above examples will output: 2000-01-10 Example #2 Further DateTime.sub(3) examples <?php $date = new DateTime('2000-01-20'); $date->sub(new DateInterval('PT10H30S')); echo $date->format('Y-m-d H:i:s') . " "; $date = new DateTime('2000-01-20'); $date->sub(new DateInterval('P7Y5M4DT4H3M2S')); echo $date->format('Y-m-d H:i:s') . " "; ?> The above example will output: 2000-01-19 13:59:30 1992-08-15 19:56:58 Example #3 Beware when subtracting months <?php $date = new DateTime('2001-04-30'); $interval = new DateInterval('P1M'); $date->sub($interval); echo $date->format('Y-m-d') . " "; $date->sub($interval); echo $date->format('Y-m-d') . " "; ?> The above example will output: 2001-03-30 2001-03-02 NOTES
DateTime.modify(3) is an alternative when using PHP 5.2. SEE ALSO
DateTime.add(3), DateTime.diff(3), DateTime.modify(3). PHP Documentation Group DATETIME.SUB(3)
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