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I couldn't find a simple way to do that in PHP, so I threw this together.

I couldn't find a simple way to do that in PHP, so I threw this together.

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Note that incorrect results will be returned for years less than 1601 or greater than 2399.

This is because the Julian calendar (from which the Easter date is calculated) deviates from the Gregorian by one day for each century-year that is NOT a leap-year, i.e. (In the old Julian reckoning, EVERY 4th year was a leap-year.) This algorithm was first proposed by the mathematician/physicist Gauss.

= 0) elseif([email protected] == 0) else Since PHP 5.6.23 en 7.0.8 the support for requesting the weeknumber for a given date, where the first day of the week is Sunday, has been removed. This is sure that the information you shared is clearly identive and fabulous for keeping it in minds.

For those of us still needing it, here is simple hack to get the job done.

The solution that I am using which I found on another site(so not taking credit) is to use this: date("Y/m/d H:i:s").

substr((string)microtime(), 1, 6);that will give you: yyyy/mm/dd hh:ii:ss.uuuuuuhope this helps someone in need! In order to define leap year you must considre not only that year can be divide by 4!The correct alghoritm is:if (year is not divisible by 4) then (it is a common year)else if (year is not divisible by 100) then (it is a leap year)else if (year is not divisible by 400) then (it is a common year)else (it is a leap year)So the code should look like this:if($year%4 == 0 && $year0 !Most spreadsheet programs have a rather nice little built-in function called NETWORKDAYS to calculate the number of business days (i.e.Monday-Friday, excluding holidays) between any two given dates.For example using a 24 hour notation without leading zeros is the option '%G' in PHP but '%k' in My SQL.When using dynamically generated date formatting string, be careful to generate the correct options for either PHP or My SQL.thanks all While this will work for the majority of years it will not work on years that are multiples of 100 but not multiples of 400 i.e.(2100).A function not using php's date() function that will also account for this small anomaly in leap years: While is_leapyear_working will not return true for the few non-leap years divisible by four I couldn't tell you if this is more or less efficient than using php's date() as an even earlier poster suggested: The following function will return the date (on the Gregorian calendar) for Orthodox Easter (Pascha)."skip July 4th in any year, skip the first Monday in September in any year"). If you see the number 86400 in a date calculation, think very hard before deciding that it is correct. In many places, some days have only 82,800 seconds and some have 90,000. Assuming that now plus 86,400 seconds is equivalent to some time tomorrow can sometimes be wrong.It might actually be the day after tomorrow or still today. Thanks to tcasparr at gmail dot com for the great idea (at least for me) ;)I changed the code a little to replicate the functionality of date_parse_from_format, once I don't have PHP 5.3.0 yet. Hope you don't mind changing your code tcasparr at gmail dot com./******************************************************* * Simple function to take in a date format and return array of associated * formats for each date element * * @return array * @param string $str Format * * Example: Y/m/d g:i:s becomes * Array * ( * [year] = In order to define leap year you must considre not only that year can be divide by 4!

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