db2 day of week

FormulaCaille (Zeller) formula, is a calculation of the week formula, casually to a date, you can use this formula to calculate the day of the week.To calculate the date on or before October 4, 1582, the formula is the example of September 3, 1582: After September 3, 1582: W = (d + 2*m+3* (m+1)/5+y+y/4-y/100+y/400)%7; September 3, 1582 before: W = (d+2*m+3* (m+1)/5+y+y/4+5)% 7;NoteIn the Year 1, February as

Today is the day of the week written in PHP // The first writing method $ da = date ( quot; w quot;); if ($ da == quot; 1 quot ;) { nbsp; echo quot; today is Monday quot; ;}elseif ($ da == quot; 2 quot ;) { n today is the day of the week written in PHP // The first method: $ da = date ("w"); if ($ da = "1") {echo

Several methods for returning the day of the week by SQLThe following describes how to use SQL statements to obtain the current day of the week. For more information, see.DECLARE @ Dt datetimeSELECT @ Dt = '2017-02-21'Select datepart (wk, @ Dt)Select datepart (wk, GETDATE ())Select datepart (weekday, GETDATE ()

Public StaticDate GetDayOfWeek (intDayOfWeek,intWeekoffset) { if(Dayofweek>calendar.saturday | | dayofweekcalendar.sunday) { return NULL; } Calendar Date=calendar.getinstance (Locale.china); //set the first day of the week to Monday, the default is SundayDate.setfirstdayofweek (Calendar.monday); //week minus one, i.e. last weekDate.add (Calendar

/*** Get start date, end date for this week, quarter, month, month*/var now = new Date (); Current datevar nowdayofweek = Now.getday (); The day of the week todayvar nowday = Now.getdate (); Current dayvar nowmonth = Now.getmonth (); Current monthvar nowyear = Now.getyear (); When the year beforeNowyear + = (Nowyear var lastmonthdate = new Date (); Last month d

Topic Description: We now use the Gregorian style of dating in Russia. The leap years are years with number divisible from 4 but not divisible by, or divisible by 400.For example, years, 2180 and 2400 are leap. Years, 2181 and 2300 are not leap.Your task is to write a program which would compute the day of week corresponding to a given date in the nearest past or in The future using today ' s agreement abou

To_days (now ())-To_days (Time field) Record of the dayWhere date (Time field) =date (now ())OrWhere To_days (Time field) = To_days (now ());Query one week:SELECT * FROM table where Date_sub (Curdate (), INTERVAL 7 day) Query one months:SELECT * FROM table where Date_sub (Curdate (), INTERVAL INTERVAL 1 MONTH) Query ' 06-03 ' to ' 07-08 ' for all birthdays during this time period:Select * from user WhereDate_format (Birthday, '%m-%d ') >= ' 06-03 ' a

yearif (month = = 5) {//month codec = 0;} else if (month = = 8) {c = 1;} else if (month = = 2 | month = = 3 | | month = = 11) {c = 2;} else if (month = = 6) {c = 3;} else if (month = = | | month = = 9) {c = 4;} else if (month = = 4 | | month = = 7) {c = 5;} else if (month = = 1 | | month = = 10) {c = 6;}} else {if (month = = 5) {c = 0;} else if (month = = 2 | | month = = 8) {c = 1;} else if (month = = 3 | | month = = 11) {c = 2;} else if (month = = 6) {c = 3;} else if (month = = | | month = = 9

Label:That day: SELECT * from T_news where DateDiff (Day,addtime,getdate ()) =0 Last three days: SELECT * from T_news where DateDiff (Day,addtime,getdate ()) Week: SELECT * from T_news WHERE (DATEPART (wk, addtime) = DATEPART (wk, GETDATE ())) and (DATEPART (yy, addtime) = DATEPART (yy, GET DATE ())) Note: You cannot

The input date is assumed to be valid correctly.Find a reference date and find the best date for Sunday. I did not think of a moment to choose today, Monday, also good. Then find out that the number of days between the input date and the reference date N,n is negative indicates that the input date precedes the reference date, and that n is positive indicates that the input date is after the reference date. Because the week is the cycle of the

PHP gets the start date and end date of the week (month) of the current date Gets the start time and end time of the week of the specified date function Getweekrange ($date) {$ret =array (); $timestamp =strtotime ($date); $w =strftime ('%u ', $timestamp); $ret [' Sdate ']=date (' y-m-d 00:00:00 ', $timestamp-($w-1) *86400); $ret [' Edate ']=date (' y-m-d 23:59:59 ', $timestamp + (7-$w) *86400)

Tags: using art ar div time new SQL CThree days select * from T_news where DateDiff (Day,addtime,getdate ()) a week select * from T_news WHERE (DATEPART (wk, addtime) = DATEPART (wk, GETDATE ())) and (DATEPART (yy, addtime) = DATEPART ( YY, GETDATE ()))Note: You cannot use the DateDiff difference at this time as 7, because DateDiff only represents the number of intervalsJanuary SELECT * from T_news WHERE (D

Don't say much about yourself on the codeThis is what I started learning JavaScript, and now it's coming out. span ID = "Clock" > span >functionTime () {varToday =NewDate (); varHou; varTime ; varDay ; if(Today.gethours () ) {Hou=today.gethours (); time= "AM"; } Else if(Today.gethours () ) {Hou= Today.gethours ()-12; time= "PM"; } varD =Today.getday (); if(d = = 0) { Day= "Day"; } Else if(d = =

Print $ week_start nbsp; date (U, mktime (0, 0, 0, date (m), date (d)-7, date (Y); $ week_end nbsp; date (U, mktime (23, 59, 59, date (m), date (d)-0, date (Y); $ date prints every day of the previous week $week_start = date("U",mktime(0,0,0,date("m"),date("d")-7,date("Y")));$week_end = date("U",mktime(23,59,59,date("m"),date("d")-0,date("Y")));$date=date('Y-m-d',$week_start);$date1=date('Y-m-d',$week_en

Print $ week_start nbsp; date (U, mktime (0, 0, 0, date (m), date (d)-7, date (Y); $ week_end nbsp; date (U, mktime (23, 59, 59, date (m), date (d)-0, date (Y); $ dateda prints every day of the previous week $week_start = date("U",mktime(0,0,0,date("m"),date("d")-7,date("Y")));$week_end = date("U",mktime(23,59,59,date("m"),date("d")-0,date("Y")));$date=date('Y-m-d',$week_start);$date1=date('Y-m-d',$week_

The server returns the date string in the following format:2015-01-20 12:03:122015-01-19 19:10:002015-01-19 12:00:002015-01-19 08:41:002015-01-16 12:23:46Ask to determine if the day of the week turns into the following format:First, convert the string into a date format and look at some of the methods below. The simplestvar str = ' 2015-01-20 12:03:12 ';d atestr = Datestr.replace (/-/g, '/'); var date = new

Resources:　　The week algorithm refers to the method defined by ISO8601, if you have questions about this, please see: http://en.wikipedia.org/wiki/ISO_week_dateJS Code implementation:var normalyear = [0,31,59,90,120,151,181,212,243,273,304,334];var leapyear = [0,33,60,91,121,152,182,213,244,274,305,335];　　　　　　var nowdate = new Date ();var datestr = Nowdate.format (' yyyy-mm-dd ');var year = parseint (datestr.substring (0,4));var mon = parseint (datest

Today, in the program era, everything is solved with a program. Of course it is very convenient. So, you ask me October 1, 2003 is the day of the week, I will probably directly use the program to tell you.echo Date (' W ', Strtotime (' 2003-10-1 '));But in fact, we all know that PHP's date function has a time range, that is, only from 1970-2038 years, so the algorithm outside the range is not allowed. How d

I'm a little bit trickery this method, only after 1900-01-01 of data, before the date needs to be recalculated.1. First remove the number of days from the date of 1900-01-01 private static int GetDays (int year, int month, Int. day) {//1900-1-1 is Monday int result = 0; Int[] month31 = {1,3,5,7,8,10,12}; Int[] month30 = {4,6,9,11}; const int year = 1900; const int MONTH = 1; cons

Weekly expression Monday-monthly MondayTuesdayWednesdayThursdayFriday-Golden WeekSaturday-SaturdaySunday-Sunday In addition, you may find it hard to remember the five elements of the day, month, and month. Here I will provide a convenient memory method:1. Orthodox memory method: With the exception of Saturday, Tuesday to Friday is the Reverse Order of the Golden wood fire. 2. The most effective and fun memory method: