teradata datediff

-- Parameter:Datepart varchar2 -- year comparison? Month? ?Startdate varchar2 -- start dateEndDate Varchar2 -- end date-- Oracle9i passed the test*/ Create Or Replace Function Datediff(Datepart In Varchar2,StartDate In Varchar2,EndDate In Varchar2) Return Number IsReallyDo Numeric;BeginSelect Case Upper (Datepart)When 'yyyy' ThenTrunc (Extract (Year From CDate (EndDate ))-Extract (Year From CDate (StartDate )))When 'M' ThenDatediff ('yyyy', StartDate,

SELECT DATEDIFF (year, ' 1986-09-01 ', ' 2016-02-01 ') as Diffdate DatePart Abbreviations Years YY, yyyy Quarter QQ, Q Month MM, M The day of the year Dy, y Day DD, D Week WK, WW Week DW, W Hours hh Minutes MI, n Seconds SS, S

Datediff (timeinterval, date1, date2 [, firstdayofweek [, firstweekofyear]) // Timeinterval indicates the time interval type, CodeIs: Yyyy Q quarter M month Y number of days in a year D Number of days per week WW week H hour N minutes S seconds Example: The Code is as follows: D1 = datediff ("D", now (), cdate (RS ("adddate") // read the date added to the item from the database If D1>-15 then

The datediff function returns the interval between two dates. Format: Datediff (interval, date1, date2 [, firstdayofweek [, firstweekofyear]) ParametersIntervalRequired. String expression indicates the time interval between date1 and date2. Date1, date2Required. Date expression. Two dates used for calculation.FirstdayofweekOptional. Specify the constant of the first day of the week. If not specified, th

, $sec, $month, $day, $year); return $ret; } /**** simulation of DATEDIFF functions in SQL Server ******* $part Type: string Value range: year,month,day,hour,min,sec Indicates: Which part of the date to increase $date 1, $date 2 type: Timestamp Represents: two dates to compare return type: Numeric End * (*************/ function DateDiff ($part, $date 1, $date 2) { $diff = $date 2-$date 1; $year 1=date ("Y"

= mktime ($ hour, $ min, $ sec, $ month, $ day, $ year );Return $ ret;}/***** Simulate the datediff function in sqlserver *******$ Part type: stringValue Range: year, month, day, hour, min, secIndicates the part of the date to be added.$ Date1, $ date2 type: timestampTwo dates to be comparedReturn type: Numeric Value* ************** End *(*************/Function datediff ($ part, $ date1, $ date2 ){// $ Dif

} $ret =mktime ($hour, $min, $sec, $month, $day, $year); return $ret; } /**** simulate the DATEDIFF function in SQL Server ******* $part Type: string Value range: year,month,day,hour,min,sec Which part of the date to increase $date 1, $date 2 type: Timestamp Indicates: two dates to compare Return type: Numeric value End * (*************/ function DateDiff ($part, $date 1, $date 2) { $diff = $date 2-$date

--getdate Current system date, timeSelect getdate()--2014-11-24 13:22:40.407--the CONVERT () function can display date/time data in different formatsSelect Convert(varchar( -),getdate(), at)--2014-11-24--DateDiff Returns the number of date and time boundaries across two specified datesSelect DateDiff( Day,'2014-11-24','2014-12-28')-- theSelect DateDiff(Month,'201

Public Enum dateinterval { Second, minute, hour, day, week, month, quarter, year } Public long datediff (dateinterval interval, system. datetime startdate, system. datetime enddate) ...{ Long lngdatediffvalue = 0; System. timespan Ts = new system. timespan (enddate. ticks-startdate. ticks ); Switch (interval) ...{ Case dateinterval. Second: Lngdatediffvalue = (long) Ts. totalseconds; Break; Case dateinterval. minute: Lngdatediffvalue = (long) Ts. tota

Original link: http://www.cnblogs.com/whz111/articles/2445777.html--convert (time) corresponds to the time format of the number representation Select CONVERT (varchar, GETDATE ()) --10 Select CONVERT (varchar (), GETDATE (), 111) --2004/09 /12 Select CONVERT (varchar) --20040912 Select CONVERT (varchar (getdate), GETDATE (), 102) - -2004.09.12 Select CONVERT (varchar (), GETDATE (), 101) --09/12/2004 Select CONVERT (varchar), GETDATE () , 103) --12/09/2004 Select CONVERT (varchar (), GETDA

(calendar.month) +calendar.get (calendar.year) *12;//month = year of the current year *12 + month of the current year, same as - Calendar.settime (endDate); - intEndmonth = Calendar.get (calendar.month) +calendar.get (calendar.year) *12; - returnEndmonth-Startmonth; - } in /*find the number of days between StartDate and EndDate*/ - if("D". Equalsignorecase (datetype) | | "DD". Equalsignorecase (Datetype)) { to LongStartTime =startdate.gettime (); + LongEndTime =enddate.gettime (); - return(int)

DateAdd function: Copy Code code as follows: function DateAdd (interval,number,date) { Switch (Interval.tolowercase ()) { Case "Y": Return new Date (Date.setfullyear (Date.getfullyear () +number)); Case "M": Return new Date (Date.setmonth (Date.getmonth () +number)); Case "D": Return new Date (Date.setdate (Date.getdate () +number)); Case "W": Return new Date (Date.setdate (Date.getdate () +7*number)); Case "H": Return new Date (Date.sethours (date.gethours () +number)

Town Field Poem:——— Dream who feel, the water month Build blog. Baiqian tribulation, only know the vicissitudes of the world.——— today holds the Buddhist language, the technology is boundless willing to learn. Willing to do what you learn, cast a conscience blog.——————————————————————————————————————————1 Code1 --View the current time for easy comparison2 Select getdate()3 Select DateDiff( Year,'1996-11-01',getdate())4 Select

What I'm going to introduce today is a function that converts time (for example: 2008-8-8 13:45:22) into seconds in ASP, and the opposite of converting seconds to time. Of course when the time is converted. Must have a relatively fixed time, that is GMT. Think about it. Because Greenwich Mean Time is "January 1, 1970 00:00:00" Start, ASP provides a function called DateDiff (), this function can return a time difference of seconds, That means we put in

Extend the time functions in PHP DateDiff and DateAdd function DateDiff ($part, $begin, $end) {$diff = Strtotime ($end)-Strtotime ($begin); switch ($part) {case "Y": $retval = Bcdiv ($diff, (60 * 60 * 24 * 365)); Break Case "M": $retval = Bcdiv ($diff, (60 * 60 * 24 * 30)); Break Case "W": $retval = Bcdiv ($diff, (60 * 60 * 24 * 7)); Break Case "D": $retval = Bcdiv ($diff, (60 * 60 * 24)); Break Case

, $sec, $month, $day, $year); return $ret; } /**** simulate the DATEDIFF function in SQL Server ******* $part Type: string Value range: year,month,day,hour,min,sec Which part of the date to increase $date 1, $date 2 type: Timestamp Indicates: two dates to compare Return type: Numeric value End * (*************/ function DateDiff ($part, $date 1, $date 2) { $diff = $date 2-$date 1; $year 1=date ("Y", $date

Requirement: Splits a string that uses a delimiter connection for a fixed number of digits, such as splitting a record 215|482|850|870 into 215,482,850,870 four recordsResult: The WITH RECURSIVE statement is capable of implementing this function and

Copy codeThe Code is as follows:

Copy codeThe Code is as follows:

Copy Code code as follows: