datediff in postgresql

= 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

-- 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

First, paste the javascript code I wrote. You can refer to it later. The Code is as follows: Date. prototype. dateAdd = function (interval, number){Var d = this;Var k = {'y': 'fullyear ', 'q': 'month', 'M': 'month', 'w': 'date', 'D ': 'date', 'H': 'urs', 'N': 'minutes ', 's': 'seconds', 'ms': 'milliseconds '};Var n = {'q': 3, 'w': 7 };Eval ('d. set '+ k [interval] +' (d. get '+ k [interval] +' () + '+ (n [interval] | 1) * number) + ')');Return d;}Date. prototype.

Oracle datediff evaluate the date, Week, and month Functions Oracledatediff date week month create or replace function datediff (type in varchar2, startTime in varchar2, endTime in varchar2, v in varchar2) return number as v_result number; SQL _1 varchar2 (200 ); begin dbms_output.put_line ('1'); if (type = 'day') then SQL _1: = 'select' | v | '(TO_NUMBER (to_date (''' | startTime | ''', ''yyyy-MM-DD '') -t

Copy codeThe Code is as follows:Date. prototype. dateAdd = function (interval, number){Var d = this;Var k = {'y': 'fullyear ', 'q': 'month', 'M': 'month', 'w': 'date', 'D ': 'date', 'H': 'urs', 'N': 'minutes ', 's': 'seconds', 'ms': 'milliseconds '};Var n = {'q': 3, 'w': 7 };Eval ('d. set '+ k [interval] +' (d. get '+ k [interval] +' () + '+ (n [interval] | 1) * number) + ')');Return d;}Date. prototype. dateDiff = function (interval, objDate2){Var d =

DateAdd function:Copy codeThe Code is 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 ));Cas

I. Use the DateDiff method to obtain the date and time intervals, Ii. Code Using System; using System. windows. forms; using Microsoft. visualBasic; namespace GetDateDiff {public partial class Frm_Main: Form {public Frm_Main () {InitializeComponent ();}/* Parameter Interval Type: Microsoft. visualBasic. dateInterval is required. The DateInterval enumerated value or String expression indicates the time interval between the units to be used as the diffe

DateTime string using the string variable in the specified format. now. toString ("D") + "\ n" + DateTime. now. toString ("f") + "\ n" + DateTime. now. toString ("F") + "\ n" + DateTime. now. toString ("g") + "\ n" + DateTime. now. toString ("G") + "\ n" + DateTime. now. toString ("R") + "\ n" + DateTime. now. toString ("y") + "\ n" + "current system time:" + DateTime. now. toString (// format the string in the custom format "yyyy MM dd HH mm ss seconds ");}}}3. Use the

Processing date in JS is relatively junk. The following functions may help you better process the date in Js.Including dateadd, datediff, and current date Current dateVaR d = new date ();VaR theday = D. getyear () + "-" + (D. getmonth (). tostring (). length> 1 )? (D. getmonth () + 1): "0" + (D. getmonth () + 1) + "-" + (D. getdate (). tostring (). length> 1 )? D. getdate (): "0" + D. getdate ()); Dateadd FunctionFunction dateadd (interval, num, da

Paste the Javascript I wrote first Code For more information, see. /* Get the date after the date, month, day, and other plus numbers */ Date. Prototype. dateadd = function (interval, number) { VaR d = this; VaR K = {'y': 'fullyear ', 'Q': 'month ', 'M': 'month ', 'W': 'date ', 'D': 'date ', 'H': 'urs ', 'N': 'minutes ', 'S ': 'seconds ', 'Ms': 'milliseconds '}; VaR n = {'q': 3, 'W': 7 }; Eval ('d. set '+ k [interval] +' (D. get '+ k [interval] +' () + '+ (n [interval] | 1) * Numb

Simulate two SQL server functions: dateadd () and datediff ().? Php file name: date. inc. php3 before using these two functions, you must first convert the date or date to the timestamp type. For example: $ todaymktime (0, 0, 0, date (m), date (d), date (Y )); // File name: date. inc. php3 // Before using these two functions, convert the date or date to the timestamp type. // For example: // $ Today = mktime (0, 0, 0, date ("m"), date ("d"), date ("Y

Copy codeThe Code is as follows: Date. prototype. dateAdd = function (interval, number) { Var d = this; Var k = {'y': 'fullyear ', 'q': 'month', 'M': 'month', 'w': 'date', 'D ': 'date', 'H': 'urs', 'N': 'minutes ', 's': 'seconds', 'ms': 'milliseconds '}; Var n = {'q': 3, 'w': 7 }; Eval ('d. set '+ k [interval] +' (d. get '+ k [interval] +' () + '+ (n [interval] | 1) * number) + ')'); Return d; } Date. prototype. dateDiff = function (interval, objDate2

DateAdd function: Copy codeCode: 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 )); Case "n": re

How to Use the datediff function in VB Datediff (interval, date1, date2 [, firstweekofyear [, firstweekofyear])Returns the value of variant (long), indicating the number of time intervals between two specified days.Interval: Set the unit of Period Calculation between two dates. For example,>Interval= "M" indicates that the calculation unit is month. >IntervalSuch:Yyyy> yearQ quarterM monthD DayW weekdayH