如果我有日期，如何计算该年内该日期的周数？
例如，2008年1月1日至1月6日是第1周，1月7日至13日是第2周，所以如果我的日期是2008年1月10日，那么我的周数就是2。
一个算法非常适合让我入门，示例代码也会有所帮助 - 我正在Windows上用C ++开发。
答案 0 :(得分：28)
请注意，虽然您对 nth 一周的定义是成立的，但它也不是'标准'。
ISO 8601定义了日期，时间和时区表示的标准。它定义了星期一开始的周。它还说，一年中的第1周是从给定年份起至少包含4天的那一年。因此，20xx年12月29日，30日和31日可能是20xy的第1周（其中xy = xx + 1），1月20日的第1，第2和第3可能都在20xx的最后一周。此外，可能会有53周。
[已添加：请注意，C标准和`strftime（）函数提供从周日开始的周数以及周一开始的周数。目前尚不清楚C标准是否规定了周日周的第0周年份数。另请参阅Emerick Rogul的答案。]
然后是有趣的测试阶段 - 你什么时候进入第53周？ 答案是2010年1月1日星期五，即2009年-W53（如， 实际上，是2010年1月3日星期日）。同样，1月1日星期六 2005年是2004年-W53，但2006年1月1日星期日是2005年-W52。
这是以下代码中的注释的摘录，实际上是在Informix SPL（存储过程语言）中，但是可读 - 尽管可能不可写 - 没有进一步解释。 '||' operator是SQL字符串连接操作，星期日是第0天，星期一是第1天，...星期六是第6天。评论中有大量的注释，包括标准中的相关文本。一行评论开始'--';可能多行注释以“{”开头，并在下一个“}”结束。
-- @(#)$Id: iso8601_weekday.spl,v 1.1 2001/04/03 19:34:43 jleffler Exp $
--
-- Calculate ISO 8601 Week Number for given date
-- Defines procedure: iso8601_weekday().
-- Uses procedure: iso8601_weeknum().
{
According to a summary of the ISO 8601:1988 standard "Data Elements and
Interchange Formats -- Information Interchange -- Representation of
dates and times":
The week notation can also be extended by a number indicating the
day of the week. For example the day 1996-12-31 which is the
Tuesday (day 2) of the first week of 1997 can also be written as
1997-W01-2 or 1997W012
for applications like industrial planning where many things like
shift rotations are organized per week and knowing the week number
and the day of the week is more handy than knowing the day of the
month.
This procedure uses iso8601_weeknum() to format the YYYY-Www part of the
date, and appends '-d' to the result, allowing for Informix's coding of
Sunday as day 0 rather than day 7 as required by ISO 8601.
}
CREATE PROCEDURE iso8601_weekday(dateval DATE DEFAULT TODAY) RETURNING CHAR(10);
DEFINE rv CHAR(10);
DEFINE dw CHAR(4);
LET dw = WEEKDAY(dateval);
IF dw = 0 THEN
LET dw = 7;
END IF;
RETURN iso8601_weeknum(dateval) || '-' || dw;
END PROCEDURE;
-- @(#)$Id: iso8601_weeknum.spl,v 1.1 2001/02/27 20:36:25 jleffler Exp $
--
-- Calculate ISO 8601 Week Number for given date
-- Defines procedures: day_one_week_one() and iso8601_weeknum().
{
According to a summary of the ISO 8601:1988 standard "Data Elements and
Interchange Formats -- Information Interchange -- Representation of
dates and times":
In commercial and industrial applications (delivery times,
production plans, etc.), especially in Europe, it is often required
to refer to a week of a year. Week 01 of a year is per definition
the first week which has the Thursday in this year, which is
equivalent to the week which contains the fourth day of January. In
other words, the first week of a new year is the week which has the
majority of its days in the new year. Week 01 might also contain
days from the previous year and the week before week 01 of a year is
the last week (52 or 53) of the previous year even if it contains
days from the new year. A week starts with Monday (day 1) and ends
with Sunday (day 7). For example, the first week of the year 1997
lasts from 1996-12-30 to 1997-01-05 and can be written in standard
notation as
1997-W01 or 1997W01
The week notation can also be extended by a number indicating the
day of the week. For example the day 1996-12-31 which is the
Tuesday (day 2) of the first week of 1997 can also be written as
1997-W01-2 or 1997W012
for applications like industrial planning where many things like
shift rotations are organized per week and knowing the week number
and the day of the week is more handy than knowing the day of the
month.
Referring to the standard itself, section 3.17 defines a calendar week:
week, calendar: A seven day period within a calendar year, starting
on a Monday and identified by its ordinal number within the year;
the first calendar week of the year is the one that includes the
first Thursday of that year. In the Gregorian calendar, this is
equivalent to the week which includes 4 January.
Section 5.2.3 "Date identified by Calendar week and day numbers" states:
Calendar week is represented by two numeric digits. The first
calendar week of a year shall be identified as 01 [...]
Day of the week is represented by one decimal digit. Monday
shall be identified as day 1 of any calendar week [...]
Section 5.2.3.1 "Complete representation" states:
When the application clearly identifies the need for a complete
representation of a date identified by calendar week and day
numbers, it shall be one of the alphanumeric representations as
follows, where CCYY represents a calendar year, W is the week
designator, ww represents the ordinal number of a calendar week
within the year, and D represents the ordinal number within the
calendar week.
Basic format: CCYYWwwD
Example: 1985W155
Extended format: CCYY-Www-D
Example: 1985-W15-5
Both the summary and the formal definition are intuitively clear, but it
is not obvious how to translate it into an algorithm. However, we can
deal with the problem by exhaustively enumerating the seven options for
the day of the week on which 1st January falls (with actual year values
for concreteness):
1st January 2001 is Monday => Week 1 starts on 2001-01-01
1st January 2002 is Tuesday => Week 1 starts on 2001-12-31
1st January 2003 is Wednesday => Week 1 starts on 2002-12-30
1st January 2004 is Thursday => Week 1 starts on 2003-12-29
1st January 2010 is Friday => Week 1 starts on 2010-01-04
1st January 2005 is Saturday => Week 1 starts on 2005-01-03
1st January 2006 is Sunday => Week 1 starts on 2006-01-02
(Cross-check: 1st January 1997 was a Wednesday; the summary notes state
that week 1 of 1997 started on 1996-12-30, which is consistent with the
table derived for dates in the first decade of the third millennium
above).
When working with the Informix DATE types, bear in mind that Informix
uses WEEKDAY values 0 = Sunday, 1 = Monday, 6 = Saturday. When the
weekday of the first of January has the value in the LH column, you need
to add the value in the RH column to the 1st of January to obtain the
date of the first day of the first week of the year.
Weekday Offset to
1st January 1st day of week 1
0 +1
1 0
2 -1
3 -2
4 -3
5 +3
6 +2
This can be written as MOD(11-w,7)-3 where w is the (Informix encoding
of the) weekday of 1st January and the value 11 is used to ensure that
no negative values are presented to the MOD operator. Hence, the
expression for the date corresponding to the 1st day (Monday) of the 1st
week of a given year, yyyy, is:
d1w1 = MDY(1, 1, yyyy) + MOD(11 - WEEKDAY(MDY(1,1,yyyy)), 7) - 3
This expression is encapsulated in stored procedure day_one_week_one:
}
CREATE PROCEDURE day_one_week_one(yyyy INTEGER) RETURNING DATE;
DEFINE jan1 DATE;
LET jan1 = MDY(1, 1, yyyy);
RETURN jan1 + MOD(11 - WEEKDAY(jan1), 7) - 3;
END PROCEDURE;
{
Given this date d1w1, we can calculate the week number of any other date
in the same year as:
TRUNC((dateval - d1w1) / 7) + 1
The residual issues are ensuring that the wraparounds are correct. If
the given date is earlier than the start of the first week of the year
that contains it, then the date belongs to the last week of the previous
year. If the given date is on or after the start of the first week of
the next year, then the date belongs to the first week of the next year.
Given these observations, we can write iso8601_weeknum as shown below.
(Beware: iso8601_week_number() is too long for servers with the
18-character limit; so is day_one_of_week_one()).
Then comes the interesting testing phase -- when do you get week 53?
One answer is on Friday 1st January 2010, which is in 2009-W53 (as,
indeed, is Sunday 3rd January 2010). Similarly, Saturday 1st January
2005 is in 2004-W53, but Sunday 1st January 2006 is in 2005-W52.
}
CREATE PROCEDURE iso8601_weeknum(dateval DATE DEFAULT TODAY) RETURNING CHAR(8);
DEFINE rv CHAR(8);
DEFINE yyyy CHAR(4);
DEFINE ww CHAR(2);
DEFINE d1w1 DATE;
DEFINE tv DATE;
DEFINE wn INTEGER;
DEFINE yn INTEGER;
-- Calculate year and week number.
LET yn = YEAR(dateval);
LET d1w1 = day_one_week_one(yn);
IF dateval < d1w1 THEN
-- Date is in early January and is in last week of prior year
LET yn = yn - 1;
LET d1w1 = day_one_week_one(yn);
ELSE
LET tv = day_one_week_one(yn + 1);
IF dateval >= tv THEN
-- Date is in late December and is in the first week of next year
LET yn = yn + 1;
LET d1w1 = tv;
END IF;
END IF;
LET wn = TRUNC((dateval - d1w1) / 7) + 1;
-- Calculation complete: yn is year number and wn is week number.
-- Format result.
LET yyyy = yn;
IF wn < 10 THEN
LET ww = '0' || wn;
ELSE
LET ww = wn;
END IF
LET rv = yyyy || '-W' || ww;
RETURN rv;
END PROCEDURE;
为了完整性，使用上面的day_one_week_one()函数也很容易编写反函数：
-- @(#)$Id: ywd_date.spl,v 1.1 2012/12/29 05:13:27 jleffler Exp $
-- @(#)Create ywd_date() and ywdstr_date() stored procedures
-- Convert a date in format year, week, day (ISO 8601) to DATE.
-- Two variants:
-- ywd_date(yyyy SMALLINT, ww SMALLINT, dd SMALLINT) RETURNING DATE;
-- ywdstr_date(ywd CHAR(10)) RETURNING DATE;
-- NB: If week 53 is supplied, there is no check that the year had week
-- 53 (GIGO).
-- NB: If year yyyy is a leap year and yyyy-01-01 falls on Wed (3) or
-- Thu (4), there are 53 weeks in the year.
-- NB: If year yyyy is not a leap year and yyyy-01-01 falls on Thu (4),
-- there are 53 weeks in the year.
CREATE PROCEDURE ywd_date(yyyy SMALLINT, ww SMALLINT, dd SMALLINT) RETURNING DATE AS date;
DEFINE d DATE;
-- Check ranges
IF yyyy < 1 OR yyyy > 9999 OR ww < 1 OR ww > 53 OR dd < 1 OR dd > 7 THEN
RETURN NULL;
END IF;
LET d = day_one_week_one(yyyy);
LET d = d + (ww - 1) * 7 + (dd - 1);
RETURN d;
END PROCEDURE;
-- Input: 2012-W52-5
CREATE PROCEDURE ywdstr_date(ywd CHAR(10)) RETURNING DATE AS date;
DEFINE yyyy SMALLINT;
DEFINE ww SMALLINT;
DEFINE dd SMALLINT;
LET yyyy = SUBSTR(ywd, 1, 4);
LET ww = SUBSTR(ywd, 7, 2);
LET dd = SUBSTR(ywd, 10, 1);
RETURN ywd_date(yyyy, ww, dd);
END PROCEDURE;
CREATE TEMP TABLE test_dates(d DATE);
INSERT INTO test_dates VALUES('2011-12-28');
INSERT INTO test_dates VALUES('2011-12-29');
INSERT INTO test_dates VALUES('2011-12-30');
INSERT INTO test_dates VALUES('2011-12-31');
INSERT INTO test_dates VALUES('2012-01-01');
INSERT INTO test_dates VALUES('2012-01-02');
INSERT INTO test_dates VALUES('2012-01-03');
INSERT INTO test_dates VALUES('2012-01-04');
INSERT INTO test_dates VALUES('2012-01-05');
INSERT INTO test_dates VALUES('2012-01-06');
INSERT INTO test_dates VALUES('2012-01-07');
SELECT d, iso8601_weeknum(d), iso8601_weekday(d), ywdstr_date(iso8601_weekday(d))
FROM test_dates
ORDER BY d;
如评论中所述，即使年份只接受52周，代码也会接受第53周的约会。
答案 1 :(得分：15)
伪代码：
int julian = getDayOfYear(myDate) // Jan 1 = 1, Jan 2 = 2, etc...
int dow = getDayOfWeek(myDate) // Sun = 0, Mon = 1, etc...
int dowJan1 = getDayOfWeek("1/1/" + thisYear) // find out first of year's day
// int badWeekNum = (julian / 7) + 1 // Get our week# (wrong! Don't use this)
int weekNum = ((julian + 6) / 7) // probably better. CHECK THIS LINE. (See comments.)
if (dow < dowJan1) // adjust for being after Saturday of week #1
++weekNum;
return (weekNum)
为了澄清，此算法假设您按照以下方式对您的周数进行编号：
S M T W R F S
1 2 3 <-- week #1
4 5 6 7 8 9 10 <-- week #2
[etc.]
getDayOfWeek（）和getDayOfYear（）是大多数语言中的标准日期对象操作。如果你的没有它们，你可以从一些已知的日期（1970年1月1日是常见的日期）进行计数，在查看它之后的星期几。
如果您打算实施自己的日期计算程序，请记住可以被100整除的年份是不闰年，除非它们也可以被400整除。所以1900年不是一个飞跃一年，但2000年。如果你要及时工作，你必须搞砸Gregorian和Julian日历等，请参阅Wikipedia了解大量信息。
This link更详细地讨论了Windows / C ++中的日期/时间函数。
答案 2 :(得分：8)
我强烈建议使用C标准库的时间函数来计算周数。具体来说，strftime函数具有在给定日期分解（struct tm）格式的情况下打印周数（在许多其他值中）的说明符。这是一个小样本程序，说明了这一点：
#include <stdio.h>
#include <string.h>
#include <time.h>
int
main(void)
{
struct tm tm;
char timebuf[64];
// Zero out struct tm
memset(&tm, 0, sizeof tm);
// November 4, 2008 11:00 pm
tm.tm_sec = 0;
tm.tm_min = 0;
tm.tm_hour = 23;
tm.tm_mday = 4;
tm.tm_mon = 10;
tm.tm_year = 108;
tm.tm_isdst = -1;
// Call mktime to recompute tm.tm_wday and tm.tm_yday
mktime(&tm);
if (strftime(timebuf, sizeof timebuf, "%W", &tm) != 0) {
printf("Week number is: %s\n", timebuf);
}
return 0;
}
此程序的输出（在Linux上使用GCC编译，在Windows上使用Microsoft Visual Studio 2005 SP1编译）是：
Week number is: 44
您可以详细了解strftime here。
答案 3 :(得分：2)
struct tm用于表示“细分时间”，至少包含以下字段：
int tm_sec Seconds [0,60]. int tm_min Minutes [0,59]. int tm_hour Hour [0,23]. int tm_mday Day of month [1,31]. int tm_mon Month of year [0,11]. int tm_year Years since 1900. int tm_wday Day of week [0,6] (Sunday =0). int tm_yday Day of year [0,365]. int tm_isdst Daylight Savings flag.
您可以使用localtime（）函数从time_t创建struct tm。
您可以使用mktime（）函数从struct tm创建time_t。
关于struct tm的最好的部分是你可以做一些事情，比如在年度成员中添加24，当你调用mktime（）时，你将获得一个未来2年的time_t（这适用于任何一个成员，所以你可以，例如，将小时增加1000，然后在未来41天获得time_t）...
答案 4 :(得分：2)
很抱歉，我是新来的，无法对答案本身发表评论，但带有复选标记的答案中的伪代码不是正确的。
伪代码：
int julian = getDayOfYear(myDate) // Jan 1 = 1, Jan 2 = 2, etc...
int dow = getDayOfWeek(myDate) // Sun = 0, Mon = 1, etc...
int dowJan1 = getDayOfWeek("1/1/" + thisYear) // find out first of year's day
int weekNum = (julian / 7) + 1 // Get our week#
if (dow < dowJan1) // adjust for being after Saturday of week #1
++weekNum;
return (weekNum)
你不应该寻找“年度第一天”，而是去年的最后一天。
getDayOfWeek("12/31/" + thisYear-1)
将是正确的，而不是
getDayOfWeek("1/1/" + thisYear)
如果你不这样做，去年的最后一个工作日（如周一）总是提前一周。
答案 5 :(得分：2)
要双向转换，请参阅此处：Wikipedia article on ISO week dates
答案 6 :(得分：1)
我的定义是非ISO 8601（足够我的目的和快速）：
// week number of the year
// (Monday as the first day of the week) as a decimal number [00,53].
// All days in a new year preceding the first Monday are considered to be in week 0.
int GetWeek(const struct tm& ts)
{
return (ts.tm_yday + 7 - (ts.tm_wday ? (ts.tm_wday - 1) : 6)) / 7;
}
答案 7 :(得分：1)
使用howardhinnant.github.io/iso_week.html中的iso_week.h很容易做到：
#include <iostream>
#include "iso_week.h"
int main() {
using namespace iso_week;
using namespace std::chrono;
// Get the current time_point and floor to convert to the sys_days:
auto today = floor<days>(system_clock::now());
// Convert from sys_days to iso_week::year_weeknum_weekday format
auto yww = year_weeknum_weekday{today};
// Print current week number of the year
std::cout << "The current week of " << yww.year() << " is: "
<< yww.weeknum() << std::endl;
// Set any day
auto any_day = 2014_y/9/28;
// Get week of `any_day`
std::cout << "The week of " << any_day.year() << " on `any day` was: "
<< any_day.weeknum() << std::endl;
}
，输出为：
The current week of 2019 is: W18
The week in 2014 on `any day` was: W09
答案 8 :(得分：0)
使用gmtime或localtime计算自星期日（即星期几）以来的天数以及自1月1日以来的天数（注意后者中1月1日为“0”）。
任意位决定第1周开始的哪一天：通常它只取决于1月1日的哪一天，当然你可以从gmtime的两条信息中计算出来。然后使用表查找7种可能性，它可能比编码规则更容易。
例如，我认为Outlook使用的标准是第1周是包含星期四的第一周。因此，如果1月1日是星期日，那么第1周的第一天是1月1日或第0天。剩下的可能性是星期一，-1;星期二，-2;星期三，-3;星期四，-4;星期五，2;星期六，1。
请注意负数：“第1周的星期日”在7个案例中有4个实际上并不存在，但如果我们假装它是前一年的一天，我们将得到正确的答案。 / p>
一旦你有了它，它和你的约会之间的天数告诉你周数：除以7并加1。
那就是说，我想在某个地方有一个Windows API可以提供与Outlook相同的周数。我只是不知道它是什么，当然如果你的第1周规则与Outlook不同，那么它可能没什么用。
未经测试的代码：
int firstdays[7] = { 0, -1, -2, -3, -4, 2, 1 }; // or some other Week 1 rule
struct tm breakdown;
time_t target = time_you_care_about();
_gmtime_s(&breakdown,&target);
int dayofweek = breakdown.tm_wday;
int dayofyear = breakdown.tm_yday;
int jan1wday = (dayofweek - dayofyear) % 7;
if (jan1wday < 0) jan1wday += 7;
int week1first = firstdays[jan1wday];
if (dayofyear < week1first) return 0;
return ((dayofyear - week1first)/7) + 1;
无论如何都是这样的。
答案 9 :(得分：0)
Boost提供gregorian :: date :: week_number（） 见http://www.boost.org/doc/libs/1_38_0/doc/html/boost/gregorian/date.html和 http://www.boost.org/doc/libs/1_38_0/boost/date_time/gregorian/greg_date.hpp
但是我找不到一种方法来获得与周数匹配的年份数（可能与该日期的日历年不同）。
答案 10 :(得分：0)
我的假设是，这一年的第一周可能包含最多7天，如Olie的回答所示。 代码不处理周日开始的文化，而不是周日，这是世界的很大一部分。
tm t = ... //the date on which to find week of year
int wy = -1;
struct tm t1;
t1.tm_year = t.tm_year;
t1.tm_mday = t1.tm_mon = 1; //set to 1st of January
time_t tt = mktime(&t1); //compute tm
//remove days for 1st week
int yd = t.tm_yday - (7 - t1.tm_wday);
if(yd <= 0 ) //first week is now negative
wy = 0;
else
wy = (int)std::ceil( (double) ( yd/7) ); //second week will be 1
答案 11 :(得分：0)
public int GetWeekOfYear(DateTime todayDate)
{
int days = todayDate.DayOfYear;
float result = days / 7;
result=result+1;
Response.Write(result.ToString());
return Convert.ToInt32(result);
}
只需将当前日期作为参数传递给此函数。 然后你会得到当前的周数。 希望它能解决你的问题。任何建议都非常受欢迎。
答案 12 :(得分：0)
time_t t = time(NULL);
tm* timePtr = localtime(&t);
double day_of_year=timePtr->tm_yday +1 ; // 1-365
int week_of_year =(int) ceill(day_of_year/7.0);
答案 13 :(得分：0)
这是我的解决方案，但它不在C ++中
NoOfDays = (CurrentDate - YearStartDate)+1
IF NoOfDays MOD 7 = 0 Then
WeekNo = INT(NoOfDays/7)
ELSE
WeekNo = INT(NoOfDays/7)+1
END
答案 14 :(得分：0)
154答案 15 :(得分：-1)
/**
* @brief WeekNo
* @param yr
* @param mon
* @param day
* @param iso
* @return
*
* Given a date, return the week number
* Note. The first week of the year begins on the Monday
* following the previous Thursday
* Follows ISO 8601
*
* Mutually equivalent definitions for week 01 are:
*
* the week with the year's first Thursday in it (the ISO 8601 definition)
* the week with the Thursday in the period 1 – 7 January
* the week starting with the Monday in the period 29 December – 4 January
* the week starting with the Monday which is nearest in time to 1 January
* the week ending with the Sunday in the period 4 – 10 January
* the week with 4 January in it
* the first week with the majority (four or more) of its days in the starting year
* If 1 January is on a Monday, Tuesday, Wednesday or Thursday, it is in week 01.
* If 1 January is on a Friday, Saturday or Sunday, it is part of week 52 or 53 of the previous year.
* the week with the year's first working day in it (if Saturdays, Sundays, and 1 January are not working days).
*** strftime has a conversion of struct tm to weeknumber. strptime fills in tm struct**
* Code uses strptime, strftime functions.
*/
int WeekNo( int yr,int mon, int day, int iso)
{
struct tm tm;
char format[32];
//memset(tm,0,sizeof(tm));
sprintf(format,"%d-%02d-%02d",yr,mon,day);
strptime(format, "%Y-%m-%d", &tm);
// structure tm is now filled in for strftime
strftime(format, sizeof(format), iso? "%V":"%U", &tm);
//puts(format);
return atoi(format);
}
作为Weekno调用（2015,12,23,1）; //对于ISO周数。 Weekno（2015,12,23,0）//非ISO周数