如果我有约会,如何计算该年内该日期的周数？
例如,在2008年,1月1日到1月6日在第1周,1月7日到13日在第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.
一个算法很适合我开始,示例代码也会有所帮助 - 我正在Windows上用C++开发.
从MS SQL Server 2005中的某个日期获取周数？
Jonathan Lef.. 29
请注意,虽然您对一年中第n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,可能有一周53.
[ 补充: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(确实是2010年1月3日星期日).同样,2005年1月1日星期六是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 - 虽然可能不可写 - 但没有进一步解释.'||' 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,使用上述函数也可以轻松编写反函数day_one_week_one():
-- @(#)$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,即使年份仅接受52周,代码也将接受第53周的约会.
请注意,虽然您对一年中第n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,可能有一周53.
[ 补充: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(确实是2010年1月3日星期日).同样,2005年1月1日星期六是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 - 虽然可能不可写 - 但没有进一步解释.'||' 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,使用上述函数也可以轻松编写反函数day_one_week_one():
-- @(#)$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,即使年份仅接受52周,代码也将接受第53周的约会.
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,此算法假设您为您的周数编号如下:
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,然后查看它是什么星期几.
如果你要实现自己的日期计算程序,请记住,可以被100整除的年份不是闰年,除非它们也可以被400整除.所以1900年不是闰年,而是2000年.如果你要及时工作,你必须搞乱格里高利与朱利安的日历等,请参阅维基百科了解大量信息.
此链接更详细地讨论了Windows/C++中的日期/时间函数.
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,该strftime函数具有在给定日期(以分解(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#include #include 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; }
该程序的输出(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
您可以在此处了解有关strftime的更多信息.
京公网安备 11010802040832号 | 京ICP备19059560号-6 