我有:
我需要得到一个天数[从1到31] - " YYYY-MM- DD " (ISO8601)。
有没有办法使用boost posix时间或使用其他一些C ++库来计算它?
即
感谢。
答案 0 :(得分:1)
我编写了一个轻量级C library,它可以做你想要的,有趣的部分是here,你可以看到算法很简单。
#include <stdio.h>
#include <stdint.h>
#include "dt_dow.h"
#include "dt_accessor.h"
const struct test {
int year;
int month; /* Month of the year [1=Jan, 12=Dec] */
int nth; /* Occurrence within month */
dt_dow_t dow; /* Day of the week [1=Mon, 7=Sun] */
int dom; /* Expected day of the month [1, 31] */
} tests[] = {
{ 2014, 3, 1, DT_SUNDAY, 2 },
{ 2014, 4, -1, DT_TUESDAY, 29 },
{ 2014, 4, -2, DT_MONDAY, 21 },
{ 2014, 4, -5, DT_TUESDAY, 1 },
{ 2014, 4, 1, DT_TUESDAY, 1 },
{ 2014, 12, 4, DT_WEDNESDAY, 24 },
};
int
main() {
int i, ntests;
ntests = sizeof(tests) / sizeof(*tests);
for (i = 0; i < ntests; i++) {
const struct test t = tests[i];
{
int dom = dt_dom(dt_from_nth_dow_in_month(t.year, t.month, t.nth, t.dow));
if (t.dom != dom) {
printf("dt_dom(dt_from_nth_dow_in_month(%d, %d, %d, %d))\n",
t.year, t.month, t.nth, t.dow);
printf(" got: %d\n", dom);
printf(" exp: %d\n", t.dom);
}
}
}
return 0;
}
如果你不想使用上面的库/代码,这是一个重新实现。
#include <stdio.h>
#include <assert.h>
#include <stdint.h>
#include <stdbool.h>
bool
leap_year(int y) {
return ((y % 4) == 0 && (y % 100 != 0 || y % 400 == 0));
}
int
days_in_month(int y, int m) {
static const int T[2][13] = {
{ 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 },
{ 0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }
};
assert(m >= 1);
assert(m <= 12);
return T[leap_year(y)][m];
}
/* Computes the day of the week [1=Mon, 7=Sun] from the given year, month, day. */
int
ymd_to_dow(int y, int m, int d) {
static const int T[13] = { 0, 6, 2, 1, 4, 6, 2, 4, 0, 3, 5, 1, 3 };
assert(y >= 1);
assert(m >= 1);
assert(m <= 12);
assert(d >= 1);
y -= m < 3;
return 1 + (y + y/4 - y/100 + y/400 + T[m] + d) % 7;
}
int
dom_from_nth_dow_in_month(int y, int m, int nth, int dow) {
int dim, dom;
assert(y >= 1);
assert(m >= 1);
assert(m <= 12);
assert(dow >= 1);
assert(dow <= 7);
dim = days_in_month(y, m);
if (nth > 0) {
dom = 1;
dom += (dow - ymd_to_dow(y, m, dom) + 7) % 7;
dom += --nth * 7;
if (dom <= dim)
return dom;
}
else if (nth < 0) {
dom = dim;
dom -= (ymd_to_dow(y, m, dom) - dow + 7) % 7;
dom -= ++nth * -7;
if (dom >= 1)
return dom;
}
return -1;
}
const struct test {
int year;
int month; /* Month of the year [1=Jan, 12=Dec] */
int nth; /* Occurrence within month */
int dow; /* Day of the week [1=Mon, 7=Sun] */
int dom; /* Expected day of the month [1, 31] */
} tests[] = {
{ 2014, 3, 1, 7, 2 },
{ 2014, 4, -1, 2, 29 },
{ 2014, 4, -2, 1, 21 },
{ 2014, 4, -5, 2, 1 },
{ 2014, 4, 1, 2, 1 },
{ 2014, 12, 4, 3, 24 },
};
int
main() {
int i, ntests;
ntests = sizeof(tests) / sizeof(*tests);
for (i = 0; i < ntests; i++) {
const struct test t = tests[i];
{
int dom = dom_from_nth_dow_in_month(t.year, t.month, t.nth, t.dow);
if (t.dom != dom) {
printf("dom_from_nth_dow_in_month(%d, %d, %d, %d))\n",
t.year, t.month, t.nth, t.dow);
printf(" got: %d\n", dom);
printf(" exp: %d\n", t.dom);
}
}
}
return 0;
}
答案 1 :(得分:0)
好工作chansen!这可能非常有用。 但是因为我可以自由使用提升 - 这是我迄今为止所做的:
// DateFromWeekNumAndDay
// year - Year YYYY.
// weekNum - From 0 to 4.
// weekDay - Week day starts from Sunday - 0 to Saturday - 6.
boost::gregorian::date DateFromWeekNumAndDay(boost::gregorian::date::year_type year, unsigned short weekNum, boost::date_time::weekdays weekDay)
{
boost::gregorian::date date(year, boost::gregorian::Apr, 1);
boost::gregorian::date::day_of_week_type d = date.day_of_week();
date += boost::gregorian::date_duration(weekNum * 7) + boost::gregorian::date_duration(weekDay - d);
return date;
}
据我所知,一切似乎都运转正常。 我从this帖子中采用了这种方法。感谢用户kebs获取该链接,并非常感谢用户Mikhail Melnik为他的&#34; GetDateFromWeekNumber&#34;功能样本。