C语言 如何在C中将Unix时间戳转换为日：月：日：年格式？
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How to convert Unix TimeStamp into day:month:date:year format in C?
提问by user2430771
How do we convert a Unix time stamp in C to day:month:date:year? For eg. if my unix time stamp is 1230728833(int), how we convert this value into this-> Thu Aug 21 2008?
我们如何将 C 中的 Unix 时间戳转换为日：月：日：年？例如。如果我的 unix 时间戳是 1230728833(int)，我们如何将此值转换为 this-> Thu Aug 21 2008？
Thanks,
谢谢，
回答by sjnarv
Per @H2CO3's proper suggestion to use strftime(3), here's an example program.
根据@H2CO3 的正确使用建议strftime(3)，这是一个示例程序。
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
static const time_t default_time = 1230728833;
static const char default_format[] = "%a %b %d %Y";
int
main(int argc, char *argv[])
{
time_t t = default_time;
const char *format = default_format;
struct tm lt;
char res[32];
if (argc >= 2) {
t = (time_t) atoi(argv[1]);
}
if (argc >= 3) {
format = argv[2];
}
(void) localtime_r(&t, <);
if (strftime(res, sizeof(res), format, <) == 0) {
(void) fprintf(stderr, "strftime(3): cannot format supplied "
"date/time into buffer of size %u "
"using: '%s'\n",
sizeof(res), format);
return 1;
}
(void) printf("%u -> '%s'\n", (unsigned) t, res);
return 0;
}
回答by Varun P
This code helps you to convert timestamp from system time into UTC and TAI human readable format.
此代码可帮助您将时间戳从系统时间转换为 UTC 和 TAI 人类可读格式。
#include <stdio.h>
#include <time.h>
#include <unistd.h>
int main(void)
{
time_t now, now1, now2;
struct tm ts;
char buf[80];
// Get current time
time(&now);
ts = *localtime(&now);
strftime(buf, sizeof(buf), "%a %Y-%m-%d %H:%M:%S", &ts);
year_pr = ts.tm_year;
printf("Local Time %s\n", buf);
//UTC time
now2 = now - 19800; //from local time to UTC time
ts = *localtime(&now2);
strftime(buf, sizeof(buf), "%a %Y-%m-%d %H:%M:%S", &ts);
printf("UTC time %s\n", buf);
//TAI time valid upto next Leap second added
now1 = now + 37; //from local time to TAI time
ts = *localtime(&now1);
strftime(buf, sizeof(buf), "%a %Y-%m-%d %H:%M:%S", &ts);
printf("TAI time %s\n", buf);
return 0;
}
