如何将以微秒为单位的字符串转换为C中的struct tm?
How to convert character string in microseconds to struct tm in C?
我有一个字符串,其中包含自纪元以来的微秒数。如何将其转换为时间结构?
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
int main ()
{
struct tm tm;
char buffer [80];
char *str ="1435687921000000";
if(strptime (str, "%s", &tm) == NULL)
exit(EXIT_FAILURE);
if(strftime (buffer,80,"%Y-%m-%d",&tm) == 0)
exit(EXIT_FAILURE);
printf("%s\n", buffer);
return 0;
}
编辑: 您可以简单地截断字符串,因为 struct tm
不会存储小于 1 秒的精度。
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main ()
{
struct tm now;
time_t secs;
char buffer [80];
char str[] ="1435687921000000";
int len = strlen(str);
if (len < 7)
return 1;
str[len-6] = 0; // divide by 1000000
secs = (time_t)atol(str);
now = *localtime(&secs);
strftime(buffer, 80, "%Y-%m-%d", &now);
printf("%s\n", buffer);
printf("%s\n", asctime(&now));
return 0;
}
程序输出:
2015-06-30
Tue Jun 30 19:12:01 2015
您可以将微秒转换为秒,然后像这样使用localtime()
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int main (void)
{
struct tm *tm;
char buffer[80];
char *str = "1435687921000000";
time_t ms = strtol(str, NULL, 10);
/* convert to seconds */
ms = (time_t) ms / 1E6;
tm = localtime(&ms);
if (strftime(buffer, 80, "%Y-%m-%d", tm) == 0)
return EXIT_FAILURE;
printf("%s\n", buffer);
return EXIT_SUCCESS;
}
请注意,在打印日期中,微秒不存在,因此您可以忽略该部分。
将字符串转换为 time_t,然后使用 gmtime(3) 或 localtime(3)。
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
int main () {
struct tm *tm;
char buffer [80];
char *str ="1435687921000000";
time_t t;
/* or strtoull */
t = (time_t)(atoll(str)/1000000);
tm = gmtime(&t);
strftime(buffer,80,"%Y-%m-%d",tm);
printf("%s\n", buffer);
return 0;
}
便携式解决方案(假设 32+ 位 int
)。以下不假设任何关于 time_t
.
使用 mktime()
,不需要将字段限制在其主要范围内。
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
int main(void) {
char buffer[80];
char *str = "1435687921000000";
// Set the epoch: assume Jan 1, 0:00:00 UTC.
struct tm tm = { 0 };
tm.tm_year = 1970 - 1900;
tm.tm_mday = 1;
// Adjust the second's field.
tm.tm_sec = atoll(str) / 1000000;
tm.tm_isdst = -1;
if (mktime(&tm) == -1)
exit(EXIT_FAILURE);
if (strftime(buffer, 80, "%Y-%m-%d", &tm) == 0)
exit(EXIT_FAILURE);
printf("%s\n", buffer);
return 0;
}
我有一个字符串,其中包含自纪元以来的微秒数。如何将其转换为时间结构?
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
int main ()
{
struct tm tm;
char buffer [80];
char *str ="1435687921000000";
if(strptime (str, "%s", &tm) == NULL)
exit(EXIT_FAILURE);
if(strftime (buffer,80,"%Y-%m-%d",&tm) == 0)
exit(EXIT_FAILURE);
printf("%s\n", buffer);
return 0;
}
编辑: 您可以简单地截断字符串,因为 struct tm
不会存储小于 1 秒的精度。
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main ()
{
struct tm now;
time_t secs;
char buffer [80];
char str[] ="1435687921000000";
int len = strlen(str);
if (len < 7)
return 1;
str[len-6] = 0; // divide by 1000000
secs = (time_t)atol(str);
now = *localtime(&secs);
strftime(buffer, 80, "%Y-%m-%d", &now);
printf("%s\n", buffer);
printf("%s\n", asctime(&now));
return 0;
}
程序输出:
2015-06-30
Tue Jun 30 19:12:01 2015
您可以将微秒转换为秒,然后像这样使用localtime()
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int main (void)
{
struct tm *tm;
char buffer[80];
char *str = "1435687921000000";
time_t ms = strtol(str, NULL, 10);
/* convert to seconds */
ms = (time_t) ms / 1E6;
tm = localtime(&ms);
if (strftime(buffer, 80, "%Y-%m-%d", tm) == 0)
return EXIT_FAILURE;
printf("%s\n", buffer);
return EXIT_SUCCESS;
}
请注意,在打印日期中,微秒不存在,因此您可以忽略该部分。
将字符串转换为 time_t,然后使用 gmtime(3) 或 localtime(3)。
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
int main () {
struct tm *tm;
char buffer [80];
char *str ="1435687921000000";
time_t t;
/* or strtoull */
t = (time_t)(atoll(str)/1000000);
tm = gmtime(&t);
strftime(buffer,80,"%Y-%m-%d",tm);
printf("%s\n", buffer);
return 0;
}
便携式解决方案(假设 32+ 位 int
)。以下不假设任何关于 time_t
.
使用 mktime()
,不需要将字段限制在其主要范围内。
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
int main(void) {
char buffer[80];
char *str = "1435687921000000";
// Set the epoch: assume Jan 1, 0:00:00 UTC.
struct tm tm = { 0 };
tm.tm_year = 1970 - 1900;
tm.tm_mday = 1;
// Adjust the second's field.
tm.tm_sec = atoll(str) / 1000000;
tm.tm_isdst = -1;
if (mktime(&tm) == -1)
exit(EXIT_FAILURE);
if (strftime(buffer, 80, "%Y-%m-%d", &tm) == 0)
exit(EXIT_FAILURE);
printf("%s\n", buffer);
return 0;
}