如何在c ++中将包含time的字符串变量转换为time_t类型？

Question

我有一个包含时间的字符串变量,格式为hh:mm:ss.如何将其转换为time_t类型？例如:string time_details ="16:35:12"

另外,如何比较包含时间的两个变量,以便确定哪个是最早的？例如:string curr_time ="18:35:21"string user_time ="22:45:31"

Answer 1

您可以使用strptime(3)解析时间,然后mktime(3)将其转换为time_t:

const char *time_details = "16:35:12";
struct tm tm;
strptime(time_details, "%H:%M:%S", &tm);
time_t t = mktime(&tm);  // t is now your desired time_t

Answer 2

使用C++ 11,您现在可以做到

struct std::tm tm;
std::istringstream ss("16:35:12");
ss >> std::get_time(&tm, "%H:%M:%S"); // or just %T in this case
std::time_t time = mktime(&tm);

请参阅std :: get_time和strftime以供参考

Answer 3

这应该工作:

int hh, mm, ss;
struct tm when = {0};

sscanf_s(date, "%d:%d:%d", &hh, &mm, &ss);


when.tm_hour = hh;
when.tm_min = mm;
when.tm_sec = ss;

time_t converted;
converted = mktime(&when);

根据需要修改.