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Unix C:在不同的区域获得时间

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我正在尝试使用C在不同的时区(PST)获得时间 .

#define PST (-8);
char* Time::getSecondSystemTime() {
    time_t rawtime;
    struct tm * timeinfo;
    char buffer[80];

    time(&rawtime);
    timeinfo = gmtime(&rawtime);

    timeinfo->tm_hour = timeinfo->tm_hour + PST;

    strftime(buffer, 80, "%I:%M %p", timeinfo);


    std::string temp = std::string(buffer); // to get rid of extra stuff
    std::string extraInfo = " Pacific Time ( US & Canada )";

    temp.append(extraInfo);

    return (char*) (temp.c_str());

}

这里的问题是,当GMT时间少于8小时(例如,现在,早上3点的时间),从它减去8小时不起作用!

在Unix中的不同时区获取时间的正确方法是什么?

3 回答

  • 4

    既然你说"UNIX",这就是使用TZ,但是, TZ=[what goes here] 你需要找出你系统上的[这里有什么] . 它可能是"America/LosAngeles"或PST的其他几个字符串之一 . 如果您的系统是POSIX:TZ = PST8PST保证可以正常工作 . 但它可能不是最佳的 .

    原始非 生产环境 代码假定TZ目前尚未使用 . 这是C,而不是C,因为你的标签是C:

    setenv("TZ", "PST8PST", 1);   // set TZ
    tzset();                // recognize TZ
    time_t lt=time(NULL);   //epoch seconds
    struct tm *p=localtime(&lt); // get local time struct tm
    char tmp[80]={0x0};
    strftime(tmp, 80, "%c", p);  // format time use format string, %c 
    printf("time and date PST: %s\n", tmp); // display time and date
    // you may or may not want to remove the TZ variable at this point.
    
  • 2

    我有以下C代码藏起来处理问题 . 效率不是第一个让人想起的词(两次调用 setenv() ,两次调用 tzset() ),但标准的C库并不能让它变得更容易:

    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    #include <time.h>
    #include <unistd.h>
    
    static void time_convert(time_t t0, char const *tz_value)
    {
        char old_tz[64];
        strcpy(old_tz, getenv("TZ"));
        setenv("TZ", tz_value, 1);
        tzset();
        char new_tz[64];
        strcpy(new_tz, getenv("TZ"));
        char buffer[64];
        struct tm *lt = localtime(&t0);
        strftime(buffer, sizeof(buffer), "%Y-%m-%d %H:%M:%S", lt);
        setenv("TZ", old_tz, 1);
        tzset();
        printf("%ld = %s (TZ=%s)\n", (long)t0, buffer, new_tz);
    }
    
    int main(void)
    {
        time_t t0 = time(0);
        char *tz = getenv("TZ");
        time_convert(t0, tz);
        time_convert(t0, "UTC0");
        time_convert(t0, "IST-5:30");
        time_convert(t0, "EST5");
        time_convert(t0, "EST5EDT");
        time_convert(t0, "PST8");
        time_convert(t0, "PST8PDT");
    }
    

    在原始代码中,您必须担心在更改小时偏移后规范化时间结构 . 您可以使用mktime()功能执行此操作 . 这是一个基于问题中的函数的程序,它是纯C并且避免了返回指向局部变量的指针(以及以分号结尾的 #define )的问题:

    #include <assert.h>
    #include <stdio.h>
    #include <string.h>
    #include <stdlib.h>
    #include <time.h>
    
    #define PST (-8)
    
    extern int getSecondSystemTime(char *buffer, size_t buflen);
    
    int getSecondSystemTime(char *buffer, size_t buflen)
    {
        time_t rawtime = time(0);;
        struct tm *timeinfo;
        char t_buff[32];
    
        timeinfo = gmtime(&rawtime);
        timeinfo->tm_hour = timeinfo->tm_hour + PST;
    
        time_t pst_time = mktime(timeinfo);
        assert(pst_time != (time_t)-1);
        int len = strftime(t_buff, sizeof(t_buff), "%Y-%m-%d %H:%M:%S", timeinfo);
        assert(len != 0);
    
        int rv = snprintf(buffer, buflen, "%ld = %s (%s)", (long)rawtime, t_buff, 
                          "Pacific Time (US & Canada)");
        assert(rv > 0);
        return rv;
    }
    
    int main(void)
    {
        char buffer[128];
    
        getSecondSystemTime(buffer, sizeof(buffer));
        printf("%s\n", buffer);
        return(0);
    }
    

    显然,更好的接口会将UTC时间值和时区偏移量(以小时和分钟为单位)作为参数传递 . 尽管我的计算机默认运行在美国/太平洋(或美国/洛杉矶)时区,但我测试TZ设置为各种值(包括US / Eastern,IST-05:30)并获得正确的值;根据过去的经验,我有理由相信计算是正确的 .

    我有另一个程序试图剖析从 mktime() 返回的 -1 是由于错误还是因为转换时间对应于 (time_t)-1

    /* Attempt to determine whether time is really 1969-12-31 23:59:59 +00:00 */
    static int unix_epoch_minus_one(const struct tm *lt)
    {
        printf("tm_sec = %d\n", lt->tm_sec);
        if (lt->tm_sec != 59)
            return(0);
        printf("tm_min = %d\n", lt->tm_min);
        /* Accounts for time zones such as Newfoundland (-04:30), India (+05:30) and Nepal (+05:45) */
        if (lt->tm_min % 15 != 14)
            return(0);
        /* Years minus 1900 */
        printf("tm_year = %d\n", lt->tm_year);
        if (lt->tm_year != 69 && lt->tm_year != 70)
            return(0);
        printf("tm_mday = %d\n", lt->tm_mday);
        if (lt->tm_mday != 31 && lt->tm_mday != 1)
            return(0);
        /* Months 0..11 */
        printf("tm_mon = %d\n", lt->tm_mon);
        if (lt->tm_mon != 11 && lt->tm_mon != 0)
            return(0);
        /* Pretend it is valid after all - though there is a small chance we are incorrect */
        return 1;
    }
    
  • 6

    这是一种更简洁的方法(此示例获取GMT时间,包括DST偏差):

    struct STimeZoneFromRegistry
    {
       long  Bias;
       long  StandardBias;
       long  DaylightBias;
       SYSTEMTIME StandardDate;
       SYSTEMTIME DaylightDate;
    };
    
    
    
    static SYSTEMTIME GmtNow()
    {
       FILETIME UTC;
       GetSystemTimeAsFileTime(&UTC); 
    
       SYSTEMTIME GMT;
    
       TIME_ZONE_INFORMATION tz = {0};
       STimeZoneFromRegistry binary_data;
       DWORD size = sizeof(binary_data);
       HKEY hk = NULL;
       TCHAR zone_key[] = _T("SOFTWARE\\Microsoft\\Windows NT\\CurrentVersion\\Time Zones\\GMT Standard Time");
       if ((RegOpenKeyEx(HKEY_LOCAL_MACHINE, zone_key, 0, KEY_QUERY_VALUE, &hk) == ERROR_SUCCESS) &&
          (RegQueryValueEx(hk, "TZI", NULL, NULL, (BYTE *) &binary_data, &size) == ERROR_SUCCESS))
       {
          tz.Bias = binary_data.Bias;
          tz.DaylightBias = binary_data.DaylightBias;
          tz.DaylightDate = binary_data.DaylightDate;
          tz.StandardBias = binary_data.StandardBias;
          tz.StandardDate = binary_data.StandardDate;
       }
    
       SystemTimeToTzSpecificLocalTime(&tz, &UTC, &GMT);
    
       return GMT;
     }
    

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