Thread: Converting Time Zones To Central Standard Time (CST)

I seem to be going in circles here and I would greatly appreciate any help I can get.




I have a date that comes in as "2019-05-16T09:30:00-04:00" as an example. (I wont know the timezone in advance)


In the example above it is -4:00 from UTC


2019-05-16 09:30:00 - 04:00


According to this chart UTC to CST Converter - Savvy Time
adding 4 hours to 09:30:00 gives me 13:30 (or 8:30 AM converted to cst)




My question is how do I get the incoming date to the actual CST time.

Code:

const char *time_details = "2019-05-16T09:30:00-04:00";
struct tm tm2;
strptime(time_details, "%Y-%m-%dT%H:%M:%S%z", &tm2);
time_t tmIncomingTime = mktime(&tm2);
struct tm * tmLocalTime = localtime(&tmIncomingTime);
time_t t3 = mktime(tmLocalTime);

The output of t3 is 9:30 which is not what I am expecting.


How can I get this converting the timezones with the correct time.


Thanks in advance.

But not RFC 822... RFC 822 uses +/-hhmm zone offset format.

Take a look at this code (pay attention to the comments).

Code:

/* test.c */
#define _GNU_SOURCE
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <time.h>

static void f( char * );

int main ( void )
{
  char **p;

  // RFC 822/ISO 8601 standard date/time format.
  // The zone offset must be in format [+|-]hhmm.
  static char *tstrs[] = { "2019-05-16T16:30:00-0300",
                           "2019-05-16T16:30:00+0300",
                           NULL };

  p = tstrs;
  while ( *p )
    f( *p++ );
}

// strips the final '\n' from string pointed by p.
#define stripnl(p) {char *tp_;if(tp_=strchr((p),'\n'))*tp_='\0';}

void f( char *tstr )
{
  struct tm tm = {0}; // strptime says tm should
                      // be initialized first. So,
                      // no garbage!
  time_t t, t2;
  long int offset;
  char *p, *q;

  // %F and %z are glibc extensions.
  // %F == %Y-%m-%d
  // %z expects a zone offset [+/-]hhmm (rfc822).
  strptime ( tstr, "%FT%H:%M:%S%z", &tm );

  // tm_gmtoff is a glibc extension.
  // mktime changes the buffer, so we cannot reuse tm.tm_gmtoff.
  offset = tm.tm_gmtoff;
  t = mktime ( &tm );
  t2 = t + offset;

  // ctime returns a pointer to a static buffer.
  // duplicate them.
  p = strdup( ctime( &t ) ); 
  q = strdup( ctime( &t2 ) ); 

  // strip final '\n' from buffers.
  stripnl( p );
  stripnl( q );

  printf( "%s -> %s (%s %s%ld secs)\n", 
          tstr, 
          q, p,
          offset >= 0 ? "+" : "", 
          offset );

  free( q );
  free( p );
}

so the data comes to me as "2019-05-16T09:30:00-04:00" and i can make it "2019-05-16T09:30:00-0400" but I am still lost on how to convert it to my current timezone

Code:

const char *time_details = "2019-05-16T09:30:00-0400";
struct tm tm2;
strptime(time_details, "%Y-%m-%dT%H:%M:%S%z", &tm2);
time_t tmIncomingTime = mktime(&tm2);
struct tm * tmLocalTime = localtime(&tmIncomingTime);
time_t t3 = mktime(tmLocalTime);