How Do I Get Yesterday's Date In C?

I am wanting to get yesterday's date into a char in the format: YYYYMMDD (with no slashes dots etc.).

I am using this code to get toda开发者_Go百科y's date:

time_t now;

struct tm  *ts;  
char yearchar[80]; 

now = time(NULL);  
ts = localtime(&now);

strftime(yearchar, sizeof(yearchar), "%Y%m%d", ts);

How would I adapt this code so that it is generating yesterday's date instead of today's?

Many Thanks.

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The mktime() function will normalise the struct tm that you pass it (ie. it will convert out-of-range dates like 2020/2/0 into the in-range equivalent 2020/1/31) - so all you need to do is this:

time_t now;
struct tm  *ts;  
char yearchar[80]; 

now = time(NULL);
ts = localtime(&now);
ts->tm_mday--;
mktime(ts); /* Normalise ts */
strftime(yearchar, sizeof(yearchar), "%Y%m%d", ts);

---

how about adding

now = now - (60 * 60 * 24)

Might fail in some VERY rare corner cases (e.g. during leapseconds) but should do what you want 99.999999% of the time.

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Simply subtracting one day's worth of seconds from time(NULL); should do. Change this line:

now = time(NULL);

to this:

now = time(NULL) - (24 * 60 * 60);

---

Please try this code

#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <string.h>
 
int main(void)
{
    char yestDt[9];
    time_t now = time(NULL);
    now = now - (24*60*60);
    struct tm *t = localtime(&now);
    sprintf(yestDt,"%04d%02d%02d", t->tm_year+1900, t->tm_mday,t->tm_mon+1);
    printf("Target String: \"%s\"", yestDt);
    return 0;
}

---

You're pretty close on. First of all, Tyler's solution will almost work -- you need to use (24*60*60*1000) since time(3) returns milliseconds. But have a look at that struct tm. It has fields for all the components of a date.

Update: Damn, my mistake -- time(3) does return seconds. I was thinking of another call. But have a look at the contents of struct tm anyway.

---

You can manipulate the contents of the ts struct before passing it to strftime. The day of the month is contained in the tm_mday member. Basic procedure:

/**
 * If today is the 1st, subtract 1 from the month
 * and set the day to the last day of the previous month
 */
if (ts->tm_mday == 1)
{
  /**
   * If today is Jan 1st, subtract 1 from the year and set
   * the month to Dec.
   */
  if (ts->tm_mon == 0)
  {
    ts->tm_year--;
    ts->tm_mon = 11;
  }
  else
  {
    ts->tm_mon--;
  }

  /**
   * Figure out the last day of the previous month.
   */
  if (ts->tm_mon == 1)
  {
    /**
     * If the previous month is Feb, then we need to check 
     * for leap year.
     */
    if (ts->tm_year % 4 == 0 && ts->tm_year % 400 == 0)
      ts->tm_mday = 29;
    else
      ts->tm_mday = 28;
  }
  else
  {
    /**
     * It's either the 30th or the 31st
     */
    switch(ts->tm_mon)
    {
       case 0: case 2: case 4: case 6: case 7: case 9: case 11:
         ts->tm_mday = 31;
         break;

       default:
         ts->tm_mday = 30;
    }
  }
}
else
{
  ts->tm_mday--;
}

Edit: Yes, days of the month are numbered from 1 whereas everything else (seconds, minutes, hours, weekdays, and days of the year) are numbered from 0.

---

time_t now;
int day;

struct tm  *ts;  
char yearchar[80]; 

now = time(NULL);  
ts = localtime(&now);
day = ts->tm_mday;

now = now + 10 - 24 * 60 * 60;
ts = localtime(&now);
if (day == ts->tm_mday)
{
  now = now - 24 * 60 * 60;
  ts = localtime(&now);
}

strftime(yearchar, sizeof(yearchar), "%Y%m%d", ts);

Will work with leap seconds too.