Question

我正在执行以下操作以获得“日月日”格式的时间。

struct tm *example= localtime(&t);
strftime(buf,sizeof(buf),"%a %b %d %Y",example);
strncpy(time_buffer,buffer,sizeof(time_buffer))   ;

但是如果日期是单个数字，例如9，则显示为9.我想将其打印为09.任何想法如何做到？

Answer 1

strftime的联机帮助页说：

%d     The day of the month as a decimal number (range 01 to 31).

这似乎是你想要的。

// compile with: gcc -o ex1 -Wall ex1.c
#include "stdio.h"
#include "sys/time.h"
#include "time.h"

int main (const int argc, const char ** argv ) {
  time_t curr_time;
  char buff[1024];
  // time(&curr_time);
  curr_time = 1359684105; // Thu Jan 31 2013
  struct tm *now = localtime(&curr_time);
  strftime(buff, sizeof(buff),  "%a %b %d %Y", now);
  printf("time: %ld\n", curr_time);
  printf("time: %s\n", buff);

  curr_time += 24 * 60 * 60; // Fri Feb 01 2013
  now = localtime(&curr_time);
  strftime(buff, sizeof(buff),  "%a %b %d %Y", now);
  printf("time: %ld\n", curr_time);
  printf("time: %s\n", buff);
  return 0;
}

产生：

time: 1359684105
time: Thu Jan 31 2013
time: 1359770505
time: Fri Feb 01 2013

看起来就像你追求的那样。如果要删除前导零，可以使用％e：

   %e     Like %d, the day of the month as a decimal number, but a leading zero is replaced by a space. (SU)

作者报告％d不遵守solaris上的联机帮助页，这是另一种直接使用sprintf的解决方案：

// compile with: gcc -o ex1 -Wall ex1.c
#include "stdio.h"
#include "sys/time.h"
#include "time.h"

int main (const int argc, const char ** argv ) {
  time_t curr_time;
  char buff[1024], daynamebuff[8], monbuff[8], daynumbuff[3], yearbuff[8];

  // time(&curr_time);
  curr_time = 1359684105; // Thu Jan 31 2013
  curr_time += 24 * 60 * 60; // Fri Feb 01 2013
  struct tm *now = localtime(&curr_time);

  strftime(daynamebuff, sizeof(daynamebuff), "%a", now);
  strftime(monbuff,     sizeof(monbuff), "%b", now);
  strftime(daynumbuff,  sizeof(daynumbuff), "%e", now);
  strftime(yearbuff,    sizeof(yearbuff), "%Y", now);

  sprintf(buff, "%s %s %02d %s", daynamebuff, monbuff, now->tm_mday, yearbuff);
  printf("%s\n", buff);
  return 0;
}

strftime和struct tm中的两位数日期

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