How to convert a date format: 2019-08-22 16:16:08] to the format: aammjjhhmmss in C language

I need to convert this date : 2019-08-22 16:16:08 to this specific form: aammjjhhmmss (aa for 'année' or year; jj for 'jour' or day).

I'm working with C language. This is my code. I don't find it good because I need to call the function three times to get what I want:

#include <stdio.h>
#include <string.h>

void removeChar(char *s, int c){ 

    int j;
    int n = strlen(s); 
    for (int i=j=0; i<n; i++) 
       if (s[i] != c) 
          s[j++] = s[i]; 

    s[j] = '\0'; 
} 

int main() 
{ 
   char s[] = "2019-08-22 16:16:08"; 
   removeChar(s, '-'); 
   printf("test 1 without '-' %s \n",s); 
   removeChar(s, ':');
   printf("test 2 without ':' %s \n",s); 
   removeChar(s, ' ');
   printf("test 3 without ' ' %s \n",s); 
   return 0; 
}

Results :

test 1 without '-' 20190822 16:16:08                                                                                                        
test 2 without ':' 20190822 161608                                                                                                          
test 3 without ' ' 20190822161608

And the format is: YYMMDDHHMMSS.

Solution

You can treat this as an exercise in date/time manipulation (hard), or as an exercise in string manipulation (easier).

If you want to do date/time manipulation, then in a POSIX environment, strptime() and strftime() are your friends. (NB: strftime() is standard C; strptime() is part of POSIX.)

Your code treats it as an exercise in string manipulation. Your result shows YYYYMMDD (4 digits for the year); your request shows AA or YY (2 digits) for the year. I wonder which is correct? If you have 4 input and 4 output digits life is simpler than if you have 4 input and 2 output. Assuming that it should be 4 digits in, 4 digits out, then all you need to do is copy the digits.

There's a subtle bug in your code:

int j;
int n = strlen(s); 
for (int i=j=0; i<n; i++) 
   if (s[i] != c) 
      s[j++] = s[i]; 

s[j] = '\0';

You have two different variables j, and the j used in the last assignment is uninitialized. If you use GCC, you could use -Wshadow as a compilation option to get warnings about such variable shadowing. You should use:

int j = 0;
int n = strlen(s); 
for (int i = 0; i < n; i++) 
{
   if (s[i] != c) 
      s[j++] = s[i];
}
s[j] = '\0';

However, you legitimately worry about calling your function three times. You could write a variant that only needs to be called once, because it keeps only the digits in the input:

#include <ctype.h>

void keepDigits(char *s)
{
    int j = 0;
    int n = strlen(s);
    for (int i = 0; i < n; i++)
    {
        if (isdigit((unsigned char)s[i]))
            s[j++] = s[i];
    }
    s[j] = '\0';
}

Indeed, you could also avoid pre-scanning the string with strlen() by using:

void keepDigits(char *s)
{
    int j = 0;
    for (int i = 0; s[i] != '\0'; i++)
    {
        if (isdigit((unsigned char)s[i]))
            s[j++] = s[i];
    }
    s[j] = '\0';
}

Now you scan the string once instead of 6 times (3 calls to your 'remove' function which scans the entire string, and each of which calls strlen() which also scans the entire string).

Test code:

int main(void) 
{ 
   char s[] = "2019-08-22 16:16:08"; 
   printf("Before: [%s]\n", s);
   keepDigits(s); 
   printf("After:  [%s]\n", s);
   return 0; 
}

Output:

Before: [2019-08-22 16:16:08]
After:  [20190822161608]

If you only want 2 digits for the year, then you could use:

int main(void) 
{ 
   char s[] = "2019-08-22 16:16:08"; 
   char *p = &s[2];
   printf("Before: [%s]\n", p);
   keepDigits(p); 
   printf("After:  [%s]\n", p);
   return 0; 
}

Output:

Before: [19-08-22 16:16:08]
After:  [190822161608]

Note, however, that the leading two digits are still in s here.