如何从c

Question

我正在使用C在程序中读取文件。在文件中，每个日期都有一些单独的行，就像这样：

20190101
20190304
20180922

现在，我希望程序将它们读取为日期，并找到当前日期与这些日期之间的差异。是否可以将这些日期转换为C可读的格式？像这样的东西：2019.01.01 目前，我使用fgets进行了阅读，但无法将其转换为上述格式。 这是我的代码：

#include <stdio.h>
#include <stdlib.h>
void FileRead(FILE *pr)
{
    char date [15];
    fgets(date,15,pr);
    printf("date : %s", date);

}
int main()
{
   char x;
   FILE *pr;
   pr=fopen("mytextfile.txt","r");
   if(pr==NULL)
   {
       printf("File can not be opened.");
       return 0;
   }
  while(scanf("%c", &x))
    {
        switch(x)
        {

        case 'v' :
            FileRead(pr);
        break;

        case 'k' :
            return 0;

        }
    }
fclose(pr);
    return 0;
}

Answer 1

一旦您通过char*拥有fgets，请使用atoi转换为整数。然后，您将获得YYYYMMDD格式的日期。然后，您可以

const int day = dateYYYYMMDD % 100;
const int month = (dateYYYYMMDD % 10000)/100;
const int year = dateYYYYMMDD / 10000;

您可以使用C Date and time functions来获取当前日期。

要使当前日期与这些日期之间有所不同，一种选择是将所有日期转换为Julian date，然后对朱利安日期进行数字差异。

如果很难获取居利安日期，请归一化为同一年（从低年级到高年级，基于年份添加365或366是否飞跃）。然后到同一个月，然后进行日差。

Answer 2

输出是，您只需循环文件行并在其中设置if条件

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

int main(void)
{
    char buff[20];
    time_t now = time(NULL);

    strftime(buff, 20, "%Y%m%d\n", localtime(&now));
    printf("Today : %s \n\n", buff);

    static const char filename[] = "file.txt";
    FILE *file = fopen(filename, "r");

    if (file != NULL)
    {
        char line[128]; /* or other suitable maximum line size */
        while (fgets(line, sizeof line, file) != NULL) /* read a line */
        {
            fputs(line, stdout); /* write the line */

            if (strcmp(line, buff) == 0)
                printf("\nThis line of date equal to current date\n");

        }
        fclose(file);
    }
    else
    {
        perror(filename); /* why didn't the file open? */
    }
    return 0;
}

file.txt

20190315
20190304
20180922

Answer 3

从现在到现在，解决差异的最简单方法是从文件中读取日期并将其解析为月，日和年（m, d, y）。如果您不打算对每次转换进行完整的验证，则分隔4位数字和2位数字的月份和日期的简单方法是使用fscanf和适当的 field-width < / em>修饰符可将转换限制为所需的位数，例如

    while (fscanf (fp, "%4d%2d%2d", &y, &m, &d) == 3) {

那么循环中所需要做的就是用年，月和日的值填充struct tm（记住从年中减去1900，并设置小时，分钟和第二个成员为零，夏令时成员为-1）。一个简单的函数可以执行此操作，并在调用time_t之后返回mktime，例如

time_t fill_broken_down_time (int y, int m, int d)
{                   /* initialize struct members */
    struct tm bdt = { .tm_sec=0, .tm_min=0, .tm_hour=0, .tm_mday=d, 
                    .tm_mon=m>0?m-1:0, .tm_year=y-1900, .tm_isdst=-1 };

    return mktime(&bdt);    /* return mktime conversion to time_t */
}

最后，您需要获取time_t值并调用difftime以获取当前时间与从文件中读取的时间之间的差值，以double的秒数表示。 。在main()中继续循环，例如

    while (fscanf (fp, "%4d%2d%2d", &y, &m, &d) == 3) {
        time_t  now = time(NULL),
                then = fill_broken_down_time (y, m, d);
        printf ("date[%d] %d/%d/%d is %.2f seconds from now.\n", 
                n++, m, d, y, difftime (now, then));
    }

完全将其放入，您可以执行以下操作：

#include <stdio.h>
#include <time.h>

time_t fill_broken_down_time (int y, int m, int d)
{                   /* initialize struct members */
    struct tm bdt = { .tm_sec=0, .tm_min=0, .tm_hour=0, .tm_mday=d, 
                    .tm_mon=m>0?m-1:0, .tm_year=y-1900, .tm_isdst=-1 };

    return mktime(&bdt);    /* return mktime conversion to time_t */
}

int main (int argc, char **argv) {
    /* use filename provided as 1st argument (stdin by default) */
    FILE *fp = argc > 1 ? fopen (argv[1], "r") : stdin;
    int y, m, d, n = 1;

    if (!fp) {  /* validate file open for reading */
        perror ("file open failed");
        return 1;
    }

    while (fscanf (fp, "%4d%2d%2d", &y, &m, &d) == 3) {
        time_t  now = time(NULL),
                then = fill_broken_down_time (y, m, d);
        printf ("date[%d] %d/%d/%d is %.2f seconds from now.\n", 
                n++, m, d, y, difftime (now, then));
    }
    if (fp != stdin) fclose (fp);   /* close file if not stdin */

    return 0;
}

（注意：，程序希望文件名从中读取日期作为程序的第一个参数（否则，默认情况下，如果未提供参数，则程序将从stdin中读取））

示例输入文件

$ cat dat/3dates.txt
20190101
20190304
20180922

使用/输出示例

$ ./bin/time_from_now dat/3dates.txt
date[1] 1/1/2019 is 6300212.00 seconds from now.
date[2] 3/4/2019 is 943412.00 seconds from now.
date[3] 9/22/2018 is 15030212.00 seconds from now.

按评论编辑修改输入文件格式

如果您的数据文件实际上不同于您最初在问题中发布的三行日期，并且在日期信息之前包含标题行，那么在处理日期行之前，您需要阅读，识别和处理这些行。由于您希望以天而不是秒为单位输出，因此您只需将秒数除以86400即可获得以天为单位的时差。

要读取句柄标题行，只需调整读数即可一次将整行读入足够大小的缓冲区中。声明一个足够大的常量，以确保缓冲区足够大，例如

#define MAXC 1024u  /* if you need a constant, #define one (or more) */
...
int main (int argc, char **argv) {
...
    char buf[MAXC];     /* buffer to hold each line read from file */

然后，您将只使用sscanf而不是fscanf对每一行进行完全相同的信息解析。如果行格式不符合yyyymmdd格式，则说明它不是日期行-可以按需要处理这些行（在下面的示例中，它们以前缀"non-date line: "输出。

结合将文件中的时间以来的秒数除以每天86400秒，您的新读取循环将类似于：

    while (fgets (buf, MAXC, fp)) {     /* read each line in file */
        /* if line isn't a date line, just output line as non-date line */
        if (sscanf (buf, "%4d%2d%2d", &y, &m, &d) != 3) {
            printf ("non-date line: %s", buf);
            continue;
        }
        time_t  now = time(NULL),
                then = fill_broken_down_time (y, m, d);
        double secs = difftime (now, then); /* get seconds between dates */
        printf ("date[%d] %02d/%02d/%04d is %11.2f sec (%g days) from now.\n", 
                n++, m, d, y, secs, secs / 86400.0);
    }

您说：

"I am not able to open the file"

程序希望您提供文件名作为程序的第一个参数读取，否则默认情况下将从stdin读取程序。这意味着您必须向程序提供文件名，例如

./yourprogram your_date_file

或者您必须通过以下方式在stdin上提供该数据：将信息从其他程序的输出传递到程序中，或者只是将文件重定向为stdin上的输入，例如

some_utility_making_dates | ./yourprogram

或

./yourprogram < your_date_file

结合所有更改，您的程序将如下所示：

#include <stdio.h>
#include <time.h>

#define MAXC 1024u  /* if you need a constant, #define one (or more) */

time_t fill_broken_down_time (int y, int m, int d)
{                   /* initialize struct members */
    struct tm bdt = { .tm_sec=0, .tm_min=0, .tm_hour=0, .tm_mday=d, 
                    .tm_mon=m>0?m-1:0, .tm_year=y-1900, .tm_isdst=-1 };

    return mktime(&bdt);    /* return mktime conversion to time_t */
}

int main (int argc, char **argv) {
    /* use filename provided as 1st argument (stdin by default) */
    FILE *fp = argc > 1 ? fopen (argv[1], "r") : stdin;
    int y, m, d, n = 1;
    char buf[MAXC];     /* buffer to hold each line read from file */

    if (!fp) {  /* validate file open for reading */
        perror ("file open failed");
        return 1;
    }

    while (fgets (buf, MAXC, fp)) {     /* read each line in file */
        /* if line isn't a date line, just output line as non-date line */
        if (sscanf (buf, "%4d%2d%2d", &y, &m, &d) != 3) {
            printf ("non-date line: %s", buf);
            continue;
        }
        time_t  now = time(NULL),
                then = fill_broken_down_time (y, m, d);
        double secs = difftime (now, then); /* get seconds between dates */
        printf ("date[%d] %02d/%02d/%04d is %11.2f sec (%g days) from now.\n", 
                n++, m, d, y, secs, secs / 86400.0);
    }
    if (fp != stdin) fclose (fp);   /* close file if not stdin */

    return 0;
}

带有标题的示例输入文件

$ cat dat/3dates-w-headers.txt
This file contains dates to read and convert to days.
The file also contains this description and dates in the format:

yyyymmdd
20190101
20190304
20180922

使用/输出示例

$ ./bin/time_from_now2 dat/3dates-w-headers.txt
non-date line: This file contains dates to read and convert to days.
non-date line: The file also contains this description and dates in the format:
non-date line:
non-date line: yyyymmdd
date[1] 01/01/2019 is  6348645.00 sec (73.4797 days) from now.
date[2] 03/04/2019 is   991845.00 sec (11.4797 days) from now.
date[3] 09/22/2018 is 15078645.00 sec (174.521 days) from now.

仔细检查一下，如果还有其他问题，请告诉我。