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/*
* SPDX-FileCopyrightText: 1991 - 1994, Julianne Frances Haugh
* SPDX-FileCopyrightText: 1996 - 1999, Marek Michałkiewicz
* SPDX-FileCopyrightText: 2003 - 2005, Tomasz Kłoczko
* SPDX-FileCopyrightText: 2008 , Nicolas François
*
* SPDX-License-Identifier: BSD-3-Clause
*/
#if !defined(__GLIBC__)
#define _XOPEN_SOURCE 500
#endif
#include <config.h>
#include <ctype.h>
#ident "$Id$"
#include "defines.h"
#include "prototypes.h"
#ifndef USE_GETDATE
#define USE_GETDATE 1
#endif
#if USE_GETDATE
#include "getdate.h"
/*
* strtoday() now uses get_date() (borrowed from GNU shellutils)
* which can handle many date formats, for example:
* 1970-09-17 # ISO 8601.
* 70-9-17 # This century assumed by default.
* 70-09-17 # Leading zeros are ignored.
* 9/17/72 # Common U.S. writing.
* 24 September 1972
* 24 Sept 72 # September has a special abbreviation.
* 24 Sep 72 # Three-letter abbreviations always allowed.
* Sep 24, 1972
* 24-sep-72
* 24sep72
*/
long strtoday (const char *str)
{
time_t t;
bool isnum = true;
const char *s = str;
/*
* get_date() interprets an empty string as the current date,
* which is not what we expect, unless you're a BOFH :-).
* (useradd sets sp_expire = current date for new lusers)
*/
if ((NULL == str) || ('\0' == *str)) {
return -1;
}
/* If a numerical value is provided, this is already a number of
* days since EPOCH.
*/
if ('-' == *s) {
s++;
}
while (' ' == *s) {
s++;
}
while (isnum && ('\0' != *s)) {
if (!isdigit (*s)) {
isnum = false;
}
s++;
}
if (isnum) {
long retdate;
if (getlong (str, &retdate) == 0) {
return -2;
}
return retdate;
}
t = get_date (str, NULL);
if ((time_t) - 1 == t) {
return -2;
}
/* convert seconds to days since 1970-01-01 */
return (long) (t + DAY / 2) / DAY;
}
#else /* !USE_GETDATE */
/*
* Old code, just in case get_date() doesn't work as expected...
*/
#include <stdio.h>
#ifdef HAVE_STRPTIME
/*
* for now we allow just one format, but we can define more later
* (we try them all until one succeeds). --marekm
*/
static const char *const date_formats[] = {
"%Y-%m-%d",
(char *) 0
};
#else
/*
* days and juldays are used to compute the number of days in the
* current month, and the cumulative number of days in the preceding
* months. they are declared so that january is 1, not 0.
*/
static const short days[13] = { 0,
31, 28, 31, 30, 31, 30, /* JAN - JUN */
31, 31, 30, 31, 30, 31
}; /* JUL - DEC */
static const short juldays[13] = { 0,
0, 31, 59, 90, 120, 151, /* JAN - JUN */
181, 212, 243, 273, 304, 334
}; /* JUL - DEC */
#endif
/*
* strtoday - compute the number of days since 1970.
*
* the total number of days prior to the current date is
* computed. january 1, 1970 is used as the origin with
* it having a day number of 0.
*/
long strtoday (const char *str)
{
#ifdef HAVE_STRPTIME
struct tm tp;
const char *const *fmt;
char *cp;
time_t result;
memzero (&tp, sizeof tp);
for (fmt = date_formats; *fmt; fmt++) {
cp = strptime ((char *) str, *fmt, &tp);
if ((NULL == cp) || ('\0' != *cp)) {
continue;
}
result = mktime (&tp);
if ((time_t) - 1 == result) {
continue;
}
return (long) (result / DAY); /* success */
}
return -1;
#else
char slop[2];
int month;
int day;
int year;
long total;
/*
* start by separating the month, day and year. the order
* is compiled in ...
*/
if (sscanf (str, "%d/%d/%d%c", &year, &month, &day, slop) != 3) {
return -1;
}
/*
* the month, day of the month, and year are checked for
* correctness and the year adjusted so it falls between
* 1970 and 2069.
*/
if ((month < 1) || (month > 12)) {
return -1;
}
if (day < 1) {
return -1;
}
if ( ((2 != month) || ((year % 4) != 0))
&& (day > days[month])) {
return -1;
} else if ((month == 2) && ((year % 4) == 0) && (day > 29)) {
return -1;
}
if (year < 0) {
return -1;
} else if (year <= 69) {
year += 2000;
} else if (year <= 99) {
year += 1900;
}
/*
* On systems with 32-bit signed time_t, time wraps around in 2038
* - for now we just limit the year to 2037 (instead of 2069).
* This limit can be removed once no one is using 32-bit systems
* anymore :-). --marekm
*/
if ((year < 1970) || (year > 2037)) {
return -1;
}
/*
* the total number of days is the total number of days in all
* the whole years, plus the number of leap days, plus the
* number of days in the whole months preceding, plus the number
* of days so far in the month.
*/
total = (long) ((year - 1970) * 365L) + (((year + 1) - 1970) / 4);
total += (long) juldays[month] + (month > 2 && (year % 4) == 0 ? 1 : 0);
total += (long) day - 1;
return total;
#endif /* HAVE_STRPTIME */
}
#endif /* !USE_GETDATE */
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