在 C++ 中比较 2 个日期

Comparing 2 dates in c++

提问by David Faizulaev

I was wondering if there is any relatively easy and short date comparison functions in C++. My dates are of type char*, and have the following format: DD\MM\YYYY

我想知道 C++ 中是否有任何相对简单和短日期的比较函数。我的日期类型为char*，格式如下：DD\MM\YYYY

Thanks.

谢谢。

回答by Potatoswatter

Parsing is usually done on streams, not strings, but you can use a stringstream.

解析通常在流上完成，而不是在字符串上完成，但您可以使用stringstream.

std::istringstream date_s( "04\10\1984" );
struct tm date_c;
date_s >> std::get_time( &date_c, "%d\%m\%Y" );
std::time_t seconds = std::mktime( & date_c );

Now you can compare seconds using <to determine which was earlier.

现在您可以比较秒数<以确定哪个更早。

Note, std::get_timeis new in C++11. It is defined in terms of strptime, which is from POSIX but not part of the C99 standard. You can use strptimeif a C++11 library is not available. If you're brave, you can also use the std::time_getfacet… it's ugly though.

注意，std::get_time在 C++11 中是新的。它是根据 定义的strptime，它来自 POSIX 但不是 C99 标准的一部分。strptime如果 C++11 库不可用，您可以使用。如果你很勇敢，你也可以使用std::time_getfacet……虽然它很丑。

If you don't want to know anything about the dates other than which is earlier, you can use std::lexicographical_compare. It would be a one-liner but the function name is so long.

如果您不想知道除更早日期以外的任何日期，您可以使用std::lexicographical_compare. 这将是一个单行但函数名太长了。

// return true if the date string at lhs is earlier than rhs
bool date_less_ddmmyyyy( char const *lhs, char const *rhs ) {
    // compare year
    if ( std::lexicographical_compare( lhs + 6, lhs + 10, rhs + 6, rhs + 10 ) )
        return true;
    if ( ! std::equal( lhs + 6, lhs + 10, rhs + 6 ) )
        return false;
    // if years equal, compare month
    if ( std::lexicographical_compare( lhs + 3, lhs + 5, rhs + 3, rhs + 5 ) )
        return true;
    if ( ! std::equal( lhs + 3, lhs + 5, rhs + 3 ) )
        return false;
    // if months equal, compare days
    return std::lexicographical_compare( lhs, lhs + 2, rhs, rhs+2 );
}

See also how to convert datetime to unix timestamp in c?.

另请参阅如何在 c 中将日期时间转换为 unix 时间戳？.

回答by Olaf Dietsche

If this is really a fixed format, you can do it with simple C string comparison

如果这确实是固定格式，则可以通过简单的 C 字符串比较来完成

int date_cmp(const char *d1, const char *d2)
{
    int rc;
    // compare years
    rc = strncmp(d1 + 6, d2 + 6, 4);
    if (rc != 0)
        return rc;

    // compare months
    rc = strncmp(d1 + 3, d2 + 3, 2);
    if (rc != 0)
        return rc;

    // compare days
    return strncmp(d1, d2, 2);
}

This works like strncmp. It returns a value less than 0, if d1is earlier than d2, 0 if both are the same date, and a value greater than 0, if d1is later than d2.

这就像strncmp. 它返回一个小于 0 的值，如果d1早于d2，如果两者是相同的日期，则返回 0，如果d1是晚于，则返回大于 0 的值d2。

Another approach would be to convert it with strptimeand mktimeto time_tand compare these with difftime

另一种方法是将其与转换strptime，并mktime以time_t和比较这些与difftime

struct tm tm;
time_t t1, t2;
strptime(d1, "%d\%m\%Y", &tm);
t1 = mktime(&tm);
// do the same with d2
double diff = difftime(t1, t2);

回答by johnnycrash

How about an efficient solution? Your fixed size dates only require 8 chars if you ignore the slashes. So with a little shifting and byte swapping you can compare them as 64 bit ints. This is faster than comparing as strings.

一个有效的解决方案怎么样？如果忽略斜线，则固定大小的日期只需要 8 个字符。因此，通过一些移位和字节交换，您可以将它们作为 64 位整数进行比较。这比作为字符串进行比较要快。

using std::cout;
using std::endl;
typedef unsigned __int16 U2;
typedef unsigned __int32 U4;
typedef unsigned __int64 U8;
#define bswap2 _byteswap_ushort
#define bswap4 _byteswap_ulong
#define bswap8 _byteswap_uint64

const int YYYYMMDD = 0;
const int YYYY_MM_DD = 1;
const int DDMMYYYY = 2;
const int DD_MM_YYYY = 3;

// compiler will optimize the if's out.
template <int FMT>
U8 DateToInt(char* sz) {
    if (FMT == YYYYMMDD) {
        return bswap8(*(U8*)sz);
    }
    if (FMT == YYYY_MM_DD) {
        U4 y = *(U4*)sz, m = *(U2*)(sz + 5), d = *(U2*)(sz + 8);
        return ((U8)bswap4(y) << 32) | (bswap2(m) << 16) | bswap2(d);
    }
    if (FMT == DD_MM_YYYY) {
        U4 y = *(U4*)(sz + 6), m = *(U2*)(sz + 3), d = *(U2*)sz;
        return ((U8)bswap4(y) << 32) | (bswap2(m) << 16) | bswap2(d);
    }
}

template<int FMT1, int FMT2 = FMT1>
__int64 CompareDate(char* sz1, char* sz2) {
    return DateToInt<FMT1>(sz1) - DateToInt<FMT2>(sz2);
}

void main() {
    cout << CompareDate<YYYYMMDD>("20151025", "20151026") << endl;
    cout << CompareDate<YYYYMMDD>("20151025", "20151024") << endl;
    cout << CompareDate<YYYYMMDD, YYYY_MM_DD>("20151025", "2015/10/26") << endl;
    cout << CompareDate<YYYYMMDD, YYYY_MM_DD>("20151025", "2015/10/24") << endl;
    cout << CompareDate<YYYYMMDD, DD_MM_YYYY>("20151025", "26/10/2015") << endl;
    cout << CompareDate<YYYYMMDD, DD_MM_YYYY>("20151025", "24/10/2015") << endl;
}

output

输出

-1
1
-1
1
-1
1

回答by eeeeasy

You need to extract the numeric data from the string. Worst case scenario is a bunch of loops and string to integer conversion functions.

您需要从字符串中提取数字数据。最坏的情况是一堆循环和字符串到整数的转换函数。

You can do it easily with sscanf and sprintf. If you're used to printfand scanfthen this is simple to understand, and you can easily adapt it to other cases. There are no secret magic function calls.

您可以使用 sscanf 和 sprintf 轻松完成。如果您习惯了printf，scanf那么这很容易理解，您可以轻松地将其调整到其他情况。没有秘密的魔法函数调用。

#include <stdio.h>
void main()
{
    char* date1 = "9\12\2012"; 
    char* date2 = "6\11\2013"; 

    int day1,month1,year1;
    int day2,month2,year2;

    sscanf(date1,"%d\%d\%d",&day1,&month1,&year1); //reads the numbers
    sscanf(date2,"%d\%d\%d",&day2,&month2,&year2); //from the string

    if (year1<year2 || month1<month2 || day1<day2) //compares 2 dates
    {
        printf("date1 < date2\n");
    }
    else
    {
        printf("date1 >= date2\n");
    }

    char newdate[15];

    sprintf(newdate,"%d\%d\%d",13,2,1998); //make a date string from numbers
    printf("%s\n",newdate);
}