如何把字符串格式的日间转换成相应的秒, 急, 急, 知道的请进来帮个忙

[Gengoo]

我想把
string time1 = "";
time1 = "2008-11-25 03:04:25"

转换成相应的秒, 请问应该如何转换

[Gengoo]

#include <time.h>
#include <iostream>

using namespace std;

int main()
{
	int seconds1 = 0;
	int seconds2 = 0;

	string time1 = "2002-01-01 05:35:41";
	string time2 = "2002-01-01 05:36:33";

	struct tm tm1;

    strptime("2002-01-01 05:35:41","%Y-%m-%d%H:%M:%s", &tm1);
	
	seconds1 = time( NULL );
	cout << "seconds1 = " << seconds1 << endl;
	return 0;
}

我经过编译,总是报
Compiling...
main.cpp
G:\Time1\main.cpp(16) : error C2065: 'strptime' : undeclared identifier
执行 cl.exe 时出错.

Time1.exe - 1 error(s), 0 warning(s)

这倒底是什么原因

[Gengoo]

#include <time.h>
#include <iostream>

using namespace std;

int main()
{
	int seconds1 = 0;
	int seconds2 = 0;

	string time1 = "2002-01-01 05:35:41";
	string time2 = "2002-01-01 05:36:33";

	struct tm tm1;

    strptime("2002-01-01 05:35:41","%Y-%m-%d%H:%M:%s", &tm1);
	
	seconds1 = time( NULL );
	cout << "seconds1 = " << seconds1 << endl;
	return 0;
}

我经过编译,总是报
Compiling...
main.cpp
G:\Time1\main.cpp(16) : error C2065: 'strptime' : undeclared identifier
执行 cl.exe 时出错.

Time1.exe - 1 error(s), 0 warning(s)

这倒底是什么原因

[fibbery]

首先，如果你的字符串时间格式标准，那么你可以把年、月日时分秒取出来，再继续算秒数。

[stanboy]

[Quote=引用 4 楼 Gengoo 的回复:]
C/C++ code#include<time.h>#include<iostream>usingnamespacestd;intmain()
{intseconds1=0;intseconds2=0;stringtime1="2002-01-01 05:35:41";stringtime2="2002-01-01 05:36:33";structtm tm1;

strptime("2002-01-01 05:35:41","%Y-%m-%d%H:%M:%s",&tm1);

seconds1=time( NULL );
cout<<"seconds1 ="<<seconds1<<endl;return0;
}

我经过编译,总是报
Compiling...
main.cpp
G:\Time1\main.cpp(16) : error …
[/Quote]
我用gcc能编过去啊

[candyice]

把这个字符串进行分割
CString strBegin = m_time.Format("%Y-%m-%d %H:%M:%S");
int nCount；
//////////////////////////时间的转换//////////////////////////////////////////
//获取年份
nCount = strBegin.Find('-');
strYear = strBegin.Left(nCount);
strBegin = strBegin.Mid(nCount+1);
//获取月份
nCount = strBegin.Find('-');
strMonth = strBegin.Left(nCount);
strBegin = strBegin.Mid(nCount+1);
//获取日期
nCount = strBegin.Find(' ');
strDay = strBegin.Left(nCount);
strBegin = strBegin.Mid(nCount+1);
//获取小时
nCount = strBegin.Find(':');
strHour = strBegin.Left(nCount);
strBegin = strBegin.Mid(nCount+1);
//获取分钟
nCount = strBegin.Find(':');
strMinute = strBegin.Left(nCount);
strBegin = strBegin.Mid(nCount+1);
//获取秒
strSecond = strBegin;

[candyice]

再转换成相应的秒，看看能不能实现

[ufouser]

Windows下好像没strptime.......

[candyice]

mktime（将时间结构数据转换成经过的秒数）

[ufouser]

strptime函数

[Gengoo]

我是想把时间转换成对应的秒, 请问strptime怎么用. 网上介绍的好像是格式化时间吧