blocks|key|1684692|text|#include+<time.h>
char+*strptime(const+char+*buf,+const+char+*format,+struct+tm+*tm);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1684693|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>#include &lt;time.h&gt;
char *strptime(const char *buf, const char *format, struct tm *tm);
</code></pre>

blocks|key|1469535|text|我没有使用strptime就解决了这个问题。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1469536|将日期分解为其组成部分，即日、月、年，然后：|1469537|struct+tm++tm;
time_t+rawtime;
time+(+&rawtime+);
tm+=+*localtime+(+&rawtime+);
tm.tm_year+=+year+-+1900;
tm.tm_mon+=+month+-+1;
tm.tm_mday+=+day;
mktime(&tm);|code-block|syntax|javascript|1469538|现在可以将tm转换为time_t并对其进行操作。|1469539|entityMap^0|5|8|0|0|0|5|2|A|6|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

I figured it out without using <code>strptime</code>.

Break the date down into its components i.e. day, month, year, then:

<pre><code>struct tm tm;
time_t rawtime;
time ( &amp;rawtime );
tm = *localtime ( &amp;rawtime );
tm.tm_year = year - 1900;
tm.tm_mon = month - 1;
tm.tm_mday = day;
mktime(&amp;tm);
</code></pre>

<code>tm</code> can now be converted to a <code>time_t</code> and be manipulated.

blocks|key|1469562|text|#include+<time.h>
#include+<iostream>
#include+<sstream>
#include+<algorithm>

using+namespace+std;

int+main+()
{
++time_t+rawtime;
++struct+tm+*+timeinfo;
++int+year,+month+,day;
++char+str[256];

++cout+<<+"Inter+date:+"+<<+endl;+
++cin.getline(str,sizeof(str));

++replace(+str,+str%2Bstrlen(str),+'/',+'+'+);++
++istringstream(+str+)+>>+day+>>+month+>>+year;

++time+(+&rawtime+);
++timeinfo+=+localtime+(+&rawtime+);
++timeinfo->tm_year+=+year+-+1900;
++timeinfo->tm_mon+=+month+-+1;
++timeinfo->tm_mday+=+day;
++mktime+(+timeinfo+);

++strftime+(+str,+sizeof(str),+"%25A",+timeinfo+);
++cout+<<+str+<<+endl;
++system("pause");
++return+0;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1469563|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>#include &lt;time.h&gt;
#include &lt;iostream&gt;
#include &lt;sstream&gt;
#include &lt;algorithm&gt;

using namespace std;

int main ()
{
 time_t rawtime;
 struct tm * timeinfo;
 int year, month ,day;
 char str[256];

 cout &lt;&lt; "Inter date: " &lt;&lt; endl; 
 cin.getline(str,sizeof(str));

 replace( str, str+strlen(str), '/', ' ' ); 
 istringstream( str ) &gt;&gt; day &gt;&gt; month &gt;&gt; year;

 time ( &amp;rawtime );
 timeinfo = localtime ( &amp;rawtime );
 timeinfo-&gt;tm_year = year - 1900;
 timeinfo-&gt;tm_mon = month - 1;
 timeinfo-&gt;tm_mday = day;
 mktime ( timeinfo );

 strftime ( str, sizeof(str), "%A", timeinfo );
 cout &lt;&lt; str &lt;&lt; endl;
 system("pause");
 return 0;
}
</code></pre>

blocks|key|1684826|text|您可以利用boost库(跨平台)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1684827|#include+<stdio.h>
#include+"boost/date_time/posix_time/posix_time.hpp"

int+main()
{++++
std::string+strTime+=+"2007-04-11+06:18:29.000";
std::tm+tmTime+=+boost::posix_time::to_tm(boost::posix_time::time_from_string(strTime));
return+0;
}|code-block|syntax|javascript|1684828|但是格式应该是前面提到的:)|1684829|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You can utilize the boost library(cross platform)

<pre><code>#include &lt;stdio.h&gt;
#include "boost/date_time/posix_time/posix_time.hpp"

int main()
{ 
std::string strTime = "2007-04-11 06:18:29.000";
std::tm tmTime = boost::posix_time::to_tm(boost::posix_time::time_from_string(strTime));
return 0;
}
</code></pre>

But the format should be as mentioned :)

blocks|key|1469641|text|为什么不使用boost来寻求更简单的解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1469642|using+namespace+boost::gregorian;
using+namespace+boost::posix_time;++++
ptime+pt+=+time_from_string("20150917");|code-block|syntax|javascript|1469643|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Why not go for a simpler solution using boost

<pre><code>using namespace boost::gregorian;
using namespace boost::posix_time; 
ptime pt = time_from_string("20150917");
</code></pre>

I know this may be simple but being C++ I doubt it will be. How do I convert a string in the form 01/01/2008 to a date so I can manipulate it? I am happy to break the string into the day month year constituents. Also happy if solution is Windows only.

Convert a string to a date in C++

我知道这可能很简单，但作为C++，我对此表示怀疑。如何将格式为01/01/2008的字符串转换为日期，以便对其进行操作？我很高兴将字符串分解为日、月、年的组成部分。如果解决方案是Windows，我也很高兴。

问将字符串转换为C++中的日期
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