subject_51770.htm

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序号：51770 发表者：navyly 发表日期：2003-09-04 09:32:35
<br>主题：向各位请教在SystemTimeToVariant中double是如何处理日期的？
<br>内容：在系统中，我们可以用doubl表示日期；但对两个double日期，我们如何计算它们的时间间隔（分别用日、时、分、秒表示）。<BR><BR>如：double dbl_Old,dbl_Cur;<BR>SYSTEMTIME sys_Time;<BR><BR>GetSystemTime(&amp;sys_Time);<BR>SystemTime(&amp;sys_Time, &amp;dbl_Old);<BR>…………<BR>…………<BR>（其他复杂操作后）<BR>GetSystemTime(&amp;sys_Time);<BR>SystemTime(&amp;sys_Time, &amp;dbl_Cur);<BR><BR>(如何计算两个日期的差值)？<BR><BR>在线等待，为盼！<BR>
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<font color=red>答案被接受</font><br>回复者：微笑的撒旦 回复日期：2003-09-05 01:22:12
<br>内容：两数之差,整数是日期,小数*24是小时;小数*24*60是分钟;<BR>比如:<BR>2003/09/04 16:09:32<BR>转换成double就是：<BR>37868.673287037040<BR>2003/09/05 06:07:54<BR>就是：<BR>37869.255486111113<BR>差为：<BR>0.58219907407328719&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 日期为0<BR>0.58219907407328719*24 = 13.972777777758893&nbsp;&nbsp; 13小时<BR>0.972777777758893*60 = 58.366666665533579&nbsp;&nbsp;&nbsp;&nbsp; 58分<BR>0.366666665533579*60 = 21.999999932014742&nbsp;&nbsp;&nbsp;&nbsp; 22秒<BR><BR>看到这个问题，研究了一下。有不对请指出来噢！<BR>这是MSDN上的一段解释<BR>A variant time is stored as an 8-byte real value (double), representing a date between January 1, 100 and December 31, 9999, inclusive. The value 2.0 represents January 1, 1900; 3.0 represents January 2, 1900, and so on. Adding 1 to the value increments the date by a day. The fractional part of the value represents the time of day. Therefore, 2.5 represents noon on January 1, 1900; 3.25 represents 6:00 A.M. on January 2, 1900, and so on. Negative numbers represent dates prior to December 30, 1899.
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