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<title>Re: 两位之解答都还是时间，不是时间差或和！</title>
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<h1>Re: 两位之解答都还是时间，不是时间差或和！</h1>
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Posted by <a href="mailto:w100@ms1.url.com.tw">小v</a> on July 07, 1999 at 15:54:47:<p>
In Reply to: <a href="15312.html">两位之解答都还是时间，不是时间差或和！</a> posted by 新手 on July 07, 1999 at 01:15:14:<p>
: 最后答案应该是例如：<br>: T1=16:30:20<br>: T2=9:10:10<br>: T3=T1-T2<br>: T3=7:20:10<br>: 时间差为7时20分10秒，而非AM 7:20:10 或是PM<p>手边有个函数TimeDiffs()给你参考!<br>若有问题请来信告知!<p>Public Function TimeDiffS(ByVal StartTim As Variant, ByVal EndTim As Variant) As String<br>Dim T1 As Long, T2 As Long<br>'计算两个时间字串(格式为hh:mm:ss)相隔的时间(hh:mm:ss)<br>If TypeName(StartTim) = "String" Then StartTim = CDate(StartTim)<br>If TypeName(EndTim) = "String" Then EndTim = CDate(EndTim)<br>T1 = Time2Sec(StartTim)<br>T2 = Time2Sec(EndTim)<br>If T2 < T1 Then T2 = T2 + 86400<br>TimeDiffS = Sec2Time(T2 - T1)<br>End Function<p>Public Function Time2Sec(ByVal TimStr As Variant) As Long<br>Dim HH As Long, mM As Long, ss As Long<br>'将时间字串(格式必须是hh:mm:ss)化成秒数<br>If TypeName(TimStr) = "String" Then<br> TimStr = CDate(TimStr)<br>End If<br>HH = Hour(TimStr)<br>mM = Minute(TimStr)<br>ss = Second(TimStr)<br>Time2Sec = (HH * 3600) + (mM * 60) + ss<br>End Function<p>Public Function Sec2Time(ByVal nSec As Variant) As String<br>Dim HH As Long, mM As Long, ss As Long<br>'将秒数化成时间字串(格式为hh:mm:ss)<br>HH = Fix(nSec / 3600)<br>nSec = nSec - (HH * 3600)<br>mM = Fix(nSec / 60)<br>nSec = nSec - (mM * 60)<br>ss = nSec<br>HH = HH Mod 24<br>Sec2Time = StrZero(HH, 2) + ":" + StrZero(mM, 2) + ":" + StrZero(ss, 2)<br>End Function<p><br>
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