练习: 时间加减的自定义函数(挑战最简代码)

zjxia889 · 发表于 2008-1-21 21:27

一个关于时间加减的问题r1=444分50秒r2=10分40秒 r1+r2结果应该是455分30秒大家来作个自定义函数吧，sjjs=(r1,r2,"+") 先不考虑速度.看看最的最简便,代码最少           -----------兰色幻想

[此贴子已经被兰色幻想于2008-1-22 15:54:50编辑过]

cn12345678 · 发表于 2008-1-22 12:51

试试，只会笨方法。Function sjjs(r1 As String, r2 As String, ys As String) As String     Dim f1 As Integer, f2 As Integer, m1 As Integer, m2 As Integer     f1 = Mid(r1, 1, Application.Find("分", r1) - 1)     f2 = Mid(r2, 1, Application.Find("分", r2) - 1)     m1 = Mid(r1, Application.Find("分", r1) + 1, Len(r1) - Application.Find("分", r1) - 1)     m2 = Mid(r2, Application.Find("分", r2) + 1, Len(r2) - Application.Find("分", r2) - 1)     If ys = "+" Then         sjjs = f1 + f2 + (m1 + m2) \ 60 & "分" & (m1 + m2) Mod 60 & "秒"     Else         If ys = "-" Then             sjjs = f1 - f2 - 1 + (m1 + 1 - m2) \ 60 & "分" & (m1 + 60 - m2) Mod 60 & "秒"         Else             sjjs = ""         End If     End If End Function

兰色幻想 · 发表于 2008-1-22 15:29

这个题目的答案我已有了答案最后贴出来.

[此贴子已经被作者于2008-1-22 15:55:11编辑过]

晓 · 发表于 2008-1-22 16:04

做好了发现思路跟楼上的一样的耶，再想想

suye1010 · 发表于 2008-1-22 16:52

Function shijian(r1 As String, r2 As String, j As String) Dim arr1, arr2 arr1 = Split(Mid(r1, 1, Len(r1) - 1), "分") arr2 = Split(Mid(r2, 1, Len(r2) - 1), "分") shijian = IIf(j = "+", --arr1(0) + --arr2(0) & "分" & --arr1(1) + --arr2(1) & "秒", arr1(0) - arr2(0) & "分" & arr1(1) - arr2(1) & "秒") End Function 我的五行代码*_^

[此贴子已经被作者于2008-1-22 16:52:56编辑过]

兰色幻想 · 发表于 2008-1-22 16:57

<div class="msgheader">QUOTE:</div><div class="msgborder">以下是引用suye1010在2008-1-22 16:52:00的发言： Function shijian(r1 As String, r2 As String, j As String) Dim arr1, arr2 arr1 = Split(Mid(r1, 1, Len(r1) - 1), "分") arr2 = Split(Mid(r2, 1, Len(r2) - 1), "分") shijian = IIf(j = "+", --arr1(0) + --arr2(0) & "分" & --arr1(1) + --arr2(1) & "秒", arr1(0) - arr2(0) & "分" & arr1(1) - arr2(1) & "秒") End Function 我的五行代码*_^ </div>不符合题义 如果第一个时间的秒数比第二个时间的秒数小,那么得出的结果是错误的

zjxia889 · 发表于 2008-1-22 17:57

Function sjjs(r1 As Range, r2 As Range, Optional fh As String = "+") '44分50秒+10分2秒=54分52秒 Dim ss& Dim r1s$, r2s$ r1s = Replace(r1, "分", "*60+") r1s = Replace(r1s, "秒", "") r2s = Replace(r2, "分", "*60+") r2s = Replace(r2s, "秒", "") ss = Evaluate(r1s & fh & r2s) sjjs = ss \ 60 & "分" & ss Mod 60 & "秒" End Function很高兴能引起兰版的注意，与大家一起学习 [em01]问题VBA中replace为什么不能向函数一样嵌套？

cn12345678 · 发表于 2008-1-22 18:11

学习了，replace可以嵌套的。Function sjjs(r1 As Range, r2 As Range, Optional fh As String = "+") '44分50秒+10分2秒=54分52秒 Dim ss& Dim r1s$, r2s$ 'r1s = Replace(r1, "分", "*60+") r1s = Replace(Replace(r1, "分", "*60+"), "秒", "") 'r2s = Replace(r2, "分", "*60+") r2s = Replace(Replace(r2, "分", "*60+"), "秒", "") ss = Evaluate(r1s & fh & r2s) sjjs = ss \ 60 & "分" & ss Mod 60 & "秒" End Function

兰色幻想 · 发表于 2008-1-22 18:15

<div class="msgheader">QUOTE:</div><div class="msgborder">以下是引用zjxia889在2008-1-22 17:57:00的发言： Function sjjs(r1 As Range, r2 As Range, Optional fh As String = "+") '44分50秒+10分2秒=54分52秒 Dim ss& Dim r1s$, r2s$ r1s = Replace(r1, "分", "*60+") r1s = Replace(r1s, "秒", "") r2s = Replace(r2, "分", "*60+") r2s = Replace(r2s, "秒", "") ss = Evaluate(r1s & fh & r2s) sjjs = ss \ 60 & "分" & ss Mod 60 & "秒" End Function很高兴能引起兰版的注意，与大家一起学习 [em01]问题VBA中replace为什么不能向函数一样嵌套？</div>你的自定义函数是不是不支持减啊? 我的代码同时支持加和减,而且代码比你少了很多行,呵呵,继续努力

[此贴子已经被作者于2008-1-22 18:15:34编辑过]

suye1010 · 发表于 2008-1-22 18:18

<div class="msgheader">QUOTE:</div><div class="msgborder">以下是引用兰色幻想在2008-1-22 16:57:00的发言： <div class="msgheader">QUOTE:</div><div class="msgborder">以下是引用suye1010在2008-1-22 16:52:00的发言： Function shijian(r1 As String, r2 As String, j As String) Dim arr1, arr2 arr1 = Split(Mid(r1, 1, Len(r1) - 1), "分") arr2 = Split(Mid(r2, 1, Len(r2) - 1), "分") shijian = IIf(j = "+", --arr1(0) + --arr2(0) & "分" & --arr1(1) + --arr2(1) & "秒", arr1(0) - arr2(0) & "分" & arr1(1) - arr2(1) & "秒") End Function 我的五行代码*_^ </div>不符合题义 如果第一个时间的秒数比第二个时间的秒数小,那么得出的结果是错误的</div> 的确是这样的。Let，s me 想想……

[此贴子已经被作者于2008-1-22 18:23:57编辑过]

		自动登录	找回密码
密码			注册

[习题] 练习: 时间加减的自定义函数(挑战最简代码)

我先发上来让大家看看

浏览过的版块