OPEN_DATA = 20151128075728 (string类型的)
EXIT_DATA = 20151128080212 (string类型的)
求他们的时间间隔。
VB求两个string类型转换date类型并求他们的时间间隔
答案:2 悬赏:60
解决时间 2021-02-16 18:07
- 提问者网友:回憶丶初
- 2021-02-15 23:06
最佳答案
- 二级知识专家网友:飘零作归宿
- 2021-02-16 00:40
dim a as string, b as date
dim h, m, s
a = 80
h = 0
m = a \ 60
s = a mod 60
b = timeserial(h, m, s)
dim h, m, s
a = 80
h = 0
m = a \ 60
s = a mod 60
b = timeserial(h, m, s)
全部回答
- 1楼网友:社会水太深
- 2021-02-16 00:45
Private Sub Command1_Click()
Dim OPEN_DATA As String, EXIT_DATA As String
Dim O_DATA As Date, E_DATA As Date
OPEN_DATA = "20151128075728" ' (string类型的)
EXIT_DATA = "20151128080212" ' (string类型的)
O_DATA = Mid(OPEN_DATA, 1, 4) & "-" & Mid(OPEN_DATA, 5, 2) & "-" & Mid(OPEN_DATA, 7, 2) & " " & Mid(OPEN_DATA, 9, 2) & ":" & Mid(OPEN_DATA, 11, 2) & ":" & Mid(OPEN_DATA, 13, 2)
E_DATA = Mid(EXIT_DATA, 1, 4) & "-" & Mid(EXIT_DATA, 5, 2) & "-" & Mid(EXIT_DATA, 7, 2) & " " & Mid(EXIT_DATA, 9, 2) & ":" & Mid(EXIT_DATA, 11, 2) & ":" & Mid(EXIT_DATA, 13, 2)
Print DateDiff("s", O_DATA, E_DATA) '使用DateDiff函数获得间隔秒数
End Sub
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