iam将一个文本框值(值为10:1:1)传递给一个字符串,当iam运行应用程序时,我收到以下错误“coversion from string”10:1:1“to double type is invalid”。请找到下面的代码并帮助我:(文本框值也应始终大于0)
Dim strEncrypt As String = txtData.Text
If strEncrypt > 0 Then // I am getting the error here
txtEncryptedData.Text = Encrypt(strEncrypt)
Else
MessageBox.Show(
"Enter the Value greater then 0:")
End If
谢谢, 公羊
答案 0 :(得分:2)
我相信你想要这个...
如果strEncrypt.Length> 0然后
答案 1 :(得分:1)
也许你想要:
Function Check(ByVal s As String) As Boolean
Dim parts As String()
parts = s.Split(":")
If parts.Length = 0 Then
Check = False
Else
Check = True
For Each sval As String In parts
Check = Check And Int32.Parse(sval) > 0
Next
End If
End Function
所以你可以像Check(txtData.Text)
一样使用它。