获取文本框以仅接受vb.net中的字符 [英] Get a textbox to accept only characters in vb.net

问题描述

我正在Vb.net中创建一个简单的应用程序，需要执行某些验证.因此，我希望名称文本框仅接受a-z和A-Z中的字符.

I'm creating a simple application in Vb.net where I need to perform certain validations. So I want a name textbox to accept only characters from a-z and A-Z for example.

为此，我编写了以下代码:

For this I wrote the following code:

   Private Sub txtname_KeyPress(ByVal sender As Object, ByVal e As System.Windows.Forms.KeyPressEventArgs) Handles TextBox5.KeyPress
    If Asc(e.KeyChar) <> 8 Then
        If Asc(e.KeyChar) > 65 Or Asc(e.KeyChar) < 90 Or Asc(e.KeyChar) > 96 Or Asc(e.KeyChar) < 122 Then
            e.Handled = True
        End If
    End If
End Sub

但是以某种方式不允许我输入字符.当我尝试输入任何字符时，它什么都不做.

But somehow it is not allowing me to enter characters. When I try to enter any character it does nothing.

什么原因导致此问题，我该如何解决?

What causes this problem and how can I resolve it?

推荐答案

或者，您可以执行类似的操作

Alternatively, you can do something like this,

Private Sub TextBox1_KeyPress(sender As Object, e As KeyPressEventArgs) _
                              Handles txtName.KeyPress

    If Not (Asc(e.KeyChar) = 8) Then
        Dim allowedChars As String = "abcdefghijklmnopqrstuvwxyz"
        If Not allowedChars.Contains(e.KeyChar.ToString.ToLower) Then
            e.KeyChar = ChrW(0)
            e.Handled = True
        End If
    End If

End Sub

，或者如果您仍然希望使用ASCII方式，请尝试此操作

or if you still want the ASCII way, try this,

Private Sub TextBox1_KeyPress(sender As Object, e As KeyPressEventArgs) Handles txtName.KeyPress

    If Not (Asc(e.KeyChar) = 8) Then
        If Not ((Asc(e.KeyChar) >= 97 And Asc(e.KeyChar) <= 122) Or (Asc(e.KeyChar) >= 65 And Asc(e.KeyChar) <= 90)) Then
            e.KeyChar = ChrW(0)
            e.Handled = True
        End If
    End If

End Sub

两者的行为相同.

这篇关于获取文本框以仅接受vb.net中的字符的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！