我已经创建了一个登录屏幕,以便在输入有效密码的情况下对excel工作簿进行身份验证。
代码如下:
Private Sub CommandButton1_Click()
name_selected = ComboBox1.Text
pwd_entered = TextBox2.Text
validation_sheet = "Z"
act_p_col_num = 3
Application.Visible = True
For validation_check = 2 To Worksheets(validation_sheet).Cells(Rows.Count, 1).End(xlUp).Row - 1
If (Worksheets(validation_sheet).Cells(validation_check, 1) = name_selected) Then
bk_pd = Worksheets(validation_sheet).Cells(validation_check, act_p_col_num).Value
If (bk_pd = pwd_entered) Then
Worksheets("INDIVIDUAL_TRACKER").Select
MsgBox ("Authentication successful")
UserForm1.Hide
'Set UserForm1.Visible = False
Else
Application.Visible = False
MsgBox ("Please enter a valid password! Account will be locked after 3 tries")
TextBox2.Text = ""
End If
End If
Next validation_check
End Sub
无论如何,也使用unload me
命令和userform1.hide
尝试过此代码,这仍然抛出运行时错误424,需要对象。
我的目标是在认证通过后进入工作簿,userform
应该自动关闭。
有人可以帮助我解决这个问题吗?
答案 0 :(得分:2)
我的猜测是for循环的下一次迭代实际上是导致问题的原因。表单卸载后,您尝试在if语句上访问TextBox。
尝试添加该行
Exit For
在Unload Me
PS:尚未解锁评论,很抱歉,如果只是作为答案而发布作为答案...
答案 1 :(得分:1)
重新组合application.visible=True
行已解决了该目的。在正确的位置使用它可以使其正常工作。使用Exit For
已成为一个附加优势。
Private Sub CommandButton1_Click()
name_selected = ComboBox1.Text
pwd_entered = TextBox2.Text
validation_sheet = "Z"
act_p_col_num = 3
Application.Visible = True
For validation_check = 2 To Worksheets(validation_sheet).Cells(Rows.Count, 1).End(xlUp).Row - 1
If (Worksheets(validation_sheet).Cells(validation_check, 1) = name_selected) Then
bk_pd = Worksheets(validation_sheet).Cells(validation_check, act_p_col_num).Value
If (bk_pd = pwd_entered) Then
Unload Me
'UserForm1.Hide
'Set UserForm1.Visible = False
Application.Visible = True
Worksheets("INDIVIDUAL_TRACKER").Select
MsgBox ("Authentication successful")
Exit For
Else
'Application.Visible = False
MsgBox ("Please enter a valid password! Account will be locked after 3 tries")
TextBox2.Text = ""
End If
End If
Next validation_check
End Sub
感谢大家 ,他为这个问题提供了一些启示。这是一个很好的集思广益。谢谢大家。这个问题现在已经解决。