Private Sub Command1_Click()
Dim a, b, i As Integer, c As String, d As String, e, f
a = Text3.Text
e = Left(a, 1)
e = Asc(e) - 40 '就是运行到这里出现了无效的过程调用和参数,这是怎么回事啊?
f = e
c = ""
For i = 2 To Len(a) - 1
b = Mid$(a, i, 1)
b = Chr(Asc(b) Xor e)
c = c + b
e = e + f
f = (-f)
Next i
a = Text4.Text
e = Left(a, 1)
e = Asc(e) - 43
f = e
d = ""
For i = 2 To Len(a) - 1
b = Mid$(a, i, 1)
b = Chr(Asc(b) Xor e)
d = d + b
e = e + f
f = (-f)
Next i
If Text1.Text = "c" And Text2.Text = "d" Then
'MsgBox "欢迎进入人事档案管理系统", vbOKOnly, "提示信息"
Form1.Hide
Form5.Show
Else
MsgBox "口令错误,请重新输入", vbCritical, "提示信息"
Text2.Text = ""
Text2.SetFocus
Exit Sub
'Else
MsgBox "用户名/口令错误,请重新输入", vbCritical, "提示信息"
Text1.Text = ""
Text2.Text = ""
Text1.SetFocus
Exit Sub
End If
Unload Me
End Sub
我用的就是VB6.0做的啊,还是一样的问题
---------------------------------
Private Sub Command1_Click()
if Text3.Text = "" OR Text4.Text = "" then exit sub
这样是出现数据访问错误----------当我填写好密码和用户名的时候
VB实时错误'5'
答案:3 悬赏:30
解决时间 2021-02-18 02:39
- 提问者网友:剪短发丝
- 2021-02-17 11:47
最佳答案
- 二级知识专家网友:懂得ㄋ、沉默
- 2021-02-17 12:09
Private Sub Command1_Click()
if Text3.Text = "" then exit sub
...
...
试试
Private Sub Command1_Click()
if Text3.Text = "" OR Text4.Text = "" then exit sub
这样
数据访问错误 是在那里报出来的?
你的代码没删干净啊,有两个连续的Exit Sub
Exit Sub
'Else
MsgBox "用户名/口令错误,请重新输入", vbCritical, "提示信息"
Text1.Text = ""
Text2.Text = ""
Text1.SetFocus
Exit Sub
还有这里
If Text1.Text = "c" And Text2.Text = "d" Then
应该是
If Text1.Text = c And Text2.Text = d Then
吧
还有是不是别的事件里的代码有问题啊?
if Text3.Text = "" then exit sub
...
...
试试
Private Sub Command1_Click()
if Text3.Text = "" OR Text4.Text = "" then exit sub
这样
数据访问错误 是在那里报出来的?
你的代码没删干净啊,有两个连续的Exit Sub
Exit Sub
'Else
MsgBox "用户名/口令错误,请重新输入", vbCritical, "提示信息"
Text1.Text = ""
Text2.Text = ""
Text1.SetFocus
Exit Sub
还有这里
If Text1.Text = "c" And Text2.Text = "d" Then
应该是
If Text1.Text = c And Text2.Text = d Then
吧
还有是不是别的事件里的代码有问题啊?
全部回答
- 1楼网友:转身后的回眸
- 2021-02-17 13:23
e = Asc(e) - 40 换成
e = Asc(CInt(e)) - 40
补充,我用的是VB2005和VB2008,如果你用的是VB6。0这样的旧版本的话我就没有试过了。
- 2楼网友:為→妳鎖鈊
- 2021-02-17 13:08
num, nam
这俩个数组没有定义……找了半天
还有数组下标越界的问题不知道你碰到没有
你出现错误5的话应该是string 函数中少了一个空格
print " "; stu(i); string(11 - len(stu(i)), ""); stu(i + 1)
双引号之间中间加一个空格
string(11 - len(stu(i)), " ")
我要举报
如以上问答内容为低俗、色情、不良、暴力、侵权、涉及违法等信息,可以点下面链接进行举报!
大家都在看
推荐资讯