我在ACCESS的VBA中写入以下代码:
Private Sub Command19_Click()
Inet1.URL = "ftp://192.168.0.101/" 'IP地址
Inet1.Protocol = icFTP
Inet1.UserName = "xxx" 'username
Inet1.Password = "xxx" 'password
Inet1.Execute , "put C:\123.txt 456.txt" '上传文件
MsgBox "done!"
End Sub
每次一运行就会有如图的错误(已经引用了MS Internet control)
(access与OLE服务器或ACTIVEX控件通讯时出现问题)
我调试了一下,发现每次程序都还没有进入此sub就报错了
请问是什么问题呢,我尝试网上说的开启了几个服务也是不行的。。谢谢了
Private Sub Command19_Click()
Inet1.URL = "ftp://192.168.0.101/" 'IP地址
Inet1.Protocol = icFTP
Inet1.UserName = "xxx" 'username
Inet1.Password = "xxx" 'password
Inet1.Execute , "put C:\123.txt 456.txt" '上传文件
MsgBox "done!"
End Sub
每次一运行就会有如图的错误(已经引用了MS Internet control)
(access与OLE服务器或ACTIVEX控件通讯时出现问题)
我调试了一下,发现每次程序都还没有进入此sub就报错了
请问是什么问题呢,我尝试网上说的开启了几个服务也是不行的。。谢谢了
Inet1.URL = "fttp://192.168.0.101/" 'IP地址
Inet1.Protocol = icFTP
Inet1.UserName = "xxx" 'username
Inet1.Password = "xxx" 'password
Inet1.Execute , "put C:\123.txt 456.txt" '上传文件
MsgBox "done!"
End Sub
似乎应为:
Inet1.URL = "Http://192.168.0.101/" 'IP地址
Private Sub Command19_Click()
Inet1.URL = "Http://192.168.0.101/" 'IP地址
Inet1.Protocol = icFTP
Inet1.UserName = "xxx" 'username
Inet1.Password = "xxx" 'password
Inet1.Execute , "put C:\123.txt 456.txt" '上传文件
MsgBox "done!"
End Sub