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'******************************************************************************************************
'函数原型:
'函数功能:判断用户是否可以登陆本系统
'函数参数:
'函数返回:True-表示可以登陆系统,False - 表示无权限登陆系统
'创建日期:2008-05-13
'修改日期:
'******************************************************************************************************
Function LoginGrant(ByVal UserID As String, ByVal SoftwareName As String) As Boolean
Dim vSqladp As New SqlDataAdapter
Dim ds As New DataSet
vSqladp.SelectCommand = New SqlCommand
Try
vSqladp.SelectCommand.CommandText = "select LJMC from T_YHLB where YHBH='" & UserID & "'"
vSqladp.SelectCommand.Connection = SqlConn
Dim vSljmcarr(), vSljmc As String
vSqladp.Fill(ds, "T_YHLB") '=====>>执行到这句的时候提示出错:提示信息为:Invalid object name "T_YHLB"
If ds.Tables("T_YHLB").Rows.Count > 0 Then
vSljmc = ds.Tables("T_YHLB").Rows(0).Item("LJMC")
vSljmcarr = vSljmc.Split(",")
For Each s As String In vSljmcarr
If s.ToLower = SoftwareName.ToLower Then
Return True
End If
Next
Else
Return False
End If
Catch ex As Exception
MsgBox(ex.Message)
Finally
vSqladp.SelectCommand.Dispose()
vSqladp.Dispose()
ds.Dispose()
End Try
End Function
Function LoginGrant(ByVal UserID As String, ByVal SoftwareName As String) As Boolean
Dim vSqladp As New SqlDataAdapter
Dim ds As New DataSet
Try
vSqladp.SelectCommand = New SqlCommand
vSqladp.SelectCommand.Connection = SqlConn
vSqladp.SelectCommand.CommandText = "Select LJMC from T_YHLB where YHBH='" & UserID & "'"
Dim vSljmcarr(), vSljmc As String
vSqladp.Fill(ds) '=====>>一样还是这里出错,错误的信息与上面的一样
If ds.Tables(0).Rows.Count > 0 Then
vSljmc = ds.Tables(0).Rows(0).Item("LJMC")
vSljmcarr = vSljmc.Split(",")
For Each s As String In vSljmcarr
If s.ToLower = SoftwareName.ToLower Then
Return True
End If
Next
Else
Return False
End If
Catch ex As Exception
Finally
vSqladp.SelectCommand.Dispose()
vSqladp.Dispose()
ds.Dispose()
End Try
End Function
vSqladp.Fill(ds, "T_YHLB")
改为
vSqladp.Fill(ds,T_YHLB) 这样是不行的