Question

我用谷歌搜索了这个错误，并没有得出结论为什么我收到这个错误。我正在尝试用一些数据填充DataGridView。这是我的代码。

Private Sub LoadGrid()
    Dim cmd As New OleDbCommand
    Dim dt As DataTable
    With cmd
        .CommandText = "project_master_query"
        .CommandType = CommandType.StoredProcedure
        .Connection = New OleDbConnection(My.Settings.cnnString)
    End With
    dt = GetData(cmd)
    dgvData.DataSource = dt
End Sub

Private Function GetData(ByVal cmd As OleDbCommand) As DataTable
    Dim dt As New DataTable
    Using cmd.Connection
        cmd.Connection.Open()
        dt.Load(cmd.ExecuteReader(CommandBehavior.CloseConnection))
    End Using
    Return dt
End Function

查询存储在Access中的“project_master_query”。

    SELECT project_master.*, location_master.LocationName,
    project_currentmilestonedef.MilestoneDefID, 
    project_currentmilestonedef.MilestoneName, project_regions.RegionName,
    owner_fullname.FullName AS OwnerFullName, designer_fullname.FullName AS DesignerFullName,
    project_issuecount.HasOpenIssues, project_updated_closedate.UpdatedCloseDate, 
    project_bonusdays.BonusDays, project_bonusdays.IsGreen, project_bonusdays.IsYellow, 
    project_bonusdays.IsRed, checklist_days_perproject_defined_1.Week1, 
    checklist_days_perproject_defined_1.Week2, checklist_days_perproject_defined_1.Week3, 
    checklist_days_perproject_defined_1.Week4, project_issueduration.ProjectIssueDurationDays, 
    project_active_status.ProjectIsOpen, project_requirement_status.RequirementStatusName, 
    priority_def.PriorityShortName
    FROM project_requirement_status 
    RIGHT JOIN (project_regions 
    RIGHT JOIN (priority_def RIGHT JOIN (location_master 
    RIGHT JOIN ((((checklist_days_perproject_defined AS checklist_days_perproject_defined_1 
    RIGHT JOIN ((((((((contacts_fullname AS designer_fullname 
    RIGHT JOIN (contacts_fullname AS owner_fullname 
    RIGHT JOIN project_master ON owner_fullname.ContactID = project_master.ContactOwner) 
    ON designer_fullname.ContactID = project_master.ContactDesigner) 
    LEFT JOIN project_issuecount 
    ON project_master.ProjectID = project_issuecount.ProjectID) 
    LEFT JOIN project_currentmilestonedef 
    ON project_master.ProjectID = project_currentmilestonedef.ProjectID) 
    LEFT JOIN project_within_benchmark_week1 
    ON project_master.ProjectID = project_within_benchmark_week1.ProjectID) 
    LEFT JOIN project_within_benchmark_week2 
    ON project_master.ProjectID = project_within_benchmark_week2.ProjectID) 
    LEFT JOIN project_within_benchmark_week3 
    ON project_master.ProjectID = project_within_benchmark_week3.ProjectID) 
    LEFT JOIN project_updated_closedate 
    ON project_master.ProjectID = project_updated_closedate.ProjectID) 
    LEFT JOIN checklist_days_perproject_defined 
    ON project_master.ProjectID = checklist_days_perproject_defined.ProjectID) 
    ON checklist_days_perproject_defined_1.ProjectID = project_master.ProjectID) 
    LEFT JOIN project_issueduration 
    ON project_master.ProjectID = project_issueduration.ProjectID) 
    LEFT JOIN project_active_status 
    ON project_master.ProjectID = project_active_status.ProjectID)    
    LEFT JOIN project_bonusdays ON project_master.ProjectID = project_bonusdays.ProjectID) 
    ON location_master.LocationID = project_master.Location) 
    ON priority_def.PriorityDefID = project_master.ProjectPriority) 
    ON project_regions.RegionID = project_master.Region) 
    ON project_requirement_status.RequirementStatusID = project_master.RequirementStatus;

正如您所看到的，查询中根本没有Nz所以我不明白为什么会发生这种情况。调用dt.Load后会发生错误。

Answer 1

Nz函数很可能出现在您在多个连接中引用的视图/查询中。你必须仔细研究它们。

由于Nz()是Access应用程序的功能而不是Access驱动程序，因此只要您尝试从Access应用程序外部使用它，它就会失败。您可以将Nz替换为IIf(IsNull())构造。

请参阅IIf和IsNull

的文档

当放在一起时：

Nz(expr, [valueifnull])

变为

IIf(IsNull(expr), valueifnull, valueifnotnull)

<强>实施例

默认值：Nz(tbl.A) => IIf(IsNull(tbl.A), '', tbl.A)

使用后备：Nz(tbl.A, tbl.B) => IIf(IsNull(tbl.A), tbl.B, tbl.A)

Answer 2

我创建了一个在这种情况下使用的函数。我叫ez。

val point: Int = someMap["foo"] ?: throw AppError.exit(code)

如您所见，它比原始函数稍远一些，因为它不仅会在第一个为空时考虑第二个值，而且还会将其为空

此外，默认值是0，而Nz是空字符串。

表达式中未定义的函数'Nz'

2 个答案: