空参考异议。你调用的对象是空的 [英] Null Reference Objection. Object reference not set to an instance of an object

问题描述

大家好，



当我将数据表链接到页脚插入的asp.net网格视图中的下拉列表时，我收到了上述错误。 />


我已经立即检查了窗口，数据表有2条记录（datatable.rows.count）。



以下是asp和vb中的代码...请建议



Asp。

 <   asp：TemplateField     HeaderText   = 输入    HeaderStyle-Horizo​​ntalAlign   = 中心  >  
 <   EditItemTemplate  >  
 <   asp：DropDownList     ID   =  ddlGridType    runat   =  server    DataTextField   =  BOMIT    DataValueField   =  BOMIT >  <   / asp：DropDownList  >  
 <   / EditItemTemplate  >  
 <   ItemTemplate  >  
 <   asp：标签    ID   =  lblGridType    runat   =  server   文字  ='  <％ #Eval（  BOMIT）％> ' >  <   / asp：Label  >  
 <   / ItemTemplate  >  
 <   FooterTemplate  >  
 <   asp：DropDownList     ID   =  ddlNewGridType    runat   =  server    DataTextField   =  BOMIT    DataValueField   =  BOMIT >  <   / asp：DropDownList  >  
 <   / FooterTemplate  >  
 <   / asp：TemplateField  >  

VB：

 受保护的  Sub  DataGridView1_RowDataBound（发件人 As  对象，e  As  GridViewRowEventArgs）句柄 DataGridView1.RowDataBound 
 
 如果 e.Row.RowType = DataControlRowType.Footer 然后 
 
  Dim  daGridTypes 作为 SqlDataAdapter = 新 SqlDataAdapter 
 daGridTypes.SelectCommand = 新 SqlCommand （ 从bom中选择不同的BOMIT，连接）
 
  Dim  tbGridTypes  As  DataTable = 新 DataTable 
 connection.Open（）
 daGridTypes.Fill（tbGridTypes）
 connection.Close（）
 
 ' 要排序DATATABLE  
  Dim  dataGridView  As  新 DataView（tbGridTypes）
 dataGridView.Sort =   BOMIT ASC 
 tbGridTypes = 新 DataTable 
 tbGridTypes = dataGridView.ToTable（）
 
 ddlNewGridType = e.Row.FindControl（< span class =code-string>  ddlNewGridype）
 
 ********** ddlNewGridType .DataSource = tbGridTypes 
 ddlNewGridType.DataTextField =   BOMIT 
 ddlNewGridType.DataValueField =   BOMIT 
 ddlNewGridType.DataBind（）
 
 < span class =code-keyword>结束 如果 
 
 结束  Sub  

解决方案

你有一个错字，在控件ID中缺少一个字母。

 ddlNewGridType = e.Row.FindControl（  ddlNewGridype）
 

应该是

 ddlNewGridType = e.Row.FindControl（   ddlNewGridType）
 

它是总是一个好主意，检查返回的值是否为null，以便捕获并记录这些错误。这可以防止.aspx文件的更改无声地破坏.aspx.cs文件。

 ddlNewGridType = e.Row.FindControl（   ddlNewGridype）
  if （ddlNewGridType ==  null ）
 logger.Error（ 哎呀，我找不到'ddlNewGridType'。）; 
  

你不能直接访问控制通过id在项目，页眉或页脚等模板中

你可以做的是找到控件并设置数据源。添加到下面的行

 如果 e.Row.RowType = DataControlRowType.Footer 那么 
   Dim  ddlNewGridType  As  DropDownList =  DirectCast （e.Row .FindControl（  ddlNewGridType），DropDownList） 

Hi All,

I am getting the above error while i have linked a datatable to a dropdownlist in asp.net Grid view in Footer for Insert.

I have checked in immediate window and datatable is having 2 records (datatable.rows.count).

following is the code in asp and vb... Please sugggest

Asp.

<asp:TemplateField HeaderText="Type" HeaderStyle-HorizontalAlign="Center">
                <EditItemTemplate>
                    <asp:DropDownList ID="ddlGridType" runat="server" DataTextField="BOMIT" DataValueField="BOMIT"> </asp:DropDownList>
                </EditItemTemplate>
                <ItemTemplate>
                    <asp:Label ID="lblGridType" runat="server" Text='<%# Eval("BOMIT")%>'></asp:Label>
                </ItemTemplate>
                <FooterTemplate>
                    <asp:DropDownList ID="ddlNewGridType" runat="server" DataTextField="BOMIT" DataValueField="BOMIT"> </asp:DropDownList>
                 </FooterTemplate>
            </asp:TemplateField>

VB:

Protected Sub DataGridView1_RowDataBound(sender As Object, e As GridViewRowEventArgs) Handles DataGridView1.RowDataBound

        If e.Row.RowType = DataControlRowType.Footer Then

            Dim daGridTypes As SqlDataAdapter = New SqlDataAdapter
            daGridTypes.SelectCommand = New SqlCommand("select distinct BOMIT from bom", connection)

            Dim tbGridTypes As DataTable = New DataTable
            connection.Open()
            daGridTypes.Fill(tbGridTypes)
            connection.Close()

            'To SORT THE DATATABLE
            Dim dataGridView As New DataView(tbGridTypes)
            dataGridView.Sort = "BOMIT ASC"
            tbGridTypes = New DataTable
            tbGridTypes = dataGridView.ToTable()

            ddlNewGridType = e.Row.FindControl("ddlNewGridype")

            **********ddlNewGridType.DataSource = tbGridTypes
            ddlNewGridType.DataTextField = "BOMIT"
            ddlNewGridType.DataValueField = "BOMIT"
            ddlNewGridType.DataBind()

        End If

    End Sub

解决方案

You have a typo, missing a letter in the control ID.

ddlNewGridType = e.Row.FindControl("ddlNewGridype")

should be

ddlNewGridType = e.Row.FindControl("ddlNewGridType")

It is always a good idea to check that the returned value is not null, so that you catch and log these errors. This prevents changes to the .aspx file from silently breaking the .aspx.cs file.

ddlNewGridType = e.Row.FindControl("ddlNewGridype")
if (ddlNewGridType == null)
    logger.Error("Oops, I can't find 'ddlNewGridType'.);

you can't directly access control by id inside the templates like item, header or footer
what you can do is Find the control and set the data source. add below line

If e.Row.RowType = DataControlRowType.Footer Then
    Dim ddlNewGridType As DropDownList = DirectCast(e.Row.FindControl("ddlNewGridType"), DropDownList)

这篇关于空参考异议。你调用的对象是空的的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！