如何使用C#在gridview中添加新行?

时间:2016-09-20 13:22:35

标签: c# asp.net gridview

我想从数据库中检索数据,修改完记录后我想在Gridview中显示它。

我已经通过Edit columns创建了列名称,这些名称为Date,Location,Job Title,Details

这是我的asp.net代码

<asp:GridView ID="GridViewRecord" runat="server" AutoGenerateColumns="False" CellPadding="4" ForeColor="#333333" GridLines="None">
                <AlternatingRowStyle BackColor="White" ForeColor="#284775" />
                <Columns>
                    <asp:BoundField HeaderText="Date" />
                    <asp:BoundField HeaderText="Location" />
                    <asp:BoundField HeaderText="Job Title" />
                    <asp:BoundField HeaderText="Experience" />
                    <asp:HyperLinkField HeaderText="Details" />
                </Columns>
                <EditRowStyle BackColor="#999999" />
                <FooterStyle BackColor="#5D7B9D" Font-Bold="True" ForeColor="White" />
                <HeaderStyle BackColor="#5D7B9D" Font-Bold="True" ForeColor="White" />
                <PagerStyle BackColor="#284775" ForeColor="White" HorizontalAlign="Center" />
                <RowStyle BackColor="#F7F6F3" ForeColor="#333333" />
                <SelectedRowStyle BackColor="#E2DED6" Font-Bold="True" ForeColor="#333333" />
                <SortedAscendingCellStyle BackColor="#E9E7E2" />
                <SortedAscendingHeaderStyle BackColor="#506C8C" />
                <SortedDescendingCellStyle BackColor="#FFFDF8" />
                <SortedDescendingHeaderStyle BackColor="#6F8DAE" />
            </asp:GridView>

然后我尝试直接添加一个样本记录。但是我收到了一个错误。 这是我在页面加载时的c#代码

protected void Page_Load(object sender, EventArgs e)
{
    if (!this.IsPostBack)
    {
        DataTable dt = new DataTable();

        DataRow dr = dt.NewRow();
        dr[0] = "12-12-12"; //Error message occured here
        dr[1] = "Jeddah";
        dr[2] = "Java";
        dr[3] = "2";
        dr[4] = "View Details";

        dt.Rows.Add(dr);
        GridViewRecord.DataSource = dt;
        GridViewRecord.DataBind();
    }
}

错误讯息:

  

System.Data.dll中出现'System.IndexOutOfRangeException'类型的异常但未在用户代码中处理
  附加信息:找不到第0列。

我是c#的新手,谢谢

3 个答案:

答案 0 :(得分:1)

您尚未向DataTable添加任何列,这就是原因。

// Create a new DataTable object.
DataTable table = new DataTable();

// Declare a DataColumn
DataColumn column;

// Create the new DataColumn, set DataType, ColumnName and add then add to DataTable.    
column = new DataColumn();
column.DataType = System.Type.GetType("System.Int32");
column.ColumnName = "id";
table.Columns.Add(column);

// I think that the line-by-line explanation is better for the purpose of this answer
// you can of course do all of this in one row, assuming that you already have a datatable
table.Columns.Add("id", typeof(int));

但是,在您的示例中,您正在创建自己的DataTable,为什么不使用数据库中的那个?这将具有与您的选择查询匹配的列。

// create DataSet
DataSet ds = new DataSet();

// your operations for filing the DataSet with data from the database which you have not shared
// ...
// ...

// check to see whether we have a DataTable in the DataSet (if the query fails, ds.Tables.Count == 0)
if (ds.Tables.Count > 0) {
   DataRow row = ds.Tables[0].NewRow();
   // add data according to the schema
   // e.g.
   row["id"] = "blah";
   // add the rest of the columns

   // and lastly add the newly created row to the DataTable;
   ds.Tables[0].Rows.Add(row);
}

// now bind
GridViewRecord.DataSource = ds.Tables[0];
GridViewRecord.DataBind();

答案 1 :(得分:-1)

您还可以使用通用列表或字符串列表来简化代码 与列表

相比,性能明智DataTable也不好

例如:

List<string> lstRecord = new List<string>
{
    "12-12-12",
    "Jeddah",
    "Java",
    "2",
    "View Details"
};
GridViewRecord.DataSource = lstRecord;
GridViewRecord.DataBind();

答案 2 :(得分:-2)

  if (!this.IsPostBack)
            {   DataTable dt = new DataTable();
                dt.Columns.Add("0");
                dt.Columns.Add("1");
                dt.Columns.Add("2");
                dt.Columns.Add("3");
                dt.Columns.Add("4");

                DataRow dr = dt.NewRow();
                dr[0] = "12-12-12"; //Error message occured here
                dr[1] = "Jeddah";
                dr[2] = "Java";
                dr[3] = "2";
                dr[4] = "View Details";

                dt.Rows.Add(dr);
                GridViewRecord.DataSource = dt;
                GridViewRecord.DataBind();
            }

你需要添加到DataTable列来处理它(0,1,2,3,4)对应于表数组,就像你写dr [0] = Add(“0”);

或者您可以轻松地管理您想要插入日期的列

dr["2"]= "Java";