链接按钮不触发动态网格视图中asp.net

Question

我有一个带有两个链接按钮的动态gridview。如果我点击链接按钮，事件不会触发。但如果我在页面加载中调用“display”方法，它就可以正常工作了。代码如下

public void display()
    {
       GridView grdv = new GridView();
        grdv.AutoGenerateColumns = false;
        DataSet ds = new DataSet();
        DataTable dt = new DataTable();
        BL.ESSBL bl = new BL.ESSBL();
        ds = bl.GetContactList();//getting datatable
        if (ds != null && ds.Tables.Count > 0 && ds.Tables[0].Rows.Count > 0)
        {
            dt = ds.Tables[0];
            grdv.RowDataBound += new GridViewRowEventHandler(grdv_RowDataBound);
            grdv.DataSource = null;
            grdv.DataBind();
            grdv.Columns.Clear();
            for (int i = 0; i < dt.Columns.Count; i++)
            {
                BoundField boundfield = new BoundField();
                boundfield.DataField = dt.Columns[i].ColumnName.ToString();
                boundfield.HeaderText = dt.Columns[i].ColumnName.ToString();
                grdv.Columns.Add(boundfield);
            }
            TemplateField tmf = new TemplateField();
            grdv.Columns.Add(tmf);
            tmf = new TemplateField();
            grdv.Columns.Add(tmf);
            grdv.DataSource = dt;
            grdv.DataBind();
            pnlupdate.Controls.Add(grdv);
           }
    }
 void grdv_RowDataBound(object sender, GridViewRowEventArgs e)
    {
        if (e.Row.RowType == DataControlRowType.DataRow)
        {    
           int count= e.Row.Cells.Count;
            LinkButton lnkupdate= new LinkButton();
            lnkupdate.ID = "Update";
            lnkupdate.Text = "Update";
            LinkButton lnkdelete = new LinkButton();
            lnkdelete.ID = "delete";
            lnkdelete.Text = "delete";
            lnkdelete.Click += new EventHandler(lnkdelete_Click);
            lnkupdate.CommandArgument = (e.Row.DataItem as DataRowView).Row[0].ToString();
            lnkdelete.CommandArgument = (e.Row.DataItem as DataRowView).Row[0].ToString();
            lnkupdate.Click += new EventHandler(lnkupdate_Click);
            e.Row.Cells[count-2].Controls.Add(lnkupdate);
            e.Row.Cells[count-1].Controls.Add(lnkdelete);

        }
    }
   protected void ddlProcess_SelectedIndexChanged(object sender, EventArgs e)
    {
      dynamicgridview();        // not working
            Clear();
    }

void lnkupdate_Click(object sender, EventArgs e)
    {
        Response.Write(@"<script language=""javascript"">alert(""update details "");</script>");
    }

如果我把“显示”方法放在页面加载中，它将调用每个回发。我不想要那个。我想在下拉选择更改事件中调用此方法。如果我把“显示”方法放在那个链接点击事件不触发。所以我必须做什么来克服这个问题。

Answer 1

而不是grdv。柱。清除（）；你能试试看grdv吗。AutoGenerateColumns=false；。因为我不知道你的代码有错误

Answer 2

如果回发后未绑定网格，则必须在加载页面之前创建gridview。GridView需要加载ViewState。

    private GridView gv;
    protected void Page_Init(object sender, EventArgs e)
    {
        gv = new GridView();
        gv.ID = "gv";
        gv.AutoGenerateColumns = false;
        gv.Columns.Add(new TemplateField());
        gv.RowCreated += gv_RowCreated;
        gv.RowDataBound += gv_RowDataBound;
        pnl.Controls.Add(gv);
    }
    protected void Page_Load(object sender, EventArgs e)
    {
        if (!IsPostBack)
        {
            gv.DataSource = new object[] {
                new object()
            };
            gv.DataBind();
        }
    }

    void gv_RowDataBound(object sender, GridViewRowEventArgs e)
    {
        var lb = e.Row.FindControl("Update") as LinkButton;
        lb.CommandArgument = "1";
    }

    void gv_RowCreated(object sender, GridViewRowEventArgs e)
    {
        // If you bind gridview after Page_Init,
        // This event will not be fired after postback.
        LinkButton lb = new LinkButton();
        lb.ID = "Update";
        lb.Text = "Update";
        lb.Click += lb_Click;
        e.Row.Cells[e.Row.Cells.Count - 1].Controls.Add(lb);
    }

    void lb_Click(object sender, EventArgs e)
    {
        var lb = (LinkButton)sender;
        string arg = lb.CommandArgument;
    }