Question

我必须显示n网格，n是可变的，然后我不知道我会有多少个网格。

我的问题是，我必须使用Visible false初始化此网格，当单击按钮显示该按钮的特定网格时，如何将按钮链接到gridview？

生成网格的代码：

    foreach (List<DataRow> lst in grids)
    {

        dt = lst.CopyToDataTable();

        GridView grv = new GridView();
        grv.AlternatingRowStyle.BackColor = System.Drawing.Color.FromName("#cccccc");
        grv.HeaderStyle.BackColor = System.Drawing.Color.Gray;
        grv.ID = "grid_view"+i;
        grv.Visible = false;
        grv.DataSource = dt;
        grv.DataBind();


        Label lblBlankLines = new Label();
        lblBlankLines.Text = "<br /><br />";


        Label lblTipo = new Label();
        string tipoOcorrencia = lst[0]["DESC_OCORRENCIA"].ToString();
        tipoOcorrencia = CultureInfo.CurrentCulture.TextInfo.ToTitleCase(tipoOcorrencia);
        int quantidade = lst.Count;
        lblTipo.Text = tipoOcorrencia + ": " + quantidade;


        LinkButton lkBtn = new LinkButton();
        lkBtn.ID = "link_button"+i;
        lkBtn.Text = "+";

        place_grids.Controls.Add(lblBlankLines);
        place_grids.Controls.Add(lkBtn);
        place_grids.Controls.Add(lblTipo);
        place_grids.Controls.Add(grv);


        place_grids.DataBind();

        i++;
    }

提前致谢。

Answer 1

修改你的foreach循环，如下所示。

private void GenerateControls()
{
    int i = 0;
    foreach (List<DataRow> lst in grids)
    {
        dt = lst.CopyToDataTable();

        GridView grv = new GridView();
        grv.AlternatingRowStyle.BackColor = System.Drawing.Color.FromName("#cccccc");
        grv.HeaderStyle.BackColor = System.Drawing.Color.Gray;
        grv.ID = "grid_view" + i;
        //grv.Visible = false;//Commented as the grid needs be generated on client side, in order to make it visible from JavaScript/jQuery
        grv.Attributes.Add("style", "display:none;");
        grv.DataSource = dt;
        grv.DataBind();

        //Adding dynamic link button
        LinkButton lnkButton = new LinkButton();
        lnkButton.Text = "button " + i;
        //lnkButton.Click += new EventHandler(lnkButton_Click);
        lnkButton.ID = "lnkButton" + i;
        lnkButton.OnClientClick = "ShowGrid('" + grv.ClientID + "');";

        Label lblTipo = new Label();
        lblTipo.Text = "text " + i;
        lblTipo.ID = "lbl" + i;

        tempPanel.Controls.Add(lblTipo);
        tempPanel.Controls.Add(grv);
        tempPanel.Controls.Add(lnkButton);

        tempPanel.DataBind();
        i++;
    }
}

如果您希望触发服务器端事件，则必须按如下所示添加链接按钮单击事件。（取消注释将事件处理程序分配给链接按钮的行。）

protected void lnkButton_Click(Object sender, EventArgs e)
{
    LinkButton lnkButton = (LinkButton)sender;
    String index = lnkButton.ID.Substring(lnkButton.ID.Length - 1);

    GridView grv = (GridView)tempPanel.FindControl("grid_view" + index);
    grv.Visible = true;
}

您需要在Page_Init事件中添加所有动态添加的控件以维护其状态。请参阅下面的链接可能很有用。

Dynamically Created Controls losing data after postback

ViewState in Dynamic Control

从GenerateControls事件中调用方法Page_Init，如下所示。

protected void Page_Init(object sender, EventArgs e)
{
    GenerateControls();
}

编辑：

JavaScript函数......

function ShowGrid(gridID) {
    document.getElementById(gridID).style.display = ''
}

我保持服务器端点击事件不变。但我已经注释了将事件处理程序分配给链接按钮的行。

链接按钮动态显示网格视图asp.net

1 个答案: