通过超链接打开新的gridview [英] Open new gridview through Hyperlink

问题描述

如何在同一页面上通过超链接打开一个新的gridview，我不想关闭现有的gridview，但想要显示另一个gridview，当用户单击此表上的任何超链接时，就会显示另一个gridview。我有一些像下面这样的数据，当我点击任何这些超链接时，我想在同一页面上打开一个新的gridview。新的gridview将有来自不同表格的数据。

我无法插入图片，因为我没有足够的声望，但可以共享我的HTML代码：

 < asp：GridView ID =GridView1runat =serverAutoGenerateColumns =FalseDataSourceID =SqlDataSource1DataKeyNames =UN_AT_Group> 
<列> 
< asp：HyperLinkField DataTextField =Group_DescriptionDataNavigateUrlFields =UN_AT_GroupDataNavigateUrlFormatString =〜/ Details.aspx？Id = {0}
< / Columns> 
< / asp：GridView> 
  

解决方案

改用LinkBut​​ton：

$ p $ lt; code>< asp：GridView ID =GridView1runat =serverAutoGenerateColumns =FalseDataSourceID =SqlDataSource1DataKeyNames =UN_AT_GroupOnRowCommand = GridView1_RowCommandVisible =True>
<列>
< asp：TemplateField>
< ItemTemplate>
< asp：LinkBut​​ton ID =GotoNextGridrunat =serverCommandArgument =NextGridCommandName =NextGridText =Show Rights>
< / asp：LinkBut​​ton>
< / ItemTemplate>
< / asp：TemplateField>
< /列>
< / asp：GridView>

为第二个 GridView 做同样的事情，但设置 Visibile =false。

然后在CodeBehind中捕捉它:(注意，无论你的 DataTextField 是什么。

  protected void GridView1_RowCommand（object sender，GridViewCommandEventArgs e）
 {
 if（e.CommandName ==NextGrid）
 {
 LinkBut​​ton lb =（LinkBut​​ton） e.CommandSource; 
 GridViewRow gvr =（GridViewRow）lb.NamingContainer; 
标签lbl = gvr.FindControl（GroupDescription）作为标签; 
字符串描述= lbl.Text; 
 GridView1.Visible = false; 
 GridView2.Visible = true; 
 FillDataForGridView2（描述）//在这里为GridView2填充数据并将描述作为参数传递
} 
} 
  

请小心使用 UpdatePanel ，th您需要添加触发器：

 <触发器> 
< asp：AsyncPostBackTrigger ControlID =GridView1EventName =RowCommand/> 
< /触发器> 
  

我希望这有助于您。

如果你有任何问题只是问。

How can I open a new gridview through hyperlink on same page, I don't want to close existing gridview but want to show another gridview adjacent to this one, when user clicks on any hyperlink on this table. I have some data like below and I want to open a new gridview on same page when I click on any of these hyperlinks. New gridview will be having data from different table.

I cannot insert image as I do not have enough reputation but can share my HTML Code:

<asp:GridView ID="GridView1" runat="server" AutoGenerateColumns="False" DataSourceID="SqlDataSource1" DataKeyNames="UN_AT_Group">
    <Columns>
           <asp:HyperLinkField DataTextField="Group_Description" DataNavigateUrlFields="UN_AT_Group" DataNavigateUrlFormatString="~/Details.aspx?Id={0}"
    </Columns>
</asp:GridView>

解决方案

Use a LinkButton instead:

<asp:GridView ID="GridView1" runat="server" AutoGenerateColumns="False" DataSourceID="SqlDataSource1" DataKeyNames="UN_AT_Group" OnRowCommand="GridView1_RowCommand" Visible="True">
    <Columns>    
        <asp:TemplateField>
            <ItemTemplate>
                 <asp:LinkButton ID="GotoNextGrid" runat="server" CommandArgument="NextGrid" CommandName="NextGrid" Text="Show Rights">
                 </asp:LinkButton>  
            </ItemTemplate>
        </asp:TemplateField>
    </Columns>  
</asp:GridView>

Do the same for you second GridView but set Visibile="false".

and then catch it in CodeBehind: (Take care, make sure that what I have as Label here can also be something else... Whatever your DataTextField is.

protected void GridView1_RowCommand(object sender, GridViewCommandEventArgs e)
{
    if (e.CommandName == "NextGrid")
    {
        LinkButton lb = (LinkButton)e.CommandSource;
        GridViewRow gvr = (GridViewRow)lb.NamingContainer;
        Label lbl = gvr.FindControl("GroupDescription") as Label;
        string description = lbl.Text;
        GridView1.Visible = false;
        GridView2.Visible = true;
        FillDataForGridView2(description) //Fill the Data for GridView2 here and pass description as parameter
    }
}

Take care if you use UpdatePanel, then you need to add an Trigger:

<Triggers>
    <asp:AsyncPostBackTrigger ControlID="GridView1" EventName="RowCommand" />
</Triggers>

I hope this helps.

If you have any questions just ask.

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