asp.net彈出資訊框——用ScriptManager

 

一、

前台代碼：

<div>
        <asp:ScriptManager ID="ScriptManager1" runat="server">
        </asp:ScriptManager>
        <asp:UpdatePanel ID="UpdatePanel1" runat="server">
        <ContentTemplate>
       
        </ContentTemplate>
        <Triggers>
            <asp:AsyncPostBackTrigger ControlID="Button1" EventName="Click"/>
            <asp:AsyncPostBackTrigger ControlID="Button2" EventName="Click" />
        </Triggers>
        </asp:UpdatePanel>
       
        <asp:Button ID="Button1" runat="server" Text="彈出確認資訊" onclick="Button1_Click" />
    &nbsp;<asp:Button ID="Button2" runat="server" Text="彈出選擇行資訊"
            onclick="Button2_Click" />
    &nbsp;
        <asp:Label ID="Label1" runat="server"></asp:Label>
    </div>

後台代碼：

protected void Button1_Click(object sender, EventArgs e)
    {
        ScriptManager.RegisterStartupScript(UpdatePanel1, UpdatePanel1.GetType(), "", "alert('確定!')", true);
    }

第二種方法：

前台代碼：

<div>
        <asp:ScriptManager ID="sm" runat="server">
        </asp:ScriptManager>
        <asp:UpdatePanel ID="up" runat="server">
        <ContentTemplate>
            <asp:Button ID="btnCommit" runat="server" Text="Button"
                onclick="btnCommit_Click" />
            &nbsp;
            <asp:Label ID="Label1" runat="server" Text="Label"></asp:Label>
        </ContentTemplate>
        </asp:UpdatePanel>
    </div>

後台代碼：

protected void btnCommit_Click(object sender, EventArgs e)
    {
        ///擷取彈出對話方塊的按鈕
        Button button = (Button)sender;
        ///註冊對話方塊的指令碼
        ScriptManager.RegisterClientScriptBlock(button, button.GetType(), button.UniqueID, "alert('這是AJAX Web環境中的對話方塊。');", true);
      
    }