Warning: file_get_contents(/data/phpspider/zhask/data//catemap/2/csharp/258.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181

Warning: file_get_contents(/data/phpspider/zhask/data//catemap/1/asp.net/30.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
C# 在ASP.NET中单击链接按钮时如何在引导模式下显示数据_C#_Asp.net_Twitter Bootstrap_Twitter Bootstrap 3 - Fatal编程技术网

C# 在ASP.NET中单击链接按钮时如何在引导模式下显示数据

C# 在ASP.NET中单击链接按钮时如何在引导模式下显示数据,c#,asp.net,twitter-bootstrap,twitter-bootstrap-3,C#,Asp.net,Twitter Bootstrap,Twitter Bootstrap 3,如何在ASP.NET中单击链接按钮时以引导模式显示数据。在我的代码中 当我点击链接按钮时,模态打开,但赋值不显示。请告诉我如何使用指定值显示模态。 谢谢 .aspx代码: <asp:LinkButton ID="linkButton8" runat="server" data-toggle="modal" data-target="#myModal" OnClick="linkButton_Click">CIS Information</asp:LinkButton>

如何在ASP.NET中单击链接按钮时以引导模式显示数据。在我的代码中 当我点击链接按钮时,模态打开,但赋值不显示。请告诉我如何使用指定值显示模态。 谢谢

.aspx代码:

  <asp:LinkButton ID="linkButton8" runat="server" data-toggle="modal" data-target="#myModal" OnClick="linkButton_Click">CIS Information</asp:LinkButton>

      <div class="modal" id="myModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true">
                <div class="modal-dialog">
                    <div class="modal-content">
                        <div class="modal-header">
                            <button type="button" class="close" data-dismiss="modal" aria-hidden="true">&times;</button>
                            <h4 class="modal-title" id="myModalLabel" align="center">Candidate Information Sheet</h4>
                        </div>
                        <div class="modal-body">

                            <table class="table table-bordered" align="center">
                                <tr>
                                    <td>Name</td>
                                    <td>
                             <asp:TextBox ID="txtName" runat="server"></asp:TextBox></td>
                                    <td>Email</td>
                                    <td>
                                        <asp:TextBox ID="txtEnail" runat="server"></asp:TextBox></td>
                                </tr>
                            </table>
                        </div>
                        <div class="modal-footer">
                      <button type="button" class="btn btn-primary">Save</button>
                <button type="button" class="btn btn-primary" data-ismiss="modal">Close</button>
                        </div>
                    </div>
                </div>
            </div>

尝试添加UseSubmitBehavior=“false”并改为使用asp:按钮

    <asp:Button ID="btnModal" runat="server" data-toggle="modal" data-target="#myModal" OnClick="linkButton_Click" UseSubmitBehavior="false">CIS Information</asp:Button>
CIS信息
UseSubmitBehavior属性确定按钮是使用浏览器提交功能还是ASP.NET回发功能。默认情况下,它设置为true

编辑


由于linkButton没有“UseSubmitBehavior”属性。

请删除数据切换,并更新代码,如下所示:

 protected void linkButton_Click(object sender, EventArgs e)
        {
            txtName.Text = "Name";
            txtEnail.Text = "Mail@gmail.com";
            Page.ClientScript.RegisterStartupScript(GetType(), "modelBox", "$("#myModal").modal('show');", true);
        }

最好是使链接按钮具有客户端单击事件,该事件通过ajax获取数据,设置textboxs值,并使用相同的
modal('show')
JavaScript函数来显示模式框,这样会更高效、更友好。

您的按钮是否在表单标记中?如果不支持,请尝试。LinkButton不支持UseSubmitBehavior:您有过解决方案吗?我也不太走运。。。
 protected void linkButton_Click(object sender, EventArgs e)
        {
            txtName.Text = "Name";
            txtEnail.Text = "Mail@gmail.com";
            Page.ClientScript.RegisterStartupScript(GetType(), "modelBox", "$("#myModal").modal('show');", true);
        }