2015-10-02 42 views
0

需要時打開密碼或用戶是不正確的一個模式,將留在否則。但是,不工作,不工作不打開模態引導模態,從代碼隱藏不工作

protected void bntLogar_Click(object sender, EventArgs e) 
    { 
     Registrar criptografia = new Registrar(); 

     if (Login.logarUsuario(txtUser.Text, criptografia.CriptografiaMD5(txtSenha.Text))) 
     { 
      //Cria um cookie do lado do servidor 
      HttpCookie cookie = new HttpCookie("estado", "conectado"); 

      //Define a validade do cookie (10 dias a partir de hoje) 
      cookie.Expires = DateTime.Now.AddMonths(12); 

      //Envia o cookie para o cliente 
      Response.Cookies.Set(cookie); 

      //Redireciona para a pagina inicial 
      Response.Redirect("Admin.aspx"); 

      //fortawsome 
     } 
     else 
     { 
      ScriptManager.RegisterClientScriptBlock(this, this.GetType(), "Pop", "openModal();", true); 
     } 

    } 

我把 JavaScript代碼

 <script type="text/javascript"> 
     function openModal() { 
      $('#modalLogin').modal('show'); 
     } 
    </script> 

莫代爾

<div class="modal fade" id="modalLogin" runat="server" role="dialog"> 
     <div class="modal-dialog"> 
      <div class="modal-content"> 
       <div class="modal-header"> 
        <h4>Warning</h4> 
       </div> 
       <div class="modal-body"> 
        <p>Watch out! Your about to be locked out.</p> 
       </div> 
       <div class="modal-footer"> 
        <a class="btn btn-primary" data-dismiss="modal">Close</a> 
       </div> 
      </div> 
     </div> 
    </div> 

在此SITEMASTER腹式<asp:ScriptManager ID="ScriptManager1" runat="server"></asp:ScriptManager>

回答

0

你可以嘗試這樣的:

ScriptManager.RegisterStartupScript(this, this.GetType(), Guid.NewGuid().ToString(), "alert('Informations invalides');", true); 

代替

ScriptManager.RegisterClientScriptBlock(this, this.GetType(), "Pop", "openModal();", true); 

,但我不知道這是否會解決您的問題。

+0

出現aletar,是另一種選擇,謝謝。 但確實需要模態,它是按設計要求。 –

+0

是的,我的意思是,使用模式,但使用RegisterStartupScript(),而不是RegisterClientScriptBlock(),我想說什麼 –