1

HTML:

<div id="divLogin">
<a> 
<img src="Image1.jpg" style="height:32px;padding-top:0px;" id="img"/></a>

<div id="subLogin" style="width:250px">        

    <asp:Panel ID="panel1" runat="server" DefaultButton="btnSend">
    <table>
        <tr>
            <td width="20%" style="color: White">
                Username
            </td>
            <td width="3%">
                :
            </td>
            <td width="27%">
                <asp:TextBox runat="server" ID="txtName"></asp:TextBox>
            </td>
        </tr>

        <tr>
            <td style="color: White">
                Password
            </td>
            <td>
                :
            </td>
            <td>
                <asp:TextBox runat="server" ID="txtPassword"></asp:TextBox>
            </td>
        </tr>


        <tr>
            <td colspan="3" align="center">
                <asp:Button Text="Send" runat="server" ID="btnSend" />
            </td>
        </tr>

    </table>
</asp:Panel>

 </div>
</div>

jQuery

$(document).ready(function () {
    $("#divLogin > a").click(function () {

        $("#subLogin").slideToggle(400);
    });
});



$(document).click(function (e) {
    var elem = (e.target || e.srcElement).id;
    if (elem != 'img') {
        $("#subShare").slideUp(400);
    }
});

当我单击图像时,我想通过 Jquery Slide 显示登录表单。我想在单击文档时上滑 subLogin div。当我单击 subLogin div 或 btnSend button.plz 帮助时,div 也会向上滑动...

更新:这里是 JsFiddle:http: //jsfiddle.net/WN2uB/

4

1 回答 1

1

这就是你想要的吗?

http://jsfiddle.net/WN2uB/2/

 $(document).click(function (e) {
     if(!$(e.target).is("#img")&&!$.contains($("#subLogin")[0],e.target)) { 
         $("#subLogin").slideUp(400);
     }
 });
于 2013-10-15T05:50:29.757 回答