jquery下拉菜单更改保持页面返回顶部

时间:2014-04-09 02:39:06

标签: javascript jquery html backbone.js

我有一个带有3个下拉菜单的html,这是

<select id="day"> <select id="month"> <select id="year">

但每当我选择其中一个值时,选择该选项后,页面将保持回到顶部

我试过跟随类似的情况 How do I stop a web page from scrolling to the top when a link is clicked that triggers JavaScript?

我在我的html中添加一个脚本

<script>
$("#day").click(function(e) {
  return false;
  e.preventDefault();
});
</script>

这是我的简短html文件

<header class="headerNavi">
    <div class="headerNaviLeft">
        <a href="#home">Back</a>
    </div>

    <div class="headerNaviRight">
        <a href="#home">Cancel</a>
    </div>
</header><!-- End Header -->
<!-- content -->

<section id="formDiv">
    <h3>Create Link</h3>

    <p class="tl01">Snap a photo or upload your own, then fill up the details
    to continue...</p>

    <div class="rowBox">
        <span class="photoBox">
            <img id="imageFile" name="imageFile" src="" style="width:100%; min-height:150px;" onclick="getPhoto(navigator.camera.PictureSourceType.PHOTOLIBRARY);" />
        </span>
    </div>

    <div class="rowBox">
        <select id="day">
            <option disabled="disabled" selected="selected" value="">
                DD
            </option>
<option value="1">
                    1
                </option>
// up to 31
        </select>
    </div>
</section>

这是我的观点

define(['jquery', 'underscore', 'backbone', 'text!modules/link/linkCreateViewTemplate.html'], function($, _, Backbone, linkCreateViewTemplate) {
    var LinkCreateView = Backbone.View.extend({
        template: _.template(linkCreateViewTemplate),
        initialize: function() {

        },
        events: {       
            "click #createButton": "create",
        },
             create: function(event) {
    window.location.replace("#linkconfirm");
},
        render: function() {
            this.$el.append(this.template());
        },
        remove: function() {
            Backbone.View.prototype.remove.call(this);
            $(window).off('scroll');
        }
    });
    return LinkCreateView;
});

1 个答案:

答案 0 :(得分:1)

尝试将代码包装在DOM就绪处理程序$(function(){...});

$(function() {
    $("#day").click(function(e) {
        return false;
        e.preventDefault();
    });
});