Javascript样式显示不起作用

时间:2010-07-13 17:19:50

标签: javascript html dom

function insert()
{
    var linkElement = document.getElementById("BackButton");
    var linkElementLnk = document.getElementById("BackButtonlnk");
    var loc_array = document.location.href.split('/');

    if (loc_array[loc_array.length-3] == "m")
    {
        linkElementLink.style.display = 'none';
    }
    else if (loc_array[loc_array.length-2] == "maps"
          || loc_array[loc_array.length-3] == "maps"
          || loc_array[loc_array.length-2] == "stations"
          || loc_array[loc_array.length-3] == "stations" )
    {
        var newT = document.createTextNode("Stations & Maps");
    }
    else
    {
        var newT = document.createTextNode(unescape(capWords(loc_array[loc_array.length-3])));
    }
    linkElement.appendChild(newT);
}

这是我的HTML

<a id="BackButtonlnk" href="#" class="white" onClick="history.go(-1); return false;">
    <div id="BackButton1"></div>
    <div id="BackButton"></div>
    <div id="BackButton3"></div>
    </a>

style.display none无效,似乎只是破坏了脚本。我错过了什么吗?

2 个答案:

答案 0 :(得分:3)

检查你的拼写

var linkElementLnk = document.getElementById("BackButtonlnk");
...
linkElementLink.style.display = 'none';

答案 1 :(得分:0)

尝试

style.display = '';