html - 如何从列表中删除点

Question

我想实现带有详细信息列表的链接，如下面的代码片段所示，但我想删除点列表我的意思是我只想显示没有点的列表，如下图所示。我想实现如下图。

.sidebarleft h2 {
    margin: 0.0em 0 0.3em;
    letter-spacing: -.001em;
    padding: 5px 10px;
    background: #461B7E url() repeat-x bottom left;
    font: bold 13px Arial;
    color: #fff;
    text-transform: uppercase;
    border-bottom: 1px solid #cccccc;
}



h2 {
    margin: 0.5em 0 .25em;
    font: normal bold 14px Arial, Tahoma, Helvetica, FreeSans, sans-serif;
    line-height: 1.4em;
    text-transform: uppercase;
    letter-spacing: .01em;
    color: #4b05bf;
}

.sidebarleft .widget-content {
    margin: 0 auto;
    padding: 5px 9px 5px 10px;
}
.sidebarleft ul {
    list-style: none;
    margin: 0 0 0;
    padding: 0 0 0;
}


.sidebarleft li {
    display: inline;
    margin: 0 auto;
    padding: 0 auto;
}





.sidebarleft a:link, .sidebarleft a:visited {
    color: #0000FF;
    text-decoration: none;
}

.sidebarleft li a {
    font: normal 12px verdana;
    text-transform: none;
    margin: 0;
    padding: 3px 0px 4px 5px;
    display: block;
    text-indent: 0px;
    border-bottom: 1px solid #aebcfd;
    line-height: 1.3em;
}
a:visited {
    color: #0000ff;
    text-decoration: none;
}
a:link {
    color: #4b05bf;
    text-decoration: none;
}

.sidebarleft ul {
    list-style: none;
    margin: 0 0 0;
    padding: 0 0 0;
}

.sidebarleft a:link, .sidebarleft a:visited {
    color: #0000FF;
    text-decoration: none;
}



.sidebarleft li a {
    font: normal 12px verdana;
    text-transform: none;
    margin: 0;
    padding: 3px 0px 4px 5px;
    display: block;
    text-indent: 0px;
    border-bottom: 1px solid #aebcfd;
    line-height: 1.3em;
}
a:visited {
    color: #0000ff;
    text-decoration: none;
}
a:link {
    color: #4b05bf;
    text-decoration: none;
}
.sidebarleft ul {
    list-style: none;
    margin: 0 0 0;
    padding: 0 0 0;
}

<div class="widget Label" data-version="1" id="Label6">
<h2>Latest IT Jobs in India - By Location</h2>
<div class="widget-content list-label-widget-content">
<ul>
<li>
<a dir="ltr" href="">Jobs in Bangalore</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Chennai</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Hyderabad</a>
</li>
<li>
<a dir="ltr" href="">Jobs in India</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Noida</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Mumbai</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Pune</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Gurgaon</a>
</li>
<li>
<a dir="ltr" href="">Jobs In Delhi</a>
</li>
<li>
<a dir="ltr" href="">Jobs in kolkata</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Coimbatore</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Ahmedabad</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Mohali</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Chandigarh</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Jaipur</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Cochin</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Haryana</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Andhra Pradesh</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Indore</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Kochi</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Tamilnadu</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Trivandrum</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Visakhapatnam</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Vijayawada</a>
</li>
<li>
<a dir="ltr" href="">Jobs in Mysore</a>
</li>
</ul>

</div>
</div>

Answer 1

li {
   list-style-type: none;
}

但是您通常应该使用一些类，为对象类型编写 css 是一个非常糟糕的主意