我从我的mysql数据库中回显这个列表,但我不想要项目符号,所以我使用引导列表组项。我觉得我的列表标签在某个地方犯了一个非常愚蠢的错误,但我不确定。我的清单旁边还有子弹。我没有包括所有连接到数据库php的内容,因为这不是问题所在。
这是我的,
<div class="panel panel-info">
<div class="panel-heading">Contents</div>
<ul class="list-group">
<li class="list-group-item">
<?php
basic connect to mysql database stuff here
}
$query = mysqli_query($dat, "SELECT * FROM Content ORDER BY ContentName") or die(mysqli_error($dat));
while($list = mysqli_fetch_array($query)){
echo"<li>";
echo"<a href = >";
echo $list['ContentName'];
echo"</a>";
echo "</li>";
}
mysqli_close($dat);
?>
</li>
</ul>
</div>
您正在将li
添加到另一个li
中。应该是
<ul class="list-group">
<?php
}
$query = mysqli_query($dat, "SELECT * FROM Content ORDER BY ContentName") or die(mysqli_error($dat));
while($list = mysqli_fetch_array($query)){
echo"<li class='list-group-item'>";
echo"<a href = >";
echo $list['ContentName'];
echo"</a>";
echo "</li>";
}
mysqli_close($dat);
?>
</ul>
使用echo "<li class='list-group-item'>";
而不是<li>
,并在php代码之前删除<li>
。
<div class="panel panel-info">
<div class="panel-heading">Contents</div>
<ul class="list-group">
<?php
basic connect to mysql database stuff here
}
$query = mysqli_query($dat, "SELECT * FROM Content ORDER BY ContentName") or die(mysqli_error($dat));
while($list = mysqli_fetch_array($query)){
echo "<li class='list-group-item'>";
echo "<a href ='www.example.com' >";
echo $list['ContentName'];
echo "</a>";
echo "</li>";
}
mysqli_close($dat);
?>
</ul>
</div>