Question

我从我的mysql数据库回显这个列表，但我不想要子弹，所以我使用bootstrap list-group-item。我觉得我在列表标签的某个地方犯了一个非常愚蠢的错误，但我不确定。我仍然在我的名单旁边获得子弹。我没有将我所有连接到我的数据库的PHP都包括在内，因为这不是问题所在。

这就是我所拥有的，

<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>

Answer 1

您要在另一个li内添加li。它应该是 -

<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>

Answer 2

使用echo "<li class='list-group-item'>";代替 <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>

我在echo list和bootstrap中做错了什么？

2 个答案: