Question

我有一个网站，其中导航栏是完全动态的，您可以根据需要添加多个孩子！逻辑很好，但我只需要添加一个分隔符，根据设计要求将它们彼此分开。现在问题是...... <li class="divider"></li>有效，但它也将分隔符添加到导航栏菜单的每次迭代的最后一个索引，因为它是自解释的。我不想在循环的最后一次迭代中添加一个分隔符！

这是我的代码：

 <ul class='nav navbar-nav'>
   <li><a href="index.php" style="color: white;"><img id="navbar-home" src="images/navbar_home.png" class="img-responsive"></a></li>
   <?php
     function display_children($parent, $level) {
       $conn=mysqli_connect("localhost","root","") or die ("not connected");
       mysqli_select_db($conn,"occ")or die ("not selected");
       $result = mysqli_query($conn,"SELECT a.id, a.title,a.page_id, Deriv1.Count FROM `menu` a  LEFT OUTER JOIN (SELECT parent, COUNT(*) AS Count FROM `menu` GROUP BY parent) Deriv1 ON a.id = Deriv1.parent WHERE a.parent=" . $parent);

       while ($row = mysqli_fetch_array($result)) {
         if ($row['Count'] > 0) {
           echo "<li><a href='" . base_url().'pages/'.$row['page_id'] . "'>" . strtoupper(str_replace('-', ' ', clean($row['title']) )). "<span class=caret></span></a>";
           echo "<ul class=\"dropdown-menu\">";
           display_children($row['id'], $level + 1);

           echo "</ul>";
           echo "</li>";
           echo "<li class=\"divider\"></li>";
         } elseif ($row['Count']==0) {
           echo "<li><a href='" . base_url().'pages/'.$row['page_id'] . "'>" . strtoupper(str_replace('-', ' ', clean($row['title'])) ) . "</a></li>";
           echo "<li class=\"divider\"></li>";
         } else;

       }

     }

     function clean($string) {
       return preg_replace('/[^A-Za-z0-9\-]/', ' ', $string); // Removes special chars.
     }

     display_children(0, 1);
   ?>
 </ul>

以下是该问题的屏幕截图，以防有人想澄清：

The issue

Answer 1

您可以使用纯CSS执行此操作，使用last-child选择器：

.nav .dropdown-menu .divider:last-child {
    display: none;
}

要使用php，您可以代替回显所有菜单项，将其添加到数组中，并检查最后一项是否等于分隔符。然后你的代码看起来像这样：

function display_children($parent, $level) {
    $conn=mysqli_connect("localhost","root","") or die ("not connected");
    mysqli_select_db($conn,"occ")or die ("not selected");
    $result = mysqli_query($conn,"SELECT a.id, a.title,a.page_id, Deriv1.Count FROM `menu` a  LEFT OUTER JOIN (SELECT parent, COUNT(*) AS Count FROM `menu` GROUP BY parent) Deriv1 ON a.id = Deriv1.parent WHERE a.parent=" . $parent);

    $returnHtml = array();
    while ($row = mysqli_fetch_array($result)) {
        if ($row['Count'] > 0) {
            $returnHtml[] = "<li><a href='" . base_url().'pages/'.$row['page_id'] . "'>" . strtoupper(str_replace('-', ' ', clean($row['title']) )). "<span class=caret></span></a>";
            $returnHtml[] = "<ul class=\"dropdown-menu\">";

            display_children($row['id'], $level + 1);

            $returnHtml[] = "</ul>";
            $returnHtml[] = "</li>";
            $returnHtml[] = "<li class=\"divider\"></li>";
        } elseif ($row['Count']==0) {
            $returnHtml[] = "<li><a href='" . base_url().'pages/'.$row['page_id'] . "'>" . strtoupper(str_replace('-', ' ', clean($row['title'])) ) . "</a></li>";
            $returnHtml[] = "<li class=\"divider\"></li>";
        } else;
    }

    if (end($returnHtml === "<li class=\"divider\"></li>")) {
        array_pop($returnHtml);
    }

    return implode('', $returnHtml);
}

function clean($string) {
    return preg_replace('/[^A-Za-z0-9\-]/', ' ', $string); // Removes special chars.
}

echo display_children(0, 1);

如何从上一个索引中删除li class =“divider”？

1 个答案: