如何在php中循环跳过一个项目从数据库显示

Question

<ul class="sub-menu">
  <?php 
    $sql = mysql_query("SELECT * FROM tbl_main_menu ORDER BY fld_main_menu_id ASC");
    while($menu_list = mysql_fetch_array($sql)){
  ?>  
  <li>
    <a href="content_editor.php?main_menu_id=<?php echo $menu_list['fld_main_menu_id']; ?>&menu_name=<?php echo $menu_list['fld_main_menu_name'];?>">
      <?php echo $menu_list['fld_main_menu_name']; ?><span class="arrow"></span>
    </a>
  </li>     
  <?php } ?>    
</ul>

此处显示数据库中的菜单：

1. 主页
2. 关于
3. 董事会
4. 管理
5. 产品
6. 权利要求
7. 职业

但是我不想在数据库中显示董事会，我想在这里放置一个静态页面链接。现在我要做什么？

Answer 1

你可以从html

这样做

    <ul class="sub-menu">
       <?php 
          $sql = mysql_query("select * from tbl_main_menu order by fld_main_menu_id asc");
          while($menu_list = mysql_fetch_array($sql)){
       ?>  

          <?php if ($menu_list['fld_main_menu_name']!='Board of directors') { // check if name is Board of directors ?>
          <li>
          <a href="content_editor.php?main_menu_id=<?php echo $menu_list['fld_main_menu_id']; ?>&menu_name=<?php echo $menu_list['fld_main_menu_name'];?>">
             <?php echo $menu_list['fld_main_menu_name']; ?><span class="arrow"></span>
          </a>
          </li>           
        <?php } //end of if 
          } //end of while ?>      
    </ul>

你可以从数据库中这样做：

$sql = mysql_query("select * from tbl_main_menu where fild_main_menu_name != 'Board of directors' order by fld_main_menu_id asc");

Answer 2

你可以在你的PHP中使用一些if语句来实现它，但是当你的MySQL正确时，为什么不这样修改：

<ul class="sub-menu">
    <!-- This is a static page -->
    <li><a href="static_page.html">Static Page</a></li>    
    <?php 
        $sql = mysql_query("SELECT * FROM tbl_main_menu WERE fld_main_menu_name != 'Board of directors' ORDER BY fld_main_menu_id asc");
        while($menu_list = mysql_fetch_array($sql)){
    ?>  
    <li>
        <a href="content_editor.php?main_menu_id=<?php echo $menu_list['fld_main_menu_id']; ?>&menu_name=<?php echo $menu_list['fld_main_menu_name'];?>">
         <?php echo $menu_list['fld_main_menu_name']; ?><span class="arrow"></span>
         </a>
    </li>           


    <?php } ?>

    <!-- This is a static page -->
    <li><a href="static_page.html">Static Page</a></li>    
</ul>

在任何一种方法中，您都需要确保链接名称的拼写和大小写是准确的。

另请注意，我在开头和结尾都包含静态链接作为示例。

修改：根据以下评论，我会如何做到这一点：

<ul class="sub-menu">   
    <?php 
        $sql = mysql_query("SELECT * FROM tbl_main_menu ORDER BY fld_main_menu_id asc");
        while($menu_list = mysql_fetch_array($sql)){
    ?>
    <?php if('fld_main_menu_name' != 'board of directors'){?>
    <li>
        <a href="content_editor.php?main_menu_id=<?php echo $menu_list['fld_main_menu_id']; ?>&menu_name=<?php echo $menu_list['fld_main_menu_name'];?>">
         <?php echo $menu_list['fld_main_menu_name']; ?><span class="arrow"></span>
         </a>
    </li>
    <?php } else { ?>
    <li>
        <a href="custom_page_path.php">Custom page name<span class="arrow"></span></a>
    </li>

    <?php } //end for if...else ?> 
    <?php } //end for while loop ?> 
</ul>