Question

我在php中创建动态菜单和子菜单....无法知道从哪里开始....我已经编写了所有查询但需要建议使用Ul和{{迭代结果集1}}带有它的元素....可能没有子菜单的子菜单，菜单可能有也可能没有子菜单......

li

任何建议....

修改

这就是我在做什么，

Menu1
  submenu1
     submenu11
     submenu12
  submenu2
     submenu21
Menu2
  submenu1
     submenu11
     submenu12

这是我的表，

alt text http://img229.imageshack.us/img229/1174/84296639.jpg
<?php $con=mysql_connect("localhost","root",""); $db=mysql_select_db("testt"); echo '<ul id="nav">'; $res=mysql_query("SELECT * FROM menu WHERE PARENT_ID=0"); while($row=mysql_fetch_array($res)) { echo '<li class="top"><a href="#" class="top_link"><span>'.$row['MENU_NAME'].'</span></a></li>'; $res1=mysql_query("SELECT * FROM menu WHERE PARENT_ID=1 AND CHILD_ID=".$row['ID'].""); echo '<ul class="sub">'; while($row1=mysql_fetch_array($res1)) { echo '<li><a href="#" class="fly">'.$row1['MENU_NAME'].'</li>'; $res2=mysql_query("SELECT * FROM menu WHERE PARENT_ID=2 AND CHILD_ID=".$row1['ID'].""); echo '<ul>'; while($row2=mysql_fetch_array($res2)) { echo '<li><a href="#">'.$row2['MENU_NAME'].'</li>'; } echo '</ul>'; } echo '</ul>'; } echo '</ul>'; ?>

Answer 1

我是这样做的：

css文件：

ul.makeMenu, ul.makeMenu ul {
    background-color:#000000;
    cursor:default;
    margin:0px 0px 0px 0px;
    padding:0px 0px 0px 0px;
    width: 180px;
}
ul.makeMenu li {
    background-image:url(../images/marker_rosu.gif);
    background-position:5px center;
    background-repeat:no-repeat;
    color:#D4020B;/*#FEBF01;*/
    list-style-type:none;
    position:relative;
    margin:0px 0px 0px 0px;
    padding:0px 5px 0px 15px;
    line-height:24px;
    text-align:left;
}
ul.makeMenu li>ul {
    display:none;
    position:absolute;
    top:0px;
    left:180px;
}
ul.makeMenu li:hover {
    background-image:url(../images/marker_negru.gif);
    background-position:5px center;
    background-repeat:no-repeat;
    background-color:#D4020B;/*#FEBF01;*/
    color:#FFFFFF;
}
ul.makeMenu {
    display:block;
}
ul.makeMenu li:hover>ul {
    display:block;
}
ul.makeMenu li a {
    color:#D4020B;/*#FEBF01;*/
    display:block;
    width:100%;
    text-decoration:none;
}
ul.makeMenu li a:hover {
    color:#000000;
}
ul.makeMenu li:hover>a {
    color:#000000;
}

产品的sql文件或其他任何类别表：

CREATE TABLE `category` (
  `id` int(10) unsigned NOT NULL auto_increment,
  `idParrent` int(10) unsigned NOT NULL,
  `name` varchar(255) NOT NULL,
  `description` varchar(255) NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=utf8;

和创建菜单的功能：

function getAFP ($id)
    {
        global $conx;
        $sql = sprintf("SELECT * FROM category WHERE idParrent = %d;", quote_smart($id));
        $rs = mysql_query($sql, $conx);
        $arrayResult = array();
        if (mysql_num_rows($rs)) {
            while ($row = mysql_fetch_array($rs)) {
                array_push($arrayResult, $row);
            }
            return $arrayResult;
        } else {
            return 0;
        }
    }

function makeListaCategory ($id)
    {
        $listaCategorii = array();
        $categorii = getAFP($id);
        if (! $categorii) {
            return 0;
        } else {
            foreach ($categorii as $categorie) {
                array_push($listaCategorii, $categorie[2] . "|" . $categorie[0]);
                $subcategorii = makeListaCategory($categorie[0]);
                if ($subcategorii) {
                    foreach ($subcategorii as $subcategorie) {
                        $valori = explode("|", $subcategorie);
                        array_push($listaCategorii, $categorie[2] . ">" . $valori[0] . "|" . $valori[1]);
                    }
                }
            }
        }
        return $listaCategorii;
    }

我希望这会对你有所帮助！

Answer 2

编辑：在您澄清问题之前发布。这是一个通用的递归菜单结构。

<?php
function print_menu($menu)
{
  echo "<ul>\n";
  foreach ($menu as $submenu)
  {
    echo "<li>\n";
    echo "<span>{$submenu['name']}</span>\n";
    if (!empty($submenu['children']))
      print_menu($submenu['children']);
    echo "</li>\n";
  }
  echo "</ul>\n";
}

print_menu(
  array(
    array(
      'name' => 'Menu 1',
      'children' => array(
        array(
          'name' => 'Submenu 11',
          'children' => array(
            array('name' => 'Submenu 111'),
            array('name' => 'Submenu 112')
          )
        ),
      )
    ),
    array(
      'name' => 'Menu 2'
    )
  )
);
?>

如何在php中迭代ul和li元素？

2 个答案: