PHP+JS三级菜单联动菜单实现方法

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PHP+JS三级菜单联动菜单实现方法_PHP

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May 27, 2016 am 10:35 AM

jsphp

本文实例讲述了PHP+JS三级菜单联动菜单实现方法。分享给大家供大家参考，具体如下：

<html>
  <head>
    <title>
      智能递归菜单-读取数据库
    </title>
    <style>
      TD { FONT-FAMILY: "Verdana", "宋体"; FONT-SIZE: 12px; LINE-HEIGHT: 130%;
      letter-spacing:1px } A:link { COLOR: #990000; FONT-FAMILY: "Verdana", "宋体";
      FONT-SIZE: 12px; TEXT-DECORATION: none; letter-spacing:1px } A:visited
      { COLOR: #990000; FONT-FAMILY: "Verdana", "宋体"; FONT-SIZE: 12px; TEXT-DECORATION:
      none; letter-spacing:1px } A:active { COLOR: #990000; FONT-FAMILY: "Verdana",
      "宋体"; FONT-SIZE: 12px; TEXT-DECORATION: none; letter-spacing:1px } A:hover
      { COLOR: #ff0000; FONT-FAMILY: "Verdana", "宋体"; FONT-SIZE: 12px; TEXT-DECORATION:
      underline; letter-spacing:1px } .Menu { COLOR:#000000; FONT-FAMILY: "Verdana",
      "宋体"; FONT-SIZE: 12px; CURSOR: hand }
    </style>
    <script language=javascript>
      function ShowMenu(MenuID) {
        if (MenuID.style.display == "none") {
          MenuID.style.display = "";
        } else {
          MenuID.style.display = "none";
        }
      }
    </script>
  </head>
  <body>
<&#63;php
// $Id:$
//基本变量设置
$GLOBALS["ID"] = 1;
//用来跟踪下拉菜单的ID号 $layer=1;
//用来跟踪当前菜单的级数
//连接数据库
$Con=mysql_connect( "localhost", "root", "123456"); mysql_select_db(
"menu");
//提取一级菜单
$sql="select * from menu where parent_id=0" ;
$result=mysql_query($sql,$Con);
//如果一级菜单存在则开始菜单的显示
if(mysql_num_rows($result)>0) ShowTreeMenu($Con, $result, $layer, $ID); 
//=============================================
//显示树型菜单函数 ShowTreeMenu($con,$result,$layer) //$con：数据库连接 
//$result：需要显示的菜单记录集
//layer：需要显示的菜单的级数 
//=============================================
function
ShowTreeMenu($Con, $result, $layer) {
//取得需要显示的菜单的项目数
$numrows=mysql_num_rows($result);
  //开始显示菜单，每个子菜单都用一个表格来表示 echo "
   < tablecellpadding = '0'cellspacing = '0'border = '0' > "; for($rows=0;$rows
        <$numrows;$rows++) { //将当前菜单项目的内容导入数组 $menu=mysql_fetch_array($result);
        //提取菜单项目的子菜单记录集 $sql="select * frommenuwhereparent_id = $menu[id]" ; $result_sub=mysql_query($sql,$Con);
        echo " < tr > "; //如果该菜单项目有子菜单，则添加JavaScript onClick语句 if(mysql_num_rows($result_sub)>
          0) { echo " < tdwidth = '20' > < imgsrc = 'tree_expand.gif'border = '0' > < / td > "; echo " < tdclass = 'Menu'onClick = 'javascript:ShowMenu(Menu".$GLOBALS["ID"].");' > "; } else { echo " < tdwidth = '20' > < imgsrc = 'tree_collapse.gif'border = '0' > < / td > "; echo " < tdclass = 'Menu' > "; } //如果该菜单项目没有子菜单，并指定了超级连接地址，则指定为超级连接， //否则只显示菜单名称 if($menu[url]!="")
              echo " < ahref = '$menu[url]' > $menu[name] < / a > "; else echo $menu[name]; echo " < / td > < / tr > "; //如果该菜单项目有子菜单，则显示子菜单 if(mysql_num_rows($result_sub)>0) { //指定该子菜单的ID和style，以便和onClick语句相对应
            echo " < trid = Menu".$GLOBALS["ID "]++ . " style='display:none'>";
  echo "<td width='20'></td>";
  echo "<td>";
  //将级数加1
  $layer++;
  //递归调用ShowTreeMenu()函数，生成子菜单 
  ShowTreeMenu($Con,$result_sub,$layer);
  //子菜单处理完成，返回到递归的上一层，将级数减1
  $layer--;
  echo "< / td > < / tr > ";
  }
  //继续显示下一个菜单项目
  } echo " < / table > "; }
  &#63;>
      <&#63;php
      $id=1 ;
      function test() {
      global $id; unset($id);
      }
      test();
      echo " < fontclass = menu > ".($id). " < / font > ";
      // 在 PHP 4 中这里会打印出 1
      &#63;>
      <&#63;php
      $a=1 ;
      $b=2 ;
      function Sum() {
      global $a, $b; $b=$ a + $b;
      }
      Sum();
      echo " < fontclass = menu > ".$b. " < / font > ";
      &#63;>
  </body>
</html>

更多关于PHP相关内容感兴趣的读者可查看本站专题：《php面向对象程序设计入门教程》、《php字符串(string)用法总结》、《php+mysql数据库操作入门教程》及《php常见数据库操作技巧汇总》

希望本文所述对大家PHP程序设计有所帮助。

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