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:
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<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>
<?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 > " ; }
?>
<?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
?>
<?php
$a =1 ;
$b =2 ;
function Sum() {
global $a , $b ; $b =$ a + $b ;
}
Sum();
echo " < fontclass = menu > " . $b . " < / font > " ;
?>
</body>
</html>
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希望本文所述对大家PHP程序设计有所帮助。