PHP+JS three-level menu linkage menu implementation method
The example in this article describes the implementation method of PHP+JS three-level menu linkage menu. Share it with everyone for your reference, the details are as follows:
<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 "]++ . ">";
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>
Readers who are interested in more PHP-related content can check out the special topics of this site: "Introduction Tutorial on PHP Object-Oriented Programming", "Summary of PHP String Usage" ", "Introduction Tutorial on PHP+MySQL Database Operation" and "Summary of Common PHP Database Operation Skills"
I hope this article will be helpful to everyone in PHP programming.
The above introduces the implementation method of PHP+JS three-level menu linkage menu, including the relevant content. I hope it will be helpful to friends who are interested in PHP tutorials.
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