


PHP JS three-level menu linkage menu implementation method, _PHP tutorial
The implementation method of PHP JS three-level menu linkage menu,
The example of 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 "]++ . " 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>
Readers who are interested in more PHP related content can check out the special topics of this site: "php object-oriented programming introductory tutorial", "php string (string) usage summary", "php mysql database operation introductory tutorial" and "php mysql database operation introductory tutorial" Summary of common database operation skills in PHP》
I hope this article will be helpful to everyone in PHP programming.
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