我正在尝试制作一个主菜单，它有 2 级深度(子菜单和子子菜单)。

我以前用 GROUP_CONCAT 做过这个，但那只有一层深。事实上，这是 2 层深，这真的让我陷入了最佳实践循环。

CREATE TABLE `database`.`main_menu` ( `main_menu_id` INT(11) NOT NULL AUTO_INCREMENT , `main_menu_name` VARCHAR(255) NOT NULL , PRIMARY KEY (`main_menu_id`)) ENGINE = InnoDB;

CREATE TABLE `database`.`sub_menu` ( `main_menu_id` INT(11) NOT NULL , `sub_menu_id` INT(11) NOT NULL AUTO_INCREMENT , `sub_menu_name` VARCHAR(255) NOT NULL , PRIMARY KEY (`sub_menu_id`)) ENGINE = InnoDB;

CREATE TABLE `database`.`sub_sub_menu` ( `main_menu_id` INT(11) NOT NULL , `sub_menu_id` INT(11) NOT NULL , `sub_sub_menu_id` INT(11) NOT NULL AUTO_INCREMENT , `sub_sub_menu_name` VARCHAR(255) NOT NULL , PRIMARY KEY (`sub_sub_menu_id`)) ENGINE = InnoDB;


INSERT INTO `main_menu` (`main_menu_id`, `main_menu_name`) VALUES (NULL, 'Food'), (NULL, 'Treats')

INSERT INTO `sub_menu` (`main_menu_id`, `sub_menu_id`, `sub_menu_name`) VALUES ('1', NULL, 'Duck'), ('1', NULL, 'Chicken')
INSERT INTO `sub_menu` (`main_menu_id`, `sub_menu_id`, `sub_menu_name`) VALUES ('2', NULL, 'Bacon Bits'), ('2', NULL, 'Dental')

INSERT INTO `sub_sub_menu` (`main_menu_id`, `sub_menu_id`, `sub_sub_menu_id`, `sub_sub_menu_name`) VALUES ('1', '1', NULL, 'In Gravy'), ('1', '2', NULL, 'in Soup')

INSERT INTO `sub_sub_menu` (`main_menu_id`, `sub_menu_id`, `sub_sub_menu_id`, `sub_sub_menu_name`) VALUES ('2', '3', NULL, 'Sticks'), ('2', '4', NULL, 'Chunks')

MySQL 查询:

SELECT ssm.sub_sub_menu_name  as    sub_sub_menu_name ,
sm.sub_menu_name  as    sub_menu_name ,
mm.main_menu_name  as   main_menu_name ,

ssm.sub_sub_menu_id  as     sub_sub_menu_id ,
sm.sub_menu_id  as      sub_menu_id ,
mm.main_menu_id  as     main_menu_id 

FROM        main_sub_sub_menu  as ssm
LEFT JOIN main_menu as mm
    ON mm.main_menu_id  = ssm.main_menu_id 
LEFT JOIN sub_menu as sm
    ON sm.sub_menu_id   = ssm.sub_menu_id  

ORDER BY ssm.sub_sub_menu_id, sm.sub_menu_id, ssm.main_menu_id

这里介绍PHP如何对信息进行分类:

<?php $cat = $getData->get_menu_categories(); //SQL Query above with PDO
$i = 1;
$j = 1;
for($a=0; $a<= count($cat); $a++){
        if($i == $cat[$a]['main_menu_id']){
            for($j=1; $j < count($cat[$a]['sub_menu_id']); $j++){ // Was getting stuck here because I can't count it unless it is an array
                 //Would show each submenu name, then I would make another for loop for the sub sub menu 
            }
        }else{
            $i++;
        }

输出为

主菜单

子菜单

子子菜单

我正在制作列表(一个 <ul><li> 样式的 html 菜单列表)，基本上是一棵树。

最佳答案

您将要遍历查询结果并打印类似这样的内容。我使用破折号来缩进，但您可以使用 html 来更干净地缩进。我为任何想提供帮助的人制作了一个 db fiddle。 https://www.db-fiddle.com/f/sbTyQbNemcEM5vX3xwuUaJ/0#&togetherjs=wkJd3lwafs

<?php 
$cat = $getData->get_menu_categories(); 
$lastmain = '';
$lastsub = '';
$lastsubsub = '';

foreach ($cat as $line) {
   $main = $line['main_menu_name'];
   $sub = $line['sub_menu_name'];
   $subsub = $line['sub_sub_menu_name'];

   if ($lastmain <> $main) {
       print "$main\n";
       $lastmain = $main;
   }

   if ($lastsub <> $sub) {
       print "---$sub\n";
       $lastsub = $sub;
   }

   if ($lastsubsub <> $subsub) {
       print "------$subsub\n";
       $lastsubsub = $subsub;
   }
}

关于php - 如何使用 MySQL PHP 以正确的方式执行多级菜单？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/50959418/