如何使用PHP和mysql表创建菜单和子菜单 [英] How to create the menu and and submenu using PHP and mysql table

问题描述

我的表中有id，refid，在我的表中列出名称

我的数据如1 0 A

2 0 B

3 0 C

4 1 A1

5 1 A2

6 2 B1

7 2 B2
8 3 C3

9 3 c4

10 3 c5

11 2 C7

12 2 C8

13 11 c9

14 11 c10



我的输出应为A

--A1

- A2

B

--B1

--B2

C

--C3

--C4

--C

到此为止它工作正常但是如果我在另一个子项目中添加子项目如何检查和打印这些值请帮我提前谢谢< br $> b $ b

我尝试了什么：



i试过这个但没有得到答案请帮助我。

i have table in that id,refid,name columns in my table
my data like 1 0 A
2 0 B
3 0 C
4 1 A1
5 1 A2
6 2 B1
7 2 B2
8 3 C3
9 3 c4
10 3 c5
11 2 C7
12 2 C8
13 11 c9
14 11 c10

my output should be A
--A1
--A2
B
--B1
--B2
C
--C3
--C4
--C5
upto this it is working fine but if i add subitems inside another subitems how to check and print the values please help me thanks in advance

What I have tried:

i tried this but not getting the answer please help me.

<pre><?php

$con = new mysqli("localhost", "root", "", "test");

$topLevelItems = "SELECT id,refid,name FROM `test` WHERE refid = 0";
$res=mysqli_query($con,$topLevelItems);

foreach($res as $item) {
 echo '<a href="...">$item["name"]</a>';
 $subItems ="SELECT id, name FROM `test` WHERE refid='$item['id']'";
 $res2=mysqli_query($con,$subItems)
 foreach($res2 as $subItem) {
    echo '<a href="...">$subItem["name"]</a>';
 }
}



 ?>

推荐答案

con = new mysqli（ localhost， root， ， test）;

con = new mysqli("localhost", "root", "", "test");

topLevelItems = SELECT id，refid，name FROM`test` WHERE refid = 0;

topLevelItems = "SELECT id,refid,name FROM `test` WHERE refid = 0";

res = mysqli_query（

res=mysqli_query(

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