How to create the menu and and submenu using PHP and mysql table

Question

0.00/5 (No votes)

See more:

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.

PHP

<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>';
 }
}



 ?>

Posted 7-Nov-17 22:55pm

Member 13153537

Updated 8-Nov-17 1:26am

v2

Add a Solution

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Member 13153537 · Answer 1 · 2017-11-07T23:15:00

i have got solution thanks.

$resultMainMenu = mysql_query("SELECT * FROM test WHERE refid=0 ORDER BY name ASC") or die(mysql_error());
while($row = mysql_fetch_array($resultMainMenu)){
    echo $row['name'] . '<br />'; // echo main menu
    $resultSubmenu = mysql_query("SELECT * FROM test WHERE refid=" . $row['id'] ." ORDER BY name ASC") or die(mysql_error());
    if(mysql_num_rows($resultSubmenu) >= 1){
        while($rowSub = mysql_fetch_array($resultSubmenu)){
            echo ' -- ' . $rowSub['name'] . '<br />'; // echo sub menu
        }
    }
}