How to make my PHP output two values from database - php

Question

Ive included a block of code below detailling how im placing sections of data from a table in my database in their own divs. However, im new to PHP and cant find out how to also output the "cup_id" from the database into their respective divs here: echo $cup["cup_name"] . "<br />"; Thanks for all the help in advance!
So in short how do i get this to work: echo $cup["cup_id", "cup_name"] . "<br />";
<?php
require_once("action/dbcon.php"); // Get the database connection
$get_cup = "SELECT * FROM cups";
$show_cup = mysqli_query($conn, $get_cup);
if (!$show_cup) {
echo "Could not load cup. " . "(" . mysqli_error($conn) . ")";
}
while ($cup = mysqli_fetch_assoc($show_cup)) {
echo '<div class="cup-info">';
echo $cup["cup_name"] . "<br />";
echo '</div>';
}
?>

Answer 1

Do you want to concatenate strings? use the dot operator:
echo $cup["cup_id"] . $cup["cup_name"];
And if you want to print it in another div, make this:
echo '<div class="cup-info">';
echo $cup["cup_id"] . "<br />";
echo '</div>';
echo '<div class="cup-info">';
echo $cup["cup_name"] . "<br />";
echo '</div>';

Answer 2

if you want the name and id to be in the same raw then you can follow this method
<?php
require_once("action/dbcon.php"); // Get the database connection
$get_cup = "SELECT * FROM cups";
$show_cup = mysqli_query($conn, $get_cup);
if (!$show_cup) {
echo "Could not load cup. " . "(" . mysqli_error($conn) . ")";
}
while ($cup = mysqli_fetch_assoc($show_cup)) {
echo '<div class="cup-info">';
echo $cup["cup_id"] . $cup["cup_name"]."<br />";
echo '</div>';
}
?>
if you want the name and id to be in different raw then you can follow this method
<?php
require_once("action/dbcon.php"); // Get the database connection
$get_cup = "SELECT * FROM cups";
$show_cup = mysqli_query($conn, $get_cup);
if (!$show_cup) {
echo "Could not load cup. " . "(" . mysqli_error($conn) . ")";
}
while ($cup = mysqli_fetch_assoc($show_cup)) {
echo '<div class="cup-info">';
echo $cup["cup_id"]."<br />";
echo '</div>';
echo '<div class="cup-info">';
echo $cup["cup_name"]."<br />";
echo '</div>';
}
?>