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I’m trying to implement a like button on a system I’m creating for testing. It already changes the value of the likes in the database, but I have to put the value in an input field that I put. So every number I put in the input and click on like it does the update in my database. I would like to know how I do not need to enter the value in the input and each time I click it add another number in the value that already has in the database. Here’s the code:
Index.php
<?php
require_once "conecta.php";
$resultado = mysqli_query($conexao, "select * from curtida");
$curtida = mysqli_fetch_assoc($resultado);
?>
<!DOCTYPE html>
<html lang="pt-br">
<head>
<meta charset="UTF-8" />
<title>Curtir</title>
<link rel="stylesheet" href="css/claudina.css" />
</head>
<body>
<section class="recipiente margem-topo-50">
<form action="envia_curtir.php" method="post">
<input type="number" name="curtida" />
<button class="icones icone-polegares-para-cima"></button>
<?=$curtida['curtidas'];?>
</form>
</section>
</body>
</html>
php.
<?php
require_once "conecta.php";
require_once "funcoes.php";
$curtida = $_POST['curtida'];
curtir($conexao, $curtida);
header("Location: index.php");
?>
funcoes.php
<?php
function curtir($conexao, $curtida) {
$query = "update curtida set curtidas= $curtida";
return mysqli_query($conexao, $query);
}
Thank you very much!
– Francis Vagner da Luz