5
Initially, I’m new to PHP and SQL, so take it easy if you have shaving in the code. I’m just testing the PHP and Mysql integration, I took an example table from a book and made a quick html page to pass the parameters.
HTML: (index.html)
<html>
<body>
<form action="add.php">
<p>Nome<input name="nome" type="text"/>
<p>Principal Ingrediente<input name="principal" type="text"/>
<p>Quantidade<input name="quant1" type="number"/>
<p>Ingrediente Secundário<input name="segundo" type="text"/>
<p>Quantidade<input name="quant2" type="number"/>
<p>Instrução<input name="instrucao" type="text"/>
<p><input type="submit"/>
</form>
</body>
</html>
PHP: (add.php)
<html>
<body>
<?php
$nome = $_GET['nome'];
$principal = $_GET['principal'];
$quant1 = $_GET['quant1'];
$segundo = $_GET['segundo'];
$quant2 = $_GET['quant2'];
$instrucao = $_GET['instrucao'];
$db = new mysqli('localhost', 'root', 'Another2/', 'drinks');
$sql = "INSERT INTO drinkfaceis(nome, principal, quant1, segundo, quant2, instrucoes)
VALUES(?, ?, ?, ?, ?, ?);";
$stmt = $db->prepare($sql);
if(!$stmt){
echo 'erro na consulta: '. $db->errno .' - '. $db->error;
}
echo $nome;
echo $principal;
echo $quant1;
echo $segundo;
echo $quant2;
echo $instrucao;
$stmt->bind_param('ssisib', $nome, $principal, $quant1, $segundo, $quant2, $instrucao);
$stmt->execute();
?>
</body>
</html>
SQL script:
CREATE DATABASE drinks;
USE drinks;
CREATE TABLE drinkfaceis
(
nome VARCHAR(20),
principal VARCHAR(20),
quant1 INT(10),
segundo VARCHAR(20),
quant2 INT(10),
instrucoes BLOB
);
When testing does not give any error, he writes the value of the variables (which I put only to test), but when I go to phpMyAdmin and look at the table the data was not inserted.
Let’s go continue this discussion in chat.
– rray