1
Good morning, I’m starting in php and Mysql. I’m trying to develop a system for registering a vehicle (with photo), and studying a little, I got into php language together with MYSQL. It turned out that I generated my scripts, but no error message appears and does not save the data in my database. Can anyone help me?
index.php file
<?php
include("classe/conexao.php");
$marca = isset( $_POST ['marca']) ? $_POST['marca']:null;
$modelo = isset( $_POST ['modelo']) ? $_POST['modelo']:null;
$ano = isset( $_POST ['ano']) ? $_POST['ano']:null;
$msg = false;
if(isset($_FILES['arquivo'])){
$extensao = strtolower(substr($_FILES['arquivo']['name'], -4)); //pega a extensao do arquivo
$novo_nome = md5(time()) . $extensao; //define o nome do arquivo
$diretorio = "upload/"; //define o diretorio para onde enviaremos o arquivo
move_uploaded_file($_FILES['arquivo']['tmp_name'], $diretorio.$novo_nome); //efetua o upload
$sql_code = "INSERT INTO veiculos (id, marca, modelo, ano, arquivo, data) VALUES(null, '$marca', '$modelo', '$ano', '$novo_nome', NOW())";
if($mysqli->query($sql_code))
$msg = "Arquivo enviado com sucesso!";
else
$msg = "Falha ao enviar arquivo.";
}
?>
HTML
<html>
<title>Cadastro de veículos</title>
<body>
<?php if(isset($msg) && $msg != false) echo "<p> $msg </p>"; ?>
<form action="index.php" method="POST" enctype="multipart/form-data">
Marca: <input name="marca" type="text">
Modelo: <input name="modelo" type="text">
Ano: <input name="ano" type="text">
Arquivo: <input type="file" required name="arquivo">
<input type="submit" value="Cadastrar">
</form>
</body>
</html>
File class/connected.php
<?php
$host = "localhost";
$usuario = "root";
$senha = "";
$bd = "sitemoraes";
$mysqli = new mysqli($host, $usuario, $senha, $bd);
if($mysqli->connect_errno)
echo "Falha na conexão: (".$mysqli->connect_errno.") ".$mysqli->connect_error;
?>
Print of the database:
-EDIT- After the changes that Jackson and anonymity recommended, the script still doesn’t work. Does not save anything in the database and returns the "Failed to send file."
Hello Guga, first thing, leave it this way:
$sql_code = "INSERT INTO veiculos (marca, modelo, ano, arquivo, data) VALUES('$marca', '$modelo', '$ano', '$novo_nome', NOW())";, like theIDis an auto increment, you should not send it in the code.– Jakson Fischer
Like the field
idis a primary key it can not beNULL. In hisINSERTchangeNULLforDEFAULTor in the manner indicated by Jackson.– anonimo
Thanks for your help with the code.. I did it Jackson’s way and yours anonymously, but in both cases now appears the message "Failed to send file." and does not send anything to the database.
– Guga
Guga, give a
var_dump($_FILES['arquivo'])and see if this guy is returning anything... I suspect that the fact that no message appears may be the fact that he is not entering theif()...– Jakson Fischer
Jacson, var dump is giving: 'name' => string 'Moraesaltini.png' (length=16) 'type' => string 'image/png' (length=9) 'tmp_name' => string 'C: wamp64 tmp phpEFD.tmp' (length=24) 'error' => int 0 'size' => int 607658 'size' => int 0
– Guga
Check the values of
$marca,$modelo,$ano,$novo_nomebefore entering into the database. All your columns are NOT NULL. Modify your script to use Prepared statements because the way you’re doing your code is susceptible to attack.– Augusto Vasques