如何从刚提交的表单中回显名称，价格和ID？

Question

I have an issue with MySQLi and PHP .

I created a form, and once I type the desired values in and hit submit, the values are right away sent to the database. Nothing wrong with this.

What I want to happen is that: after hitting the submit button, PHP shall echo the result of the just-submitted entry. That is to say:

   `INSERT INTO table VALUES (x, x, y) -> SELECT x, x, y FROM table ORDER BY id DESC LIMIT 1`

I have tried many methods to do this, but all of them either echo the previous entry (the one before the one just submitted) or plainly don't work.

I have tried mysqli_insert_id($conn) but this returns nothing.

This is where my code rests at at the moment:

    $conn = mysqli_connect($server, $user, $pw, $BD);
    if (!$conn) {
        die ('<span style="color: #FF0000;">"connection failed: "</span>' .  mysqli_connect_error());
    }

    $nome = $_POST['nome'];
    $preco = $_POST['preco'];

    $query = "INSERT INTO produtos(nome, preco) VALUES ('$nome', '$preco')";
    $result = mysqli_insert_id($conn);
    var_dump ($result);

    if (mysqli_query($conn, $query)){
        echo '<br>'."Succeeded!";
    } else {
        echo '<br>'."ERROR!" .'<br>'. $query ."<br>". mysqli_error($conn) .'<br><br>'. '<span style="color: #FF0000;">You have to fill all the fields.</span>';
    }

    mysqli_close($conn); 

to note, if of any help, var_dump outputs int(0) at the moment. Thanks in advance. I've been struggling like mad with this.

Answer 1

You can't get mysqli_insert_id without executing the query. Better use prepare statement to prevent from sql injection

 $stmt = $conn->prepare("INSERT INTO produtos(nome, preco) VALUES (?,?)");
    $stmt->bind_param('ss', $nome, $preco);
    $stmt->execute();// execute query
    $conn->insert_id;// get last insert id

Answer 2

Please see that you haven't even executed your query. On a side note, you should be aware of SQL injections and follow the below pattern:

$nome  = mysqli_real_escape_string($conn, $_POST['nome']);
$preco = mysqli_real_escape_string($conn, $_POST['preco']);

$sql    = "INSERT INTO produtos (nome, preco) VALUES ('".$nome."', '".$preco."')";
$query  = mysqli_query($conn, $sql) or die(mysqli_error($conn));
$result = mysqli_insert_id($conn);

echo $result; // Check your result.

Answer 3

Use this:

$query = "INSERT INTO produtos(nome, preco) VALUES ('$nome', '$preco')";
$res=mysqli_query($conn,$query);
$result = mysqli_insert_id($conn);
var_dump ($result);`