Problem with call in Database

Question

Next person I have several calls that update a database as the user fills the fields but specifically in the following error call in the database and not saved for anything, and all use the same calls just changing the data. Can someone fill up some error in this call that I’m not seeing?

if($op == 4) { 

$rg = $_POST["RG"];
$orgao = $_POST["orgao_rg"];
$cpf = $_POST["CPF"];
$fotofrente = $_POST["foto-id-frente"];
$fotoverso = $_POST["foto-id-verso"];

if($_POST["foto-id-frente"] && $_POST["foto-id-verso"]){
    usleep(enviar());
}

if(isset($_FILES['foto-id-frente']) && isset($_FILES['foto-id-verso'])){

    $extensao = strtolower(substr($_FILES['foto-id-frente']['name'],-4));
    $novonome = md5(time()).$extensao;
    $extensaoverso = strtolower(substr($_FILES['foto-id-verso']['name'],-4));
    $novonomeverso = md5(time()).'verso'.$extensaoverso;
    $diretorio = "documentos/";

    move_uploaded_file($_FILES['foto-id-frente']['tmp_name'], $diretorio.$novonome);
    move_uploaded_file($_FILES['foto-id-verso']['tmp_name'], $diretorio.$novonomeverso);

    $inserir = "UPDATE usuarios ";

    $inserir .= "SET ";

    $inserir .= "identidade = '$rg', orgao-id = '$orgao', foto-id-frente = '$novonome', foto-id-verso = '$novonomeverso', CPF = '$cpf' where id = '$id'";



    $operacao_inserir = mysqli_query($conexao,$inserir);

    if(!$operacao_inserir) {

        echo "<script language='javascript' type='text/javascript'>alert('Opss, erro no cadastro do banco de dados, estamos tentando corrigir!');</script>";

            die();

    }


}

}

Answer 1

I was able to solve the problem by putting the query in the following format:

    $inserir = "UPDATE `usuarios` ";

    $inserir .= " SET ";

    $inserir .= " `identidade` = '$rg', `orgao-id` = '$orgao', `foto-id-frente` = '$novonome', `foto-id-verso` = '$novonomeverso', `CPF` = '$cpf' WHERE `usuarios`.`id` = $id ";

    $operacao_inserir = mysqli_query($conexao,$inserir);

Answer 2

in the following line:

$operacao_inserir = mysqli_query($conexao,$inserir);

add the mysqli_error to know what’s wrong so:

$operacao_inserir = mysqli_query($conexao,$inserir) or die(mysqli_error());

But one thing, I saw that you have a bank field separated by a hyphen, in this case orgao-id, when using a field like this, you need to put between single quotes or crase: 'orgao-id', or instead of hyphenating, use underline.