Get id of a previous recording [duplicate]

1

Good afternoon. I have a simple form of a calendar that writes the location data, event name and etc, then I have another one that writes the participating users, I would like to know how to get the ID of this previous form (event recording). To write to the participants form, and then you can relate.

Write First Form

Case 'salvar';

    $data = '';
    $data = implode('-', array_reverse(explode('/', $_POST['data_agenda'])));

                    if (app::$key) {
                        $sql = "update agenda set compromisso_agenda='" . $_POST["compromisso_agenda"] . "'
                                                ,data_agenda='$data',hora_agenda='" . $_POST["hora_agenda"] . "',pessoa_agenda='" . $_POST["pessoa_agenda"] . "',local_agenda='" . $_POST["local_agenda"] . "' where id_agenda=" . app::$key;
                    } else {
                        $sql = "insert into agenda (compromisso_agenda,data_agenda,hora_agenda,pessoa_agenda,local_agenda)
                                       values ( '" . $_POST["compromisso_agenda"] . "','$data','" . $_POST["hora_agenda"] . "','" . $_POST["pessoa_agenda"] . "','" . $_POST["local_agenda"] . "')";
                    }

                    $dados = connection::exec($sql);

                    header('Location: ' . URL . 'agenda/pessoas');
                    break;

Recording according to the form

case 'salvar-pessoa':

$agendap_id = fetchColumn(0);

                if (app::$key) {
                    $sql = "update agendap set agendap_id='$agendap_id',agendap_pessoa='" . $_POST["agendap_pessoa"] . "' where id_agenda=" . app::$key;
                } else {
                    $sql = "insert into agendap (agendap_id,agendap_pessoa)
                                   values ( '$agendap_id','" . $_POST["agendap_pessoa"] . "')";
                }

                $dados = connection::exec($sql);

                header('Location: ' . URL . 'agenda/exibir');
                break;
    
asked by anonymous 04.03.2016 / 20:27

1 answer

-2

If you are using MySQL, you can use the mysql_insert_id () function. link

    
04.03.2016 / 20:38