php function does not insert records in the database

Question

I have a php function that inserts two types of record in a Mysql table, according to the user’s choice in a combobox. The first combo option enters the records normally, but when I choose the second one it does not insert and does not show any error. In the Inspect element I see that all the form parameters are being passed correctly. Can someone help me ? Thank you

Follows the code:

if($funcao == 'email') {

$sql = "select * from mensageminterna where mensagem like '%<a h%';";
$rst = my_query($connR, $sql);

    if($option == 'corretor'){

        $sql = "insert into mensageminterna (codempresa, codusuarioremetente, codusuario, mensagem, icone, datacriacao)
                select u.codempresa, u1.codusuario, u.codusuario, '$texto', 'cool', now()  from usuario u
                inner join usuario u1 on u.codempresa=u1.codempresa and u1.email like 'admin@%' and u1.indadministrador=1
                where u.codsituacaousuario=1;";
        $rst = my_execute($connW, $sql);

} elseif($option == 'gestor'){

        $sql = "insert into mensageminterna (codempresa, codusuarioremetente, codusuario, mensagem, icone, datacriacao)
            select u.codempresa, u1.nome remetente, u.nome destinatario, u.codtipousuario, '$texto', 'cool', now()  from usuario u
            inner join usuario u1 on u.codempresa=u1.codempresa and u1.email like 'admin@%' and u1.indadministrador=1
            inner join tipousuario tu on tu.codtipousuario = u.codtipousuario
            where u.codsituacaousuario=1 and tu.nome like '%geren%' or tu.nome like '%diret%' or tu.nome like '%coord%' or tu.nome like '%adm%'  or tu.nome like '%super%';";
        $rst = my_execute($connW, $sql);

} else {
        echo "Não enviado";
    }

exit;
}

<script language="JavaScript">
    $(document).on('click', '#btnEnviar', function(event) {
        event.preventDefault();
        $("#funcao").val("email");
        var self = $(this);
        $.ajax({
            url: "/email-broadcast.php",
            type: "POST",
            timeout:default_timeout,
            data: $('#formemail').serialize(),
            beforeSend: function(){
                self.attr('disabled', 'true');
            },
            success: function() {
                alert("Enviado com sucesso !");
            },
            error: function(jqXHR, textStatus){
                console.log(textStatus, jqXHR);
            },
            complete: function(){
                self.removeAttr('disabled');
            }
        });
    });

</script>


<form method="post" name="formemail" action="/email-broadcast.php" id="formemail">
        <input type="hidden" name="funcao" id="funcao" value="email"/>


            <label style="margin-left: 7px">Destinatário:</label>
            <select class="form-control" name="destinatario" id="destinatario" style="width: 250px;">
                <option value="corretor" id="corretor" name="corretor">Corretores</option>
                <option value="gestor" id="gestor" name="gestor">Gestores</option>
            </select>



        <textarea cols="86" rows="15" id="scriptenvio" name="scriptenvio" style=" margin-top: 10px;">
            Blá blá blá
        </textarea><br>


    <button type="button" class="btn btn-default" id="btnEnviar" name="btnEnviar">Enviar</button>
    </form>

---

I took the select of the second query (the one I wasn’t inserting) and it returned 464 records.

Answer 1

Solved!

Actually, I was trying to insert varchar data into int fields in the second query. This was the correct one:

} elseif($option == 'gestor'){

    $sql = "insert into mensageminterna (codempresa, codusuarioremetente, codusuario, mensagem, icone, datacriacao)
        select u.codempresa, u1.codusuario, u.codusuario, '$texto', 'cool', now()  from usuario u
        inner join usuario u1 on u.codempresa=u1.codempresa and u1.email like 'admin@%' and u1.indadministrador=1
        inner join tipousuario tu on tu.codtipousuario = u.codtipousuario
        where u.codsituacaousuario=1 and tu.nome like '%geren%' or tu.nome like '%diret%' or tu.nome like '%coord%' or tu.nome like '%adm%'  or tu.nome like '%super%';";
    $rst = my_execute($connW, $sql);

}

Thank you all so much for your help !