syntax SQL v PgSQL

zdravim. preco ked napisem INSERT INTO ludia VALUES(20, "ja", "ty", "on") mi PgSQL vrati

SQL chyba:



ERROR:  column "ja" does not exist



Vo výraze:

INSERT INTO ludia VALUES(20, "ja", "ty", "on")

predtym som vytvoril tabulku

CREATE TABLE ludia (

id smallint NOT NULL PRIMARY KEY,

Meno char(255),

Priezvisko char(255),

Vek char(255))

este jedna otazocka - ako mam zistit najvacsie id (primary key) ? aby som potom k nemu mohol pridat 1 a zapisat. skusal som
"SELECT Max(id)+1 FROM ludia" ale vracialo to nejake divne hodnoty. Nakoniec som to vyriesil tak :

  $pg_connection = Pg_Connect("host=$pg_host dbname=$pg_dbname user=$pg_user password=$pg_pass");

  $pg_result = Pg_Exec($pg_connection, "SELECT * FROM ludia");

    

  if ($_GET[action] == "add")

  {   

      $pg_biggest_id = 0;

      

      for ($i = 0; $i < Pg_NumRows($pg_result); $i++)

      {

          $pg_entry = Pg_Fetch_Array($pg_result, $i);

          if ($pg_entry["id"] > $pg_biggest_id)

          {

            $pg_biggest_id = $pg_entry["id"];

          }

      }  

      

      $pg_id = $pg_biggest_id + 1;    

      Pg_Exec($pg_connection, "INSERT INTO ludia VALUES ($pg_id, '" . $_GET["meno"] . "', '$_GET[priezvisko]', '$_GET[vek]')");

      Pg_Close($pg_connection);

      Header("Location: ./");

ale to mi pride trochu tazkopadne a pomale. neda sa to vyriesit nejakym SQL dotazom alebo nejako rychlejsie ? dik