Tek bir formda iki tabloları içine yerleştirin. Birinci insert ikinci bu hatayı ne oluyor Duplicate entry '0' for key 1 herhangi bir fikir üretir ince gitmek?

$connection=mysql_connect ("localhost", "foo", "bar") or die ("I cannot connect to the database.");
$db=mysql_select_db ("database", $connection) or die (mysql_error());
$query = "INSERT INTO worklog (id, newtime, datetime, clientname, clientcode, startmo, startday, startyr, endmo, endday, endyr, duemo, dueday, dueyr, market, job, allTypes, spec, status, designer, dsgnemail, adrep, ademail, frame1, frame2, frame3, rush) VALUES ('$id', $newtime, now(), '$clientname', '$clientcode', '$startmo', '$startday', '$startyr', '$endmo', '$endday', '$endyr', '$duemo', '$dueday', '$dueyr', '$market', '$job', '$allTypes', '$spec', '$status', '$designer', '$dsgnemail', '$adrep', '$ademail', '$frame1', '$frame2', '$frame3', '$rush')";
$sql_result = mysql_query($query, $connection) or die (mysql_error());

$worklog_id=mysql_insert_id($connection); 

$connection2=mysql_connect ("localhost", "foo", "bar") or die ("I cannot connect to the database.");
$db2=mysql_select_db ("database", $connection2) or die (mysql_error());
$query2 = "INSERT INTO worklognotes (worklog_id, spec) VALUES ('$worklog_id', '$spec')";

$sql_result = mysql_query($query2, $connection2) or die (mysql_error());