Question

I keep getting the following error message when trying to UPDATE one of my database tables:

idrr = 167

age_notes = 'Enumerate, Louisvilleluminary, Nacho Friend, Bulls and Bears, Bricklayer, Activity Report, Interactif, Soundman).\r\n

\r\nBashford Manor S (gr-III, 6f, defeating Flatter Than Me, Brassy Boy, Grand Slam Andre, Soundman, Westrock Gold, Vito Filitto, Even Wilder).\r\n

\r\nA maiden special weight race at Churchill Downs (5f, by 2 3/4, defeating Thiskyhasnolimit, Red Rally, Dublin, Criminal Offense, Congar, Horst, Prospective Union, Victorystart, Harley\'s Heat).'

Warning: sprintf() [function.sprintf]: Too few arguments in /data/19/1/60/63/1875552/user/2038041/htdocs/vinery/Admin/upload_stallion.php on line 305
Query was empty

Right above the Warning is an echo of the two variables used in the sprintf().

Here is the section from my php file:

        $idrr = GetSQLValueString($_POST['id1'], "int");
        $associated_horse = GetSQLValueString($_POST['associated_horse1'], "int");
        $year = GetSQLValueString($_POST['year1'], "text");
        $age = GetSQLValueString($_POST['age1'], "int");
        $starts = GetSQLValueString($_POST['starts1'], "int");
        $first = GetSQLValueString($_POST['first1'], "int");
        $first_sw = GetSQLValueString($_POST['first_sw1'], "int");
        $second = GetSQLValueString($_POST['second1'], "int");
        $second_sp = GetSQLValueString($_POST['second_sp1'], "int");
        $third = GetSQLValueString($_POST['third1'], "int");
        $third_sp = GetSQLValueString($_POST['third_sp1'], "int");
        $age_notes = GetSQLValueString($_POST['age_notes1'], "text");
        $age_text = GetSQLValueString($_POST['age_text1'], "text");
        $earned = GetSQLValueString($_POST['earned1'], "text");
        echo ("idrr = " . $idrr . "<br/>");
        echo ("age_notes = " . $age_notes);




        $insertSQL = sprintf("UPDATE race_records SET age_notes = $age_notes WHERE rr_id = %s", GetSQLValueString($idrr, "int"));

        mysql_select_db($database_XXXXXX, XXXXXX);
        $Result = mysql_query($insertSQL, $HDAdave) or die(mysql_error())

I can’t figure out why this particular UPDATE wont work.
Can anyone see what I’m doing wrong?

Answer 1

You might want to put backticks or single quotes around $age_notes and %s

$insertSQL = sprintf("UPDATE race_records SET age_notes = '$age_notes' WHERE rr_id = '%s'", GetSQLValueString($idrr, "int"));