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POGenerating SQL with sprintf(): "Too few arguments"
 

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  1. POGenerating SQL with sprintf(): "Too few arguments"
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    <p>I keep getting the following error message when trying to UPDATE one of my database tables:</p> <pre><code>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).' </code></pre> <p><strong>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</strong></p> <p>Right above the Warning is an echo of the two variables used in the <code>sprintf()</code>.</p> <p>Here is the section from my php file:</p> <pre><code> $idrr = GetSQLValueString($_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 . "&lt;br/&gt;"); 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()) </code></pre> <p>I can't figure out why this particular UPDATE wont work. Can anyone see what I'm doing wrong?</p>
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