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devans

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  1. hi.  I've changed your code a little... [code] $sql = 'SELECT id, name FROM clubs'; $result = mysql_query($sql) or die (mysql_errno().": select ".mysql_error()."<BR>" . $sql); $numRows = mysql_num_rows($result); while ($row = mysql_fetch_array($result)) { $select1_drop .= '<OPTION VALUE="' . $row['id'] . '">' . $row['name'] . '</OPTION>'; } [/code] which should get you something in your dropdown, but you might want to also alter your second query such that it only returns appropiate items for that club... daniel
  2. Thanks nogray, you set me on the right track!  For any others having the same issue, this was the final solution that worked for me... [code] <?php define("TEST_VAL","test value"); $test = "TEST_VAL"; echo $test . "<br/>"; // echos TEST_VAL eval("\$test = $test;"); echo $test . "<br/>"; // echos test value ?> [/code]
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