up vote 1 down vote favorite

So basicly this is my code:

echo "<SELECT>";
foreach($arr_res as $op) {
   $q3 = mysql_query(mysql_fetch_array("SELECT SL_TIME FROM SLOTS WHERE SL_ID='$op'"));
   echo "<OPTION value=".$op.">".$q3."</OPTION>";
}
echo "</SELECT>";

$arr_res is the resulting array of an array_diff. It has only numerical values which are the SL_ID from my timeslots.. I want to show SL_TIME but the result I get is the $q3 producing empty result.. Why isn't this working?

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1 Answer

up vote 1 down vote accepted

You don't query mysql_fetch_array. Mysql_query gives an result on wich you use mysql_fetch_array. So the correct code would be:

echo "<SELECT>";
foreach($arr_res as $op){
$result = mysql_query("SELECT SL_TIME FROM SLOTS WHERE SL_ID='$op'");
$q3=mysql_fetch_array($result);
echo "<OPTION value=".$op.">".$q3."</OPTION>";
}
echo "</SELECT>";

However, $q3 will be an array, which is probably not what you want. In addition it would be a lot better if use an abstraction such as PDO (http://www.php.net/manual/en/class.pdo.php). This will maximize performance and increase the security.

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Allthough copy and pasting your solution didn't directly work with a few minor adaptations it did fix the problem! Thank you so much! :) (too low to upvote sorry) – Jake Rowsell Oct 7 '12 at 15:54

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