假设我们有以下查询:
SELECT DISTINCT COUNT(`users_id`) FROM `users_table`;
该查询将返回表中的用户数。我需要将此值传递给PHP变量。我正在使用这个:
$sql_result = mysql_query($the_query_from_above) or die(mysql_error()); if($sql_result) { $nr_of_users = mysql_fetch_array($sql_result); } else { $nr_of_users = 0; }
请在您认为必要的地方更正我的代码。
最好的方法。您如何建议这样做?
像这样:
// Changed the query - there's no need for DISTINCT // and aliased the count as "num" $data = mysql_query('SELECT COUNT(`users_id`) AS num FROM `users_table`') or die(mysql_error()); // A COUNT query will always return 1 row // (unless it fails, in which case we die above) // Use fetch_assoc for a nice associative array - much easier to use $row = mysql_fetch_assoc($data); // Get the number of uses from the array // 'num' is what we aliased the column as above $numUsers = $row['num'];