假设我们有以下查询:
SELECT DISTINCT COUNT(`users_id`) FROM `users_table`;
此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:
this query will return the number of the users from a table. I need to pass this value to a PHP variable. I'm using this:
$sql_result = mysql_query($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.
please correct my code where you think is necessary.
这是最好的方法.您建议如何执行此操作?
Which is the best approach. How do you recommend to do this ?
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'];
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