php mysql计入PHP变量
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mysql count into PHP variable
提问by Duncan Benoit
Let say that we have the following query:
假设我们有以下查询：
SELECT DISTINCT COUNT(`users_id`) FROM `users_table`;
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:
此查询将从表中返回用户数。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;
}
please correct my code where you think is necessary.
请在您认为必要的地方更正我的代码。
Which is the best approach. How do you recommend to do this ?
这是最好的方法。你建议如何做到这一点？
回答by Greg
Like 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'];
回答by millimoose
Also, an alternative using mysqli, which you should be using anyway for parameter interpolation:
此外，使用 mysqli 的替代方法，无论如何您都应该使用它进行参数插值：
$statement = $connection->prepare($the_query_from_above);
$statement->execute();
$statement->bind_result($nr_of_users);
$statement->fetch();
