我希望mysql select语句按$combine_count排序，但$combine_count是根据该语句计算的。你有办法实现我的目标吗？学徒。
$stm =$db->prepare("SELECT id , COUNT(user_id) as 'count'FROM sign WHERE term IN (:term_0,:term_10)");
$term_0="$term[0]";
$term_1="$term[1]";
$stm->bindParam(":term_0", $term_0);
$stm->bindParam(":term_1", $term_1);
$stm->execute();
$rows = $stm->fetchALL(PDO::FETCH_ASSOC);
foreach ($rows as $rows) {
$count=$rows['count'];
$count_percentage=round(($count/$count_user_diff)*100);
$count_key_match=round(($count/$term_match_number)*100);
$combine_count=round(($count_percentage+$count_key_match)/2);
echo $combine_count ;
}//foreach您的代码没有显示$term_match_number和$count_user_diff来自哪里，但肯定不是来自SQL。所以，不，在这种特殊情况下你不能这样做。
最好的方法不是直接回显，而是在PHP端对数组进行排序。
$array = array();
foreach ($rows as $rows) {
$count=$rows['count'];
$count_percentage=round(($count/$count_user_diff)*100);
$count_key_match=round(($count/$term_match_number)*100);
$combine_count=round(($count_percentage+$count_key_match)/2);
$array[] = $combine_count ;
}
sort($array); // sorts ascending (low to high)
rsort($array); // sorts descending (high to low)
我不确定MySQL是否能处理它，但值得一试：
SELECT s.id, s.count
FROM (SELECT id, COUNT(user_id) as 'count' FROM sign WHERE term IN (:term_0,:term_10) GROUP BY id) AS s
ORDER BY s.count