php – MySQL – 如何做得更好？

$activeQuery = mysql_query("SELECT count(`status`) AS `active` FROM `assignments` WHERE `user` = $user_id AND `status` = 0");
$active = mysql_fetch_assoc($activeQuery);

$failedQuery = mysql_query("SELECT count(`status`) AS `failed` FROM `assignments` WHERE `user` = $user_id AND `status` = 1");
$failed = mysql_fetch_assoc($failedQuery);

$completedQuery = mysql_query("SELECT count(`status`) AS `completed` FROM `assignments` WHERE `user` = $user_id AND `status` = 2");
$completed = mysql_fetch_assoc($completedQuery);

必须有更好的方法来做到这一点,对吧？我不知道我需要详细说明,因为你可以看到我正在尝试做什么,但有没有办法在一个查询中完成所有这些？我需要能够输出活动,失败和完成的分配,最好是在一个查询中. 最佳答案 您可以尝试这样的查询

SELECT Status,COUNT(*) StatusCount 
FROM assignments
WHERE Status IN (0,1,2)
AND User = $user_id 
GROUP BY Status

（编辑：源码网）

【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容!