php – MySQL – 怎样做得更好？

$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);

必需有更好的要领来做到这一点,对吧？我不知道我必要具体声名,由于你可以看到我正在实行做什么,但有没有步伐在一个查询中完成全部这些？我必要可以或许输出勾当,失败和完成的分派,最好是在一个查询中. 最佳谜底 您可以实行这样的查询

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

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