$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);
必須有更好的方法來做到這一點,對吧?我不知道我有多少需要詳細說明,因爲您可以看到我想要做的事情,但有什麼辦法可以在一個查詢中完成所有這些工作?我需要能夠輸出活動,失敗和完成的任務,最好在一個查詢中輸出。MySQL - 如何更好地做到這一點?
*(參考)* HTTP:// EN。 wikipedia.org/wiki/SQL_injection – Gordon 2010-04-22 16:39:50
雖然我沒有接受任何用戶輸入。 – Andrew 2010-04-22 16:42:04
取決於您獲取$ user_id的方式。它不是直接來自cookie,是嗎? – Tom 2010-04-22 16:49:43