我已經寫了一個代碼,我經常跳到數據庫有點尷尬,我使用代碼來計算評論表中批准,待處理和垃圾郵件的數量。如何改進此代碼以使我的代碼更少地訪問數據庫?
這是我的代碼。
$query_approved = "SELECT COUNT(*) as approved FROM comments WHERE approve = '1'";
$result_approved = mysql_query($query_approved);
$row_approved = mysql_fetch_array($result_approved);
$query_unapproved = "SELECT COUNT(*) as unapproved FROM comments WHERE approve = '0'";
$result_unapproved = mysql_query($query_unapproved);
$row_unapproved = mysql_fetch_array($result_unapproved);
$query_spam = "SELECT COUNT(*) as spam FROM comments WHERE spam = '1'";
$result_spam = mysql_query($query_spam);
$row_spam = mysql_fetch_array($result_spam);
雖然該代碼工作正常,但它看起來很醜。無論如何要改善?
到底什麼是c? – 2010-09-08 18:05:09
@Ibrahim Azhar Armar:這是一個表別名。不需要單個表格,但是我是在習慣之外做的,所以我知道什麼是列和什麼表與函數等。 – 2010-09-08 18:10:16
我測試了人們所提出的所有查詢,而我發現的最可行的解決方案是你的,它的工作沒有任何問題。和別名是我發現玩的東西。謝謝你的回答。我將與您的代碼:) – 2010-09-08 18:17:19