php - PHP MYSQL Query AS の大きな問題

Question

ASクエリが必要ですが、例:

このクエリでは、rat ='10' または rat >='10'。

$minimalEntry = mysql_query("SELECT author,count(id) AS rat 
                               FROM sds_posts 
                           GROUP BY author 
                           ORDER BY rat ASC 
                              LIMIT 0,20");

ありがとうございました..

Answer 1

HAVING句が必要です：

SELECT author, COUNT(id) AS rat 
FROM sds_posts 
GROUP BY author 
HAVING COUNT(id) >= 10
ORDER BY rat ASC 
  LIMIT 0,20

Answer 2

WHERE 条件を追加するだけです。

$minimalEntry = mysql_query("SELECT author,count(id) AS rat 
                             FROM sds_posts 
                             WHERE count(id)  >= 10 
                             GROUP BY author 
                             ORDER BY rat ASC 
                             LIMIT 0,20"
                           );

編集：これは間違っています。このクエリには HAVING 句が必要です。

Answer 3

カウント集計が 2 倍になるのを避けるために、内部選択を使用します。

SELECT author, rat
FROM
(
    SELECT author, count(id) AS rat
    FROM sds_posts
) 
WHERE rat >= 10
ORDER BY rat
LIMIT 0,20