MySQL を使用してこれまでに作成した最長のクエリを最適化しようとしていますEXPLAIN
が、これが初めてなので、結果を理解できないようです。クエリとEXPLAIN
コマンドを実行した結果は次のとおりです。
EXPLAIN SELECT pb.name, s1.MessageFrom, s1.MessageText, s1.SendTime, s1.is_unread, s1.Id, s1.autoreply_sent FROM sol_inbound s1
JOIN sol_contactnum c ON s1.MessageFrom = c.number
JOIN sol_phonebk_contactnum USING (contactnum_id)
JOIN sol_phonebk pb USING (phonebk_id)
JOIN sol_message_folder mf ON s1.Id = mf.message_id
WHERE (MessageFrom, SendTime) IN (SELECT MessageFrom, MAX(SendTime) FROM sol_inbound inb
JOIN sol_message_folder mf WHERE inb.Id = mf.message_id
AND mf.folder_id=1 AND mf.direction='inbound' AND mf.user_id=1
GROUP BY MessageFrom)
AND mf.folder_id=1 AND mf.direction='inbound' AND mf.user_id=1
UNION
SELECT NULL `name`, s1.MessageFrom, s1.MessageText, s1.SendTime, s1.is_unread, s1.Id, s1.autoreply_sent FROM sol_inbound s1
LEFT JOIN sol_contactnum c ON s1.MessageFrom = c.number
JOIN sol_message_folder mf ON s1.Id = mf.message_id
WHERE c.number IS NULL
AND mf.folder_id=1 AND mf.direction='inbound' AND mf.user_id=1
AND (MessageFrom, SendTime) IN (SELECT MessageFrom, MAX(SendTime) FROM sol_inbound inb
JOIN sol_message_folder mf WHERE inb.Id = mf.message_id
AND mf.folder_id=1 AND mf.direction='inbound' AND mf.user_id=1
GROUP BY MessageFrom)
ORDER BY SendTime DESC LIMIT 100
EXPLAIN
結果は次のとおりです。
id select_type table type possible_keys key key_len ref rows Extra
------ ------------------ ---------------------- ------ ------------------------------------------------------------- ---------------- ------- ---------------------------------------------------- ------ ------------------------
1 PRIMARY pb ALL PRIMARY (NULL) (NULL) (NULL) 303
1 PRIMARY sol_phonebk_contactnum ref PRIMARY,phonebk_id1_idx,contactnum_id1_idx,phonebk_contactnum PRIMARY 4 googlep1_solane.pb.phonebk_id 1 Using index
1 PRIMARY c eq_ref PRIMARY,number_idx PRIMARY 4 googlep1_solane.sol_phonebk_contactnum.contactnum_id 1
1 PRIMARY s1 ref PRIMARY,message_from_idx message_from_idx 243 googlep1_solane.c.number 1 Using where
1 PRIMARY mf eq_ref PRIMARY PRIMARY 22 const,googlep1_solane.s1.Id,const,const 1 Using where; Using index
2 DEPENDENT SUBQUERY inb index PRIMARY message_from_idx 243 (NULL) 1
2 DEPENDENT SUBQUERY mf eq_ref PRIMARY PRIMARY 22 const,googlep1_solane.inb.Id,const,const 1 Using where; Using index
3 UNION s1 ALL PRIMARY (NULL) (NULL) (NULL) 877 Using where
3 UNION c ref number_idx number_idx 243 googlep1_solane.s1.MessageFrom 1 Using where; Using index
3 UNION mf eq_ref PRIMARY PRIMARY 22 const,googlep1_solane.s1.Id,const,const 1 Using where; Using index
4 DEPENDENT SUBQUERY inb index PRIMARY message_from_idx 243 (NULL) 1
4 DEPENDENT SUBQUERY mf eq_ref PRIMARY PRIMARY 22 const,googlep1_solane.inb.Id,const,const 1 Using where; Using index
(NULL) UNION RESULT <union1,3> ALL (NULL) (NULL) (NULL) (NULL) (NULL) Using filesort
クエリのUNION
真ん中にある は、電話帳に番号が表示されている人と表示されていない人を結合します (したがってLEFT JOIN
)。
編集:
このクエリが行うことは、各番号の最新のインバウンド メッセージを取得して返すことです。GROUP BY
最も古いメッセージを返すので使用できます...最新のメッセージが必要です。次に、電話帳に存在しない番号を結合します。これが、私がチェックする理由ですWHERE c.number IS NULL.