0

基本的に私はこのクエリを実行しようとしています

UPDATE communication_relevance SET score = (SELECT ((ces.EXPERT_SCORE * cirm.CONSUMER_RATING) + (12.5 * scs.SIMILARITY)* (1 - EXP(-0.5 * (cal.TIPS_AMOUNT / AT.AVG_TIPS)) + .15))AS ANSWER_SCORE
FROM COMMUNICATION_RELEVANCE AS cr
JOIN network_communications AS nc
ON cr.COMMUNICATION_ID=nc.COMMUNICATIONS_ID
JOIN consumer_action_log AS cal
ON cr.ACTION_LOG_ID=cal.ACTION_LOG_ID
JOIN communication_interest_mapping AS cim
ON nc.PARENT_COMMUNICATIONS_ID=cim.COMMUNICATION_ID
JOIN consumer_interest_rating_mapping AS cirm
ON cr.CONSUMER_ID=cirm.CONSUMER_ID
    AND cim.CONSUMER_INTEREST_EXPERT_ID=cirm.CONSUMER_INTEREST_ID
JOIN consumer_expert_score AS ces
ON nc.SENDER_CONSUMER_ID=ces.CONSUMER_ID
    AND cim.CONSUMER_INTEREST_EXPERT_ID=CONSUMER_EXPERT_ID
JOIN survey_customer_similarity AS scs
ON cr.CONSUMER_ID=scs.CONSUMER_ID_2
    AND cal.SENDER_CONSUMER_ID=scs.CONSUMER_ID_1 
OR cr.CONSUMER_ID=scs.CONSUMER_ID_1
    AND cal.SENDER_CONSUMER_ID=scs.CONSUMER_ID_2
CROSS JOIN
(SELECT AVG(TIPS_AMOUNT) AS AVG_TIPS
FROM CONSUMER_ACTION_LOG
JOIN COMMUNICATION_RELEVANCE
ON CONSUMER_ACTION_LOG.SENDER_CONSUMER_ID=COMMUNICATION_RElEVANCE.consumer_id) AT)
;

しかし、私はこのエラーが発生します:

Error:1/25/2011 1:03:20 PM 0:00:00.135: Lookup Error - MySQL Database Error: You can't specify target table 'communication_relevance' for update in FROM clause

どんな助けでも大歓迎です！

4

4 に答える 4

2

UPDATE (.. JOIN ..) SET 構文を使用します

UPDATE communication_relevance X
JOIN (
    SELECT cr.COMMUNICATION_ID, ((ces.EXPERT_SCORE * cirm.CONSUMER_RATING)
        + (12.5 * scs.SIMILARITY)
        * (1 - EXP(-0.5 * (cal.TIPS_AMOUNT / AT.AVG_TIPS)) + .15)) AS ANSWER_SCORE
    FROM COMMUNICATION_RELEVANCE AS cr
    JOIN network_communications AS nc ON cr.COMMUNICATION_ID=nc.COMMUNICATIONS_ID
    JOIN consumer_action_log AS cal ON cr.ACTION_LOG_ID=cal.ACTION_LOG_ID
    JOIN communication_interest_mapping AS cim ON nc.PARENT_COMMUNICATIONS_ID=cim.COMMUNICATION_ID
    JOIN consumer_interest_rating_mapping AS cirm ON cr.CONSUMER_ID=cirm.CONSUMER_ID
        AND cim.CONSUMER_INTEREST_EXPERT_ID=cirm.CONSUMER_INTEREST_ID 
    JOIN consumer_expert_score AS ces ON nc.SENDER_CONSUMER_ID=ces.CONSUMER_ID
        AND cim.CONSUMER_INTEREST_EXPERT_ID=CONSUMER_EXPERT_ID
    JOIN survey_customer_similarity AS scs ON
        cr.CONSUMER_ID=scs.CONSUMER_ID_2 AND cal.SENDER_CONSUMER_ID=scs.CONSUMER_ID_1 
     OR cr.CONSUMER_ID=scs.CONSUMER_ID_1 AND cal.SENDER_CONSUMER_ID=scs.CONSUMER_ID_2
    CROSS JOIN
    (
        SELECT AVG(L.TIPS_AMOUNT) AS AVG_TIPS
        FROM CONSUMER_ACTION_LOG L
        JOIN COMMUNICATION_RELEVANCE R ON L.SENDER_CONSUMER_ID=R.consumer_id
    ) AT
) ON X.COMMUNICATION_ID = AT.COMMUNICATION_ID
SET X.score = AT.ANSWER_SCORE;

これを読んでいる他の人のための概念実証として、作成して構文を試すことができる表を次に示します。

create table user_news(
    user_id int, article_id int, article_date timestamp,
    primary key(user_id, article_id));
insert into user_news select 1,2,'2010-01-02';
insert into user_news select 1,3,'2010-01-03';
insert into user_news select 1,4,'2010-01-01';
insert into user_news select 2,1,'2010-01-01';
insert into user_news select 2,2,'2010-01-02';
insert into user_news select 2,3,'2010-01-02';
insert into user_news select 2,4,'2010-01-02';
insert into user_news select 4,5,'2010-01-05';

更新を実行します (すべてのレコードの article_date を同じユーザーの MAX article_date に設定します)。

update user_news a
join (
  select b.user_id, max(b.article_date) adate
  from user_news b
  group by b.user_id) c
  on a.user_id=c.user_id
set a.article_date = c.adate;

最後に中身をチェック

select * from user_news;
于 2011-01-25T19:03:06.887 に答える
0

これを行う場合は、一時テーブルを使用する必要があります。

あなたが何をしているのか、なぜ、そしてどのようなリスクがあるのか​​を考える時が来たと思います:)

于 2011-01-25T18:08:22.850 に答える
0

テーブル名

COMMUNICATION_RELEVANCE

キャップですか、それともタイプミスですか？

于 2011-01-25T18:09:31.307 に答える
0

基本的なデータベースの正規化は、テーブルに計算フィールドがあると正規化のルールが破られることを示します....必要なときにオンザフライでクエリでこの計算を実行できるはずです。または、計算フィールドを含むビューを作成します。

于 2011-01-25T18:26:02.997 に答える