Вы используете UPDATE (.. JOIN ..) SET синтаксис
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;
В качестве доказательства концепции для всех, кто читает это, вот таблица, которую вы можете создать и попробовать синтаксис на
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';
Теперь запустите обновление (оно устанавливает для article_date всех записей значение MAX article_date от того же пользователя)
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;
Наконец, проверьте содержимое
select * from user_news;