Mysql 基于列自身表中的值更新列

基本上，我正在尝试执行这个查询

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)
;

但我得到了这个错误：

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

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任何帮助都将不胜感激

如果要执行此操作，必须使用临时表

我想说是时候考虑一下你在做什么、为什么以及风险是什么了：）

表名

COMMUNICATION_RELEVANCE

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是大写还是打字错误？

基本数据库规范化表示表中有计算字段违反了规范化规则……您应该能够在需要时在查询中动态执行此计算。或者创建包含计算字段的视图。

您可以使用更新（…连接..）集语法

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;

作为其他阅读本文的人的概念证明，这里有一个表，您可以创建并尝试语法

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';

现在运行更新（它将所有记录的文章日期设置为同一用户的最大文章日期）

最后，检查内容

select * from user_news;

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我是SQL新手。您所说的临时表是什么意思？您必须创建一个表（使用查询），用相关数据填充它，然后将临时表用作您试图运行的查询中的源。Google“临时表mysql”：）mysql不允许您在同一查询中更新正在读取的表。在多雾的天气里，在结冰的条件下以200公里/小时的速度在高速公路上行驶时，你不妨尝试同时更换汽车上的所有4个轮胎。这确实有效。您只需将其表述为rightTable名称在OS X和Windows上不区分大小写。他可能是在用那个来运行数据库？：）

select * from user_news;