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我在 MySQL 中使用 UNION 来将两个单独查询的结果组合在一起。这两个查询使用了很多相同的表。有没有办法利用这一点来优化查询？

    SELECT  2 AS RELEVANCE_SCORE_TYPE,
        VIEWER_ID, 
        QUESTION_ID, 
        sum(ANSWER_SCORE) AS RELEVANCE_SCORE
FROM(SELECT  cr.COMMUNICATIONS_ID AS ANSWER_ID, 
        cr.CONSUMER_ID as VIEWER_ID,
        nc.PARENT_COMMUNICATIONS_ID AS QUESTION_ID,
        case when  cr.CONSUMER_ID= nc.SENDER_CONSUMER_ID then 3*((24/(((UNIX_TIMESTAMP(NOW())-UNIX_TIMESTAMP(cal.LAST_MOD_TIME)+3600)/3600))*(ces.EXPERT_SCORE * cirm.CONSUMER_RATING) + (12.5 * scs.SIMILARITY)* (1 - EXP(-0.5 * (cal.TIPS_AMOUNT / ATV.AVG_TIPS)) + .15)))
            else ((24/(((UNIX_TIMESTAMP(NOW())-UNIX_TIMESTAMP(cal.LAST_MOD_TIME)+3600)/3600))*(ces.EXPERT_SCORE * cirm.CONSUMER_RATING) + (12.5 * scs.SIMILARITY)* (1 - EXP(-0.5 * (cal.TIPS_AMOUNT / ATV.AVG_TIPS)) + .15)))
        end as ANSWER_SCORE
FROM (SELECT 238 AS CONSUMER_ID, 
             ACTION_LOG_ID, 
            COMMUNICATIONS_ID 
     FROM consumer_action_log 
     WHERE COMM_TYPE_ID=4) AS cr
JOIN network_communications AS nc 
    ON cr.COMMUNICATIONS_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=ces.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(cal.TIPS_AMOUNT) AS AVG_TIPS
        FROM CONSUMER_ACTION_LOG AS cal
        JOIN (SELECT 234 AS CONSUMER_ID, 
                     ACTION_LOG_ID, 
                     COMMUNICATIONS_ID 
              FROM consumer_action_log 
              WHERE COMM_TYPE_ID=4) AS cr 
        ON cal.SENDER_CONSUMER_ID=cr.consumer_id
    ) ATV) AS ASM
GROUP BY ANSWER_ID
UNION
SELECT 1 AS RELEVANCE_SCORE_TYPE,
       qcr.CONSUMER_ID AS Viewer_ID, 
       qcr.COMMUNICATIONS_ID, 
       case when reply.replies IS NOT NULL AND qcr.CONSUMER_ID <> qcr.SENDER_CONSUMER_ID then
       24/((UNIX_TIMESTAMP(NOW())-UNIX_TIMESTAMP(qcr.LAST_MOD_TIME)+3600)/3600)*(ces.EXPERT_SCORE+2.5*scs.SIMILARITY)*(EXP(-reply.replies))
       when reply.replies IS NULL AND qcr.CONSUMER_ID <> qcr.SENDER_CONSUMER_ID then
       24/((UNIX_TIMESTAMP(NOW())-UNIX_TIMESTAMP(qcr.LAST_MOD_TIME)+3600)/3600)*(ces.EXPERT_SCORE+2.5*scs.SIMILARITY)*(EXP(0))
       when reply.replies IS NULL AND qcr.CONSUMER_ID = qcr.SENDER_CONSUMER_ID then
       24/((UNIX_TIMESTAMP(NOW())-UNIX_TIMESTAMP(qcr.LAST_MOD_TIME)+3600)/3600)*(7.5)*(1-EXP(0))
       when reply.replies IS NOT NULL AND qcr.CONSUMER_ID = qcr.SENDER_CONSUMER_ID then
       24/((UNIX_TIMESTAMP(NOW())-UNIX_TIMESTAMP(qcr.LAST_MOD_TIME)+3600)/3600)*(7.5)*(1-EXP(-reply.replies))
            else null
       end as QUESTION_SCORE
FROM (SELECT 238 AS CONSUMER_ID,
            SENDER_CONSUMER_ID, 
            COMMUNICATIONS_ID,
            LAST_MOD_TIME
     FROM network_communications
     WHERE NETWORK_COMM_TYPE_ID=1) AS qcr
JOIN communication_interest_mapping AS cim 
    ON qcr.COMMUNICATIONS_ID=cim.COMMUNICATION_ID
JOIN consumer_expert_score AS ces
    ON ces.CONSUMER_ID=qcr.CONSUMER_ID
    AND cim.CONSUMER_INTEREST_EXPERT_ID=ces.CONSUMER_EXPERT_ID
JOIN survey_customer_similarity AS scs 
    ON qcr.CONSUMER_ID=scs.CONSUMER_ID_2 
    AND qcr.SENDER_CONSUMER_ID=scs.CONSUMER_ID_1 
    OR qcr.CONSUMER_ID=scs.CONSUMER_ID_1 
    AND qcr.SENDER_CONSUMER_ID=scs.CONSUMER_ID_2
LEFT JOIN (SELECT COUNT(*) AS replies, 
           PARENT_COMMUNICATIONS_ID
           FROM network_communications AS nc1
           WHERE NETWORK_COMM_TYPE_ID=2
GROUP BY PARENT_COMMUNICATIONS_ID) AS reply
    ON qcr.COMMUNICATIONS_ID=reply.PARENT_COMMUNICATIONS_ID
ORDER BY RELEVANCE_SCORE DESC;
FROM (SELECT 234 AS CONSUMER_ID, 
             ACTION_LOG_ID, 
            COMMUNICATIONS_ID 
     FROM consumer_action_log 
     WHERE COMM_TYPE_ID=4) AS cr
JOIN network_communications AS nc 
    ON cr.COMMUNICATIONS_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=ces.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(cal.TIPS_AMOUNT) AS AVG_TIPS
        FROM CONSUMER_ACTION_LOG AS cal
        JOIN (SELECT 234 AS CONSUMER_ID, 
                     ACTION_LOG_ID, 
                     COMMUNICATIONS_ID 
              FROM consumer_action_log 
              WHERE COMM_TYPE_ID=4) AS cr 
        ON cal.SENDER_CONSUMER_ID=cr.consumer_id
    ) ATV) AS ASM
GROUP BY ANSWER_ID
ORDER BY ANSWER_SCORE_SUMMED DESC;

它很长，所以不要觉得有义务阅读整本书。要点很简单，联合两边的查询使用了很多相同的表。

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2 回答 2

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我的第一个想法是不值得尝试超越 MySQL 优化器。特别是因为您正在进行 18 个连接和 2 个交叉连接。

UNION 两侧的查询使用许多相同的表并不罕见。这是我希望优化器能够处理的那种情况。

为了获得更好的性能，您需要查看执行配置文件，并且可能需要重写查询以消除交叉连接。要查看执行配置文件，请运行这两个之一。

EXPLAIN <your query>
EXPLAIN EXTENDED <your query>
于 2011-01-27T22:47:31.360 回答
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看起来顶部和底部之间存在一些细微的差异。所以不要以为没有UNION. 但是，如果您确定来自顶部和底部查询的数据不会重叠使用UNION ALL，UNION则尝试使结果不同，从而使引擎做额外的工作，如果有很多记录，这可能是相当可观的。

根据@Catcall 的建议，使用EXPLAIN将验证该UNION ALL方法。

于 2011-01-27T23:11:46.213 回答