Question

我想创建一个查询，创建一个计算字段ANSWER_SCORE然后按顺序排序查询结果。我还想选择另外两个字段QUESTION_ID和ANSWER_ID。

我只想选择不同的QUESTION_ID。我还想选择与QUESTION_ID相关的其他一些信息（这对于返回QUESTION_ID的所有查询都是通用的。我不知道如何做到这一点。

这是我的问题：

SELECT  QUESTION_ID,
        ANSWER_ID,
        sum(ANSWER_SCORE) AS ANSWER_SCORE_SUMMED
FROM(SELECT
        nc.PARENT_COMMUNICATIONS_ID AS QUESTION_ID,
        cr.COMMUNICATIONS_ID AS ANSWER_ID,
        case when  cr.CONSUMER_ID= nc.SENDER_CONSUMER_ID then 2*(1 - EXP(-0.5 * (cal.REAL_TIPS_AMOUNT / ATV.AVG_TIPS) + .15))*(24/(((UNIX_TIMESTAMP(NOW())-UNIX_TIMESTAMP(cal.LAST_MOD_TIME)+3600)/3600)))*(IFNULL(ces.EXPERT_SCORE,2) * IFNULL(cirm.CONSUMER_RATING,2) + (12.5 * IFNULL((bit_count((conv(scr_tip.survey_string, 2, 10) & conv(scr_view.survey_string, 2, 10)) | ((0xFFFFFFFF>>(32-6))&~(conv(scr_tip.survey_string, 2, 10)|conv(scr_view.survey_string, 2, 10))))/6),.3)))
            else (1 - EXP(-0.5 * (cal.REAL_TIPS_AMOUNT / ATV.AVG_TIPS) + .15))*(24/(((UNIX_TIMESTAMP(NOW())-UNIX_TIMESTAMP(cal.LAST_MOD_TIME)+3600)/3600)))*(IFNULL(ces.EXPERT_SCORE,2) * IFNULL(cirm.CONSUMER_RATING,2) + (12.5 * IFNULL((bit_count((conv(scr_tip.survey_string, 2, 10) & conv(scr_view.survey_string, 2, 10)) | ((0xFFFFFFFF>>(32-6))&~(conv(scr_tip.survey_string, 2, 10)|conv(scr_view.survey_string, 2, 10))))/6),.3)))
        end as ANSWER_SCORE
  FROM (SELECT 232 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 network_communications AS nc1
    ON nc.PARENT_COMMUNICATIONS_ID=nc1.COMMUNICATIONS_ID
JOIN (SELECT ACTION_LOG_ID,
             LAST_MOD_TIME,
             CONSUMER_ID,
             SENDER_CONSUMER_ID,
             sum(tips_amount) AS real_tips_amount
      FROM consumer_action_log
      WHERE COMM_TYPE_ID=4
      GROUP BY COMMUNICATIONS_ID, SENDER_CONSUMER_ID) 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
LEFT 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 network_communications_message AS ncm
  ON nc.PARENT_COMMUNICATIONS_ID=ncm.COMMUNICATIONS_ID
LEFT JOIN consumer_expert_score AS ces 
    ON nc.SENDER_CONSUMER_ID=ces.CONSUMER_ID
    AND cim.CONSUMER_INTEREST_EXPERT_ID=ces.CONSUMER_EXPERT_ID
LEFT JOIN surveycustomerrelation AS scr_tip
     ON cal.SENDER_CONSUMER_ID=scr_tip.Cust_ID
LEFT JOIN surveycustomerrelation AS scr_view
    ON cr.CONSUMER_ID=scr_view.CUST_ID
STRAIGHT_JOIN
    (
        SELECT AVG(cal.TIPS_AMOUNT) AS AVG_TIPS,
                   cal.SENDER_CONSUMER_ID AS CONSUMER_ID
        FROM CONSUMER_ACTION_LOG AS cal
        WHERE COMM_TYPE_ID=4
        GROUP BY cal.SENDER_CONSUMER_ID
    ) AS ATV
    ON cal.SENDER_CONSUMER_ID=ATV.CONSUMER_ID
    ) AS ASM
GROUP BY ANSWER_ID
ORDER BY ANSWER_SCORE_SUMMED DESC
LIMIT 0,20;

Answer 1

如果您返回的每个QUESTION_ID和ANSWER_ID的其他字段确实相同，您只需将它们添加到您的选择列表并将它们放在GROUP BY子句中即可。 e.g。

SELECT QUESTION_ID,
    ANSWER_ID,
    FOO,
    BAR,
    SUM(ANSWER_SCORE) AS ANSWER_SCORE_SUMMED
FROM...
<snip>
...GROUP BY QUESTION_ID, ANSWER_ID, FOO, BAR

这样做的另一个好处就是检查您对每个返回的QUESTION_ID的其他数据的假设是否相同，因为不同的组合将全部用自己的ANSWER_SCORE_SUMMED进行布局。

MySQL Distinct Entries忽略ORDER_BY之后的所有重复

1 个答案: