[英]MySQL Distinct Entries disregarding all repeats after ORDER_BY
I want create a query that creates a calculated field ANSWER_SCORE which it then orders the results of the query by. 我想创建一个查询,该查询创建一个计算字段ANSWER_SCORE,然后按该字段排序查询结果。 I also want to select two other fields QUESTION_ID and ANSWER_ID. 我还想选择另外两个字段QUESTION_ID和ANSWER_ID。
I only want to select distinct QUESTION_ID's. 我只想选择不同的QUESTION_ID。 I also however want to select some other information related to that QUESTION_ID (which will be common to all queries which returned that QUESTION_ID. I am unsure of how to do this. 但是,我还想选择与该QUESTION_ID相关的其他信息(这对于返回该QUESTION_ID的所有查询都是通用的。我不确定如何执行此操作。
Here is my query: 这是我的查询:
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;
If your other fields really are the same for every QUESTION_ID and ANSWER_ID you return, you can simply add them to your select list and put them in the GROUP BY clause. 如果您返回的每个QUESTION_ID和ANSWER_ID的其他字段确实都相同,则只需将它们添加到选择列表中,然后将它们放在GROUP BY子句中即可。 eg 例如
SELECT QUESTION_ID,
ANSWER_ID,
FOO,
BAR,
SUM(ANSWER_SCORE) AS ANSWER_SCORE_SUMMED
FROM...
<snip>
...GROUP BY QUESTION_ID, ANSWER_ID, FOO, BAR
This has the added benefit of checking your assumption about the other data all being the same for each returned QUESTION_ID, as the different combinations will all be laid out for you with their own ANSWER_SCORE_SUMMED. 这样做还有一个好处,即检查您对每个返回的QUESTION_ID的其他数据都相同的假设,因为不同的组合都将通过自己的ANSWER_SCORE_SUMMED为您布置。
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