MySQL Союз и Средний

Question

Я понятия не имею, если я использую UNION правильно в этом случае - вполне может быть лучше/легкий путь, и я открыт для предложений:

У меня есть следующий код :

SELECT COUNT(*), AVG (q1) AS q1, AVG (q2) AS q2, AVG (q3) AS q3, AVG (q4) AS q4, AVG (q5) AS q5, AVG (q6) AS q6, AVG (q7) AS q7, AVG (q8) AS q8, AVG (q9) AS q9, AVG (q10) AS q10, AVG (q11) AS q11, AVG (q12) AS q12, AVG (q13) AS q13, AVG (q14) AS q14, AVG (q15) AS q15, AVG (q16) AS q16, AVG (q17) AS q17, AVG (q18) AS q18, AVG (q19) AS q19, AVG (q20) AS q20, AVG (q21) AS q21, AVG (q22) AS q22 FROM thotels_results WHERE brand = 'EFG' AND date = 'NOV2010' GROUP BY brand 
UNION 
SELECT COUNT(*), AVG (q1) AS q1, AVG (q2) AS q2, AVG (q3) AS q3, AVG (q4) AS q4, AVG (q5) AS q5, AVG (q6) AS q6, AVG (q7) AS q7, AVG (q8) AS q8, AVG (q9) AS q9, AVG (q10) AS q10, AVG (q11) AS q11, AVG (q12) AS q12, AVG (q13) AS q13, AVG (q14) AS q14, AVG (q15) AS q15, AVG (q16) AS q16, AVG (q17) AS q17, AVG (q18) AS q18, AVG (q19) AS q19, AVG (q20) AS q20, AVG (q21) AS q21, AVG (q22) AS q22 FROM thotels_results WHERE brand = 'XYC' AND date = 'NOV2010' GROUP BY brand 
UNION 
SELECT COUNT(*), AVG (q1) AS q1, AVG (q2) AS q2, AVG (q3) AS q3, AVG (q4) AS q4, AVG (q5) AS q5, AVG (q6) AS q6, AVG (q7) AS q7, AVG (q8) AS q8, AVG (q9) AS q9, AVG (q10) AS q10, AVG (q11) AS q11, AVG (q12) AS q12, AVG (q13) AS q13, AVG (q14) AS q14, AVG (q15) AS q15, AVG (q16) AS q16, AVG (q17) AS q17, AVG (q18) AS q18, AVG (q19) AS q19, AVG (q20) AS q20, AVG (q21) AS q21, AVG (q22) AS q22 FROM thotels_results WHERE brand = 'ABC' AND date = 'NOV2010' GROUP BY brand 

Он выводит следующее:

 q1  q2  q3 etc.                       
140 8.7714 8.8429 8.1643 8.7500 8.7571 8.9000 9.4071 9.1214 8.5714 8.7643 9.5143 8.9429 9.1643 8.9857 7.9500 8.9286 8.7000 9.0429 9.0143 8.7214 9.1214 9.3071 
29 8.1724 8.2414 8.2414 7.8966 8.5862 8.5517 9.0000 8.5862 8.1724 7.9655 8.8966 8.6207 8.2414 8.3793 7.8276 8.3793 7.9310 8.4138 8.6897 8.3448 8.8621 8.5172 
897 8.6009 8.5686 7.8528 8.3133 8.3423 8.6410 9.0301 8.6912 8.3233 8.3389 9.2029 8.3969 8.6856 8.5017 7.8071 8.4816 8.3512 8.6789 8.6789 8.3913 8.6388 8.8986 

Все, что я бы 'хотел бы сделать' является СРЕДНЯЯ каждый столбцов q1, q2, q3 или SUM их разделите и разделите на 3.

Как я уже сказал, если есть лучший способ, который не использует JOIN, все в порядке со мной !!!

Заранее благодарен,

Гомер.

Answer 1

(1) Вам не нужно вычислять результаты отдельно, а затем UNION им подобные. (2) Думаю, WITH ROLLUP, вероятно, сделает все, что вам нужно.

SELECT COUNT(*), AVG (q1) AS q1,... 
FROM thotels_results WHERE brand in ('ABC','EFG','XYZ') AND date = 'NOV2010' 
GROUP BY brand 
WITH ROLLUP 

Но, как указано в комментариях, это не означает, что требуется усреднение среднего значения. Лучшее, что я мог придумать для этого, это

CREATE TEMPORARY TABLE results 
SELECT 
     brand, 
     COUNT(*) AS cnt, 
     AVG (q1) AS q1 
     ... 
FROM thotels_results 
WHERE brand in ('ABC','EFG','XYZ') AND date = 'NOV2010' 
GROUP BY brand; 

CREATE TEMPORARY TABLE results2 
SELECT cast(NULL as char), AVG(cnt), AVG(q1) 
FROM results r2; 


/*MySQL doesn't allow the same temp table to be accessed twice in a UNION!*/ 
SELECT * FROM results r 
UNION ALL 
SELECT * 
FROM results2; 

DROP TEMPORARY TABLE results; 
DROP TEMPORARY TABLE results2; 

Answer 2

Без JOIN, почему бы не просто использовать переменную temp или таблицу. Так вставить СУММ:

INSERT INTO TempTable (SELECT COUNT (*), SUM (q1) AS q1 .... FROM thotels_results WHERE бренд = 'EFG' и дата = GROUP 'NOV2010' BY бренда)

INSERT INTO TempTable (SELECT COUNT (*), СУММА (q1) КАК q1 .... ОТ thotels_results ГДЕ бренд = 'XYC' и дата = ГРУППА '' NOV2010 по марке)

INSERT INTO TempTable (SELECT COUNT (*), SUM (q1) AS q1 .... FROM thotels_results WHERE бренд = 'ABC' и дата = GROUP 'NOV2010' BY бренда)

Затем принести adverage

SELECT SUM (q1)/3, как q1Adverage, .... От TempTable