SQL从其他表中选择平均值

我花了很多时间来构建这个选择，但我无法解决它。我有两张桌子。第一个表称为汽车并具有PK（主键）id_car和另一列名称等。第二个表称为评级，并具有col_id_rating（PK），id_car（FK）和rating_value（整数）。正如你怀疑的那样，一辆车可以有一次以上的咆哮。我想选择所有的汽车，我想知道每辆车的平均评分。最后，我想通过这个平均desc来排序结果。我正在尝试这样的事情：SQL从其他表中选择平均值

SELECT id_car, name, average 
    FROM car C, rating R 
    WHERE C.id_car = R.id_car 
    ORDER BY (average) (
     SELECT AVG(rating_value) AS average 
     FROM rating R 
     WHERE C.id_car = R.id_car)

但它不起作用。

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2012-11-28 matlos

对于SQL Server;此外，我建议你使用JOIN代替WHERE table1, table2..

SELECT C.id_car, name, AVG(rating_value) AS average 
FROM car C JOIN rating R 
     ON C.id_car = R.id_car 
GROUP By C.id_car, name 
ORDER BY average DESC

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2012-11-28 11:48:13 Kaf

只是添加更多信息：如果您想要包括尚未评级的汽车，请使用LEFT JOIN。 –

@ShantanuGupta +1表示:) – Kaf

谢谢先生！奇迹般有效 – matlos

SELECT C.name, AVG(R.rating_value) 
FROM car C, rating R 
WHERE C.id_car = R.id_car 
GROUP BY C.name 
ORDER BY AVG(R.rating_value) DESC

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2012-11-28 11:48:31 Emyl

这实现了聚合函数AVG()然后GROUP BY的车厢ID名称：

select c.id_car, c.name, avg(r.rating_value) aver 
from car c 
left join rating r 
    on c.id_car = r.id_car 
group by c.id_car, c.name 
order by aver desc

使用LEFT JOIN将包括所有汽车在结果中甚至还没有被评级。

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2012-11-28 11:49:07 Taryn

SELECT c.id_car, c.name,avg(r.rating_value) as rating 
FROM car c 
join rating r 
on c.id_car = r.id_car 
group by r.id_car 
order by rating

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2012-11-28 11:49:58

SQL从其他表中选择平均值

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