Question

我在下面有以下代码，我正在尝试为3列获取AVG。

 SELECT 
    (SUM(score) * .3) As score_a,
    (SUM(score) * .6) As score_b,
    (SUM(score) * .8) As score_c

     --now I want to get the AVG of the above scores
    AVG(score_a + score_b + score_c) As avg_score
 FROM score_table

但这不起作用。我得到的错误是“无效的列名称score_a”。我正在使用SQL Server 2008

Answer 1

AVG()是一个聚合函数，它从多个行中获取值并给出它们的平均值。

您尝试平均3 列。

我的具体情况，可以使用代数来完成。

SELECT 
  (SUM(score) * .3)                       As score_a,
  (SUM(score) * .6)                       As score_b,
  (SUM(score) * .8)                       As score_c,
  (SUM(score) * ((0.3 + 0.6 + .8) / 3.0)) As score_avg
FROM
  score_table

在更一般化的情况下，您受限于以下事实：您无法引用刚才在另一列中定义的列...

SELECT
  a + 1      AS inc_a,
  inc_a * 2  AS this_is_invalid
FROM
  your_table

您需要重复自己，或使用子查询......

的重复 的

SELECT (SUM(score) * .3) As score_a, (SUM(score) * .6) As score_b, (SUM(score) * .8) As score_c, ((SUM(score) * .3) + (SUM(score) * .3) + (SUM(score) * .3)) / 3.0 As score_avg FROM score_table

子查询

SELECT score_a, score_b, score_c, (score_a + score_b + score_c) / 3.0 AS score_avg FROM ( SELECT (SUM(score) * .3) As score_a, (SUM(score) * .6) As score_b, (SUM(score) * .8) As score_c FROM score_table ) AS data

Answer 2

你可以尝试一下：

 SELECT 
    (SUM(score) * .3) As score_a,
    (SUM(score) * .6) As score_b,
    (SUM(score) * .8) As score_c,
    AVG(score_a + score_b + score_c) As avg_score
 FROM score_table

看看它是否适合你？

Answer 3

有手动执行平均值的强力方法，但它不会像添加列一样进行缩放。：）

SELECT CAST((score_a + score_b + score_c) / 3 AS DECIMAL(20,2)) As avg_score,
       score_a,
       score_b,
       score_c
FROM (
    SELECT (SUM(score) * .3) As score_a,
           (SUM(score) * .6) As score_b,
           (SUM(score) * .8) As score_c
    FROM score_table
    GROUP BY userid, gameid --Whatever you're grouping by here!
)

如何在AVERAGE函数中使用别名

3 个答案: