Question

您好
我在mysql表中有以下列： rating1，rating2，price，cond，approved

是否可以选择这样的结果：

select 
average rating1 + rating2 as total_rating,  
average rating1 as rating1,  
average rating2 as rating2,  
average price if cond = '1' as price_used  
average price if cond = '2' as price_new  
where approved = '1'

到目前为止，我有：

SELECT
(AVG(t.rating1) + AVG(t.rating2)) / 2 AS total_rating  
AVG(t.rating1) AS rating1,  
AVG(t.rating2) AS rating2,  
---- price statements?? ----  
FROM t  
WHERE 1=1  
AND t.approved = '1'

非常感谢和原谅我的英语

Answer 1

试试这个：

SELECT (AVG(t.rating1) + AVG(t.rating2)) / 2 AS total_rating, 
                AVG(t.rating1) AS rating1, 
                AVG(t.rating2) AS rating2, 
                AVG(IF(cond='1', price, NULL)) price_used, 
                AVG(IF(cond='2', price, NULL)) price_new
FROM t
WHERE 1=1 
 AND t.approved = '1'

编辑：更新了查询以获得所需的结果。

Answer 2

我不认为在If语句中做一个平均值会起作用。但这是IF的语法。

IF（condition，value_to_display_if_true，value_to_display_if_false）

因此，例如，IF（1 = 1，'true'，'false'）对于此列始终显示'true'，因为1确实= 1。

Answer 3

标准SQL，适用于“所有”主要dbms：

select (avg(t.rating1) + avg(t.rating2)) / 2 as total_rating  
      ,avg(t.rating1) as rating1
      ,avg(t.rating2) as rating2
      ,avg(case when cond = '1' then price end) as price_used
      ,avg(case when cond = '2' then price end) as price_new
  from t   
 where t.approved = '1'

单个查询中的MYSQL条件不同

3 个答案: