通过添加值进行分组 - group by with adding the values

我有这个查询

select ts.name as my_name, ss.step_number, p.specs, p.price, 
ssp.class_id from optional_system_step 
as ss join system as s on s.system_id=ss.system_id join category_description 
as cd on cd.category_id=ss.category_id join optional_system_step_product as 
ssp on ss.system_step_id=ssp.system_step_id join product as p on 
p.product_id=ssp.product_id join product_description as pd on 
pd.product_id=p.product_id join template_step as ts on 
(ts.template_id=s.optional_template_id and ts.step_number=ss.step_number)
where s.system_id = '15'  order by ss.step_number, ssp.class_id;

它返回这个

admin   1       999.0000    1   
admin   1       1349.0000   1   
admin   1       1699.0000   1   
pay 1       479.0000    2   
pay 1       149.0000    2   
pay 1       269.0000    3

看起来不错，但问题是我需要按class_id分组，但在价格字段中，我需要添加三个价格，因此例如，我会返回这两行

admin   1       4047.0000   1   
pay 1   897.0000    2

所以基本上我想把这三个数字加在一起，并在价格字段

中返回该值

将聚合函数SUM()与GROUP BY:一起使用

select ts.name as my_name, ss.step_number, p.specs, SUM(p.price),  ssp.class_id
from optional_system_step  as ss
join system as s on s.system_id=ss.system_id
join category_description  as cd on cd.category_id=ss.category_id
join optional_system_step_product as  ssp on ss.system_step_id=ssp.system_step_id
join product as p on  p.product_id=ssp.product_id
join product_description as pd on  pd.product_id=p.product_id
join template_step as ts on  (ts.template_id=s.optional_template_id and ts.step_number=ss.step_number)
where s.system_id = '15' 
GROUP BY ts.name, ss.step_number, p.spects, ssp.class_id
order by ss.step_number, ssp.class_id;

可能是SUM还是GROUP BY的SUM？

如果按class_id分组，则上面实际上将返回3行，因为您有1,2和3。

我认为您需要的查询如下，但它假设您可以按ts.name、ss.step_number、p.specs和ssp.class_id 进行分组

SELECT
    ts.name AS my_name
,   ss.step_number
,   p.specs
,  SUM( p.price)
,   ssp.class_id
FROM
    optional_system_step AS ss
    JOIN system AS s
    ON s.system_id = ss.system_id
    JOIN category_description AS cd
    ON cd.category_id = ss.category_id
    JOIN optional_system_step_product AS ssp
    ON ss.system_step_id = ssp.system_step_id
    JOIN product AS p
    ON p.product_id = ssp.product_id
    JOIN product_description AS pd
    ON pd.product_id = p.product_id
    JOIN template_step AS ts
    ON ( ts.template_id = s.optional_template_id
         AND ts.step_number = ss.step_number
       )
WHERE
    s.system_id = '15'
GROUP BY
ts.NAME,
ss.step_number,
p.specs,
ssp.class_id
ORDER BY
    ss.step_number
,   ssp.class_id ;

查询的输出与SELECT中的列数不匹配，因此我不确定是否缺少任何内容。

但这应该能解决你的目的：

select ts.name as my_name, ss.step_number, p.specs, SUM(p.price) as price, ssp.class_id 
from optional_system_step as ss 
join system as s on s.system_id=ss.system_id 
join category_description as cd on cd.category_id=ss.category_id 
join optional_system_step_product as ssp on ss.system_step_id=ssp.system_step_id 
join product as p on p.product_id=ssp.product_id 
join product_description as pd on pd.product_id=p.product_id 
join template_step as ts on (ts.template_id=s.optional_template_id and ts.step_number=ss.step_number)
where s.system_id = '15' 
GROUP BY ssp.class_id;

我还想补充一点，您不需要在其他列上使用GROUP BY，因为它们似乎都有相同的值，所以ssp.class_id上的GROUP BY应该可以。

此外，虽然与您的问题没有直接关系，但我认为如果删除category_description和product_description联接，您的查询应该仍然可以正常工作，看起来也会更干净一些。我不能证实这一点，因为我不了解你数据库的结构。