通过添加值来分组

如何解决通过添加值来分组

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;

这返回这个

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，例如，我将返回这两行

admin   1       4047.0000   1   
pay 1   897.0000    2

所以基本上我想将三个数字加起来并在价格字段中返回该值

解决方法

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,p.spects,ssp.class_id
order by ss.step_number,ssp.class_id;

, 求和还是与GROUP BY求和？ , 如果您将1,2和3按class_id分组，则上面的内容实际上将返回3行。我认为您需要的查询在下面，但它假定您可以按ts.name，ss.step_number，p.specs和ssp.class_id进行分组

SELECT
    ts.name AS my_name,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,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，因此我不确定是否可能缺少任何内容。但这应该可以解决您的目的：

select ts.name as my_name,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，因为它们似乎都具有相同的值，因此在ssp.class_id上使用GROUP BY应该没问题。另外，尽管与您的问题没有直接关系，但我认为，如果删除category_description和product_description连接，您的查询应该仍然可以正常工作，并且看起来也会更简洁。由于无法理解您的数据库结构，因此无法确认。