您需要一个聚合（groupby）和一个sum（t1.score）

SELECT t1.customer, sum(t1.score) as summed_score
  FROM yourTable t1
WHERE EXISTS (SELECT 1 
                FROM yourTable t2 
               WHERE t2.customer = t1.customer 
                 AND t2.product = 'a')
GROUP BY t1.customer

一种选择是使用 HAVING 带聚合的子句：

SELECT customer, SUM(score) AS score
  FROM t
 GROUP BY customer 
HAVING SUM(CASE WHEN product = 'a' THEN 1 END)>0

演示

这只是一个简单的聚合

select customer, sum(score)
from tablename t where exists (select 1 from tablename t1 where t.customer=t2.customer and t1.product='a')
group by customer

现在只需使用group by（）&sum（）

SELECT t1.customer, sum(t1.score) 
FROM  my_table t1 
WHERE EXISTS (SELECT 1 FROM my_table t2 WHERE t2.customer = t1.customer AND t2.product = 'a') 
group by t1.customer;

或

SELECT t1.customer, sum(t1.score) 
FROM  my_table t1 
WHERE EXISTS (SELECT 1 FROM my_table t2 WHERE t2.customer = t1.customer AND t2.product = 'a') 
group by t1.customer 
order by t1.customer;

请使用下面的查询，只需检查同一表的子查询中product='a'的customer列

select customer, sum(score) as score from table_name where customer in
(select customer from table_name where product = 'a')
group by customer;