我认为您需要以下sql：

select id, product, price
  from table t1
 where (select count(t2.id) from table t2 where t2.id = t1.id) = 1;

尝试此查询。您应该考虑到您正在使用的dbms的语法来格式化它。。。

SELECT q AS id, product, price
FROM
(SELECT COUNT(id) AS q, product, price
FROM table
GROUP BY id)
WHERE q = 1

我以为你指的是重复的身份证。这是你需要问题中提到的结果的唯一方法。

试试这个：

SELECT my_table.id, product, price
FROM my_table
LEFT JOIN
  (SELECT id,
          count(product) AS p_count
   FROM my_table
   GROUP BY id) AS t ON t.id = my_table.id
WHERE t.p_count = 1

最好使用 not exists :

select t.*
from t
where not exists (select 1
                  from t t2
                  where t2.id = t.id and t2.product <> t.product
                 );

这应该比其他方法更快。它会很好地利用 (id, product) .