Question

我有一个UNION语句来合并来自两个不同表（MySQL 5.6）的学生的分数。返回的结果不一致。当我独立运行每条语句时，它们会为所有学生产生正确的结果。当我将它们与UNION结合使用时，除一名学生外，所有其他人都正确。

即使是陌生人，如果我从第一个查询中删除了要添加的任何项目，则它对所有记录的求和都是正确的。

例如，从查询中删除sum(ifnull(owl_tracker.pre_req_points,0))或sum(ifnull(owl_tracker.bonus_points,0))会使查询返回所有学生正确的结果。

select first_name, last_name, location, sum(total_points) as total_points from (
select first_name, last_name, location, (
                sum(ifnull(owl_tracker.pre_req_points,0)) + 
                sum(ifnull(owl_tracker.bonus_points,0)) +
                sum(ifnull(owl_tracker.a_points,0))) AS total_points
from products, students_products, students

在这里拨弄：http://sqlfiddle.com/#!9/7ea891/1

学生A正常工作，但学生B不能正常工作。

Answer 1

默认情况下，UNION是UNION DISTINCT，这意味着任何重复的行都将在子查询中被过滤掉。将其更改为UNION ALL。参见What is the difference between UNION and UNION ALL?

Answer 2

您需要使用全部联合，请尝试以下查询，如果您使用显式联接，则更好

http://sqlfiddle.com/#!9/7ea891/7

select first_name, last_name, location, sum(total_points)
from
(
select first_name, last_name, location, (
                sum(ifnull(owl_tracker.pre_req_points,0)) + 
                sum(ifnull(owl_tracker.bonus_points,0)) +
                sum(ifnull(owl_tracker.a_points,0))) AS total_points
from products left join students_products on products.product_id = students_products.product_id
left join students on students_products.student_id = students.unique_id
left join  owl_tracker on students.unique_id = owl_tracker.student_id
where products.product_type in ('Group_Course','Full_Course') and products.test_date_id = '19' and unique_id in ('4833' ,'4956')
group by first_name, last_name, location

union all        

        select first_name, last_name, location, 
                sum(ifnull(owl.points,0)) AS total_points
from products left join students_products on 
products.product_id = students_products.product_id
left join students on students_products.student_id = students.unique_id
left join owl on students.unique_id = owl.student_id 
where products.product_type  In ('Group_Course' ,'Full_Course') and
        products.test_date_id = '19' and
        unique_id in( '4833' , '4956')
        group by first_name, last_name, location) t_union
        group by first_name, last_name, location
        order by  location, last_name

MySQL Union返回不一致的结果

2 个答案: