题解 | #每个学校的平均年龄和平均绩点及整体情况#
每个学校的平均年龄和平均绩点及整体情况
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#不解释 select "总体" as university ,sum(age)/count(*) avg_age,sum(gpa)/count(*) avg_gpa from user_profile union all select university,sum(age)/count(*) avg_age,sum(gpa)/count(*) avg_gpa from user_profile group by university
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