我有一个数据库，其中包含具有上述关系的下表。我想获得每个客户给出的每个产品的平均分数，如所需的输出所示。
“grp”
g_id g_name
1 Ag.f
2 Gh.h
“customers”，其中 c_g_id 引用“grp”中的 g_id
c_id c_name g_id
"1" "DA" "1"
"2" "AŠ" "1"
"3" "EK" "1"
"4" "PK" "1"
"5" "DD" "2"
"6" "AA" "2"
"7" "EE" "2"
“产品”，其中 g_id 引用“grp”中的 g_id
m_id sequence_no g_id name
"1" "1" "1" "product1"
"2" "2" "1" "product2"
"3" "3" "1" "product3"
"4" "1" "2" "p1"
"5" "2" "2" "p2"
“分数”
c_id 引用“客户”中的 c_id m_id 引用“产品”中的 m_id
score_id customer_name c_id m_id score
"1" "DA" "1" "1" "2"
"2" "DA" "1" "2" "2"
"3" "DA" "1" "3" "2"
"4" "AŠ" "2" "1" "3"
"5" "AŠ" "2" "2" "2"
"6" "AŠ" "2" "3" "3"
"7" "EK" "3" "1" "2"
"8" "EK" "3" "2" "3"
"9" "EK" "3" "3" "1"
"10" "PK" "4" "1" "2"
"11" "PK" "4" "2" "3"
"12" "PK" "4" "3" "1"
"13" "DD" "5" "4" "2"
"14" "DD" "5" "5" "2"
"15" "AA" "6" "4" "3"
"16" "AA" "6" "5" "2"
"17" "EE" "7" "4" "2"
"18" "EE" "7" "5" "3"
期望的输出:
sequence_no g_id name avg.score
1 1 "product1" (2+3+2+2)/4
2 1 "product2" (2+2+3+3)/4
3 1 "product3" (2+3+1+1)/4
最佳答案
您可以使用 avg 函数和分组依据
select products.name, products.g_id, avg(score)
from scores
inner join products on products.m_id = scores.m_id
group by products.name, products.g_id
如果您需要一些相关列，最终您可以加入其他表
关于python - SQLite - 查询返回相关记录，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/43310163/