我在一个django应用程序上工作,使用几个复杂的查询返回列组、比率和其他复杂的输出。我跟踪了一些来自web的示例,这些示例帮助我找到了放置它并从返回的查询集中检索数据的最佳方法
但是,我想找到一种方法,将自定义详细信息注入到给定的记录中,例如,在我的例子中,尝试将get_absolute_url()值关联到返回的记录集中
下面是一个返回最常用兴趣的示例,此查询将始终返回一个有限的queryset,是否有方法使用模型的get_absolute_url()值扩展返回的字典?
def most_used_interests(self, limit_by=10):
cursor = connection.cursor()
cursor.execute("""
SELECT
i.name,
i.name_ar,
i.name_en,
ij.interest_id,
SUM (ij.C) item_count
FROM
(
SELECT
C .interest_id,
COUNT (b. ID) C
FROM
bargain_bargain b,
bargain_bargain_bargain_target C
WHERE
b. ID = C .bargain_id
GROUP BY
C .interest_id
UNION
SELECT
x.interest_id,
COUNT (P . ID) C
FROM
promotion_promotion P,
promotion_promotion_promo_target x
WHERE
x.promotion_id = P . ID
GROUP BY
x.interest_id
) ij, list_interest i
WHERE i.id=ij.interest_id
GROUP BY
ij.interest_id,
i.name,
i.name_ar,
i.name_en
ORDER BY
item_count DESC
LIMIT %s
""", [limit_by, ])
desc = cursor.description
if cursor.rowcount:
return [
dict(zip([col[0] for col in desc], row))
for row in cursor.fetchall()
]
return None
最佳答案
我找到了一种更简单的方法来处理这种情况,对于面临类似问题的其他人,我包装查询以始终返回对象,然后使用objects.raw来返回查询
def most_used_interests(self, limit_by=10):
return self.raw("""
SELECT
*
FROM
list_interest l,
(
SELECT
ij.interest_id,
SUM (ij. C) item_count
FROM
(
SELECT
C .interest_id,
COUNT (b. ID) C
FROM
bargain_bargain b,
bargain_bargain_bargain_target C
WHERE
b. ID = C .bargain_id
GROUP BY
C .interest_id
UNION
SELECT
x.interest_id,
COUNT (P . ID) C
FROM
promotion_promotion P,
promotion_promotion_promo_target x
WHERE
x.promotion_id = P . ID
GROUP BY
x.interest_id
) ij
GROUP BY
ij.interest_id
LIMIT %s
) tl
WHERE
l. ID = tl.interest_id
ORDER BY
item_count DESC
""", [limit_by, ]
关于sql - 将get_absolute_url组合成Django中的原始SQL语句,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/18408901/