Python peewee select()返回SQL查询,而不是实际数据
我尝试将两列中的值相加,并按天截断日期字段。我构造了SQL查询来实现这一点(这是可行的): 但当我试图将其格式化为peewee时,我遇到了一些问题:Python peewee select()返回SQL查询,而不是实际数据,python,sql,postgresql,peewee,Python,Sql,Postgresql,Peewee,我尝试将两列中的值相加,并按天截断日期字段。我构造了SQL查询来实现这一点(这是可行的): 但当我试图将其格式化为peewee时,我遇到了一些问题: Bike_Count.select(fn.date_trunc('day', Bike_Count.date).alias('Day'), fn.SUM(Bike_Count.fremont_bridge_nb).alias('Sum_NB'), fn.SUM(Bike_Count.fremont_bridge_sb).alias('Sum_SB'
Bike_Count.select(fn.date_trunc('day', Bike_Count.date).alias('Day'),
fn.SUM(Bike_Count.fremont_bridge_nb).alias('Sum_NB'),
fn.SUM(Bike_Count.fremont_bridge_sb).alias('Sum_SB'))
.group_by('Day').order_by('Day')
我没有收到任何错误,但当我打印出存储在其中的变量时,它显示:
<class 'models.Bike_Count'> SELECT date_trunc(%s, "t1"."date") AS
Day, SUM("t1"."fremont_bridge_nb") AS Sum_NB,
SUM("t1"."fremont_bridge_sb") AS Sum_SB FROM "bike_count" AS t1 ORDER
BY %s ['day', 'Day']
如果您只是在groupby/orderby中插入一个字符串,Peewee将尝试将其参数化为一个值。这是为了避免SQL注入haxx 要解决此问题,您可以在
groupby()
和order\u by()调用中使用SQL('Day')
代替'Day'
另一种方法是将函数调用粘贴到GROUP BY
和ORDER BY
中。以下是您将如何做到这一点:
day = fn.date_trunc('day', Bike_Count.date)
nb_sum = fn.SUM(Bike_Count.fremont_bridge_nb)
sb_sum = fn.SUM(Bike_Count.fremont_bridge_sb)
query = (Bike_Count
.select(day.alias('Day'), nb_sum.alias('Sum_NB'), sb_sum.alias('Sum_SB'))
.group_by(day)
.order_by(day))
或者,如果您愿意:
query = (Bike_Count
.select(day.alias('Day'), nb_sum.alias('Sum_NB'), sb_sum.alias('Sum_SB'))
.group_by(SQL('Day'))
.order_by(SQL('Day')))
谢谢,这帮了大忙。我还对如何访问数据感到有点困惑,最后用for循环迭代查询。
day = fn.date_trunc('day', Bike_Count.date)
nb_sum = fn.SUM(Bike_Count.fremont_bridge_nb)
sb_sum = fn.SUM(Bike_Count.fremont_bridge_sb)
query = (Bike_Count
.select(day.alias('Day'), nb_sum.alias('Sum_NB'), sb_sum.alias('Sum_SB'))
.group_by(day)
.order_by(day))
query = (Bike_Count
.select(day.alias('Day'), nb_sum.alias('Sum_NB'), sb_sum.alias('Sum_SB'))
.group_by(SQL('Day'))
.order_by(SQL('Day')))