The result should be Date object
结果应该是Date对象
Since the day cannot be 'removed', set it to say, 1st day of the month.
由于当天无法“删除”,因此请将其设置为月份的第1天。
Leaving only Month, Year
只留下一个月,一年
2 个解决方案
#1
23
You can use following constructs to filter the Date
column using either year
or month
:
您可以使用以下构造来使用年份或月份过滤Date列:
.filter(extract('year', Foo.Date) == 2012)
.filter(extract('month', Foo.Date) == 12)
And group_by
is also possible:
而group_by也是可能的:
.group_by(sqlalchemy.func.year(Foo.Date), sqlalchemy.func.month(Foo.Date))
Now I haven't tested it, but I assume that this might be slow because these queries result in full table scans, therefore I suggest you invest some time and learn how to use composite columns.
现在我还没有对它进行测试,但我认为这可能会很慢,因为这些查询会导致全表扫描,因此我建议您花一些时间学习如何使用复合列。
#2
0
if you declare your schema like this,
如果您声明这样的架构,
schema.Column('created', types.TIMESTAMP(), default=now()),
the function now() is defined in the same module which can be like this
函数now()在同一模块中定义,可以是这样的
def now():
now_date = datetime.datetime.now()
return now_date.replace(day=1)
as per group_by requirement, here is a method mentioned.
按照group_by的要求,这是一个提到的方法。
you can modify hour and other field similarly.
您可以类似地修改小时和其他字段。
#1
23
You can use following constructs to filter the Date
column using either year
or month
:
您可以使用以下构造来使用年份或月份过滤Date列:
.filter(extract('year', Foo.Date) == 2012)
.filter(extract('month', Foo.Date) == 12)
And group_by
is also possible:
而group_by也是可能的:
.group_by(sqlalchemy.func.year(Foo.Date), sqlalchemy.func.month(Foo.Date))
Now I haven't tested it, but I assume that this might be slow because these queries result in full table scans, therefore I suggest you invest some time and learn how to use composite columns.
现在我还没有对它进行测试,但我认为这可能会很慢,因为这些查询会导致全表扫描,因此我建议您花一些时间学习如何使用复合列。
#2
0
if you declare your schema like this,
如果您声明这样的架构,
schema.Column('created', types.TIMESTAMP(), default=now()),
the function now() is defined in the same module which can be like this
函数now()在同一模块中定义,可以是这样的
def now():
now_date = datetime.datetime.now()
return now_date.replace(day=1)
as per group_by requirement, here is a method mentioned.
按照group_by的要求,这是一个提到的方法。
you can modify hour and other field similarly.
您可以类似地修改小时和其他字段。