[英]How do I group these objects into a hash?
I have an event modal, which has a datetime
field titled scheduled_time
. 我有一个事件模式,它有一个名为
scheduled_time
的datetime
字段。 I need to create a hash that has a day name in a certain format ('mon', 'tue' etc) as the key, and the count of events that take place on that day as the value. 我需要创建一个哈希,该哈希以特定格式(“ mon”,“ tue”等)作为日期名称,并以该天发生的事件计数作为值。 How can I do this?
我怎样才能做到这一点?
{
'mon' => 2,
'tue' => 4,
'wed' => 3,
'thu' => 5,
'fri' => 12,
'sat' => 11,
'sun' => 7,
}
I'm using Rails 3.2.0 and Ruby 1.9.2 我正在使用Rails 3.2.0和Ruby 1.9.2
The easiest would be to use count
with a :group
option: 最简单的方法是将
count
与:group
选项一起使用:
h = Model.count(:group => %q{to_char(scheduled_time, 'dy')})
The specific function that you'd GROUP BY would, as usual, depend on the database; 通常,您要GROUP BY的特定功能取决于数据库;具体取决于数据库。 the
to_char
approach above would work with PostgreSQL, with MySQL you could use date_format
and lower
: 上面的
to_char
方法适用于PostgreSQL,而对于MySQL,则可以使用date_format
及lower
:
h = Model.count(:group => %q{lower(date_format(scheduled_time, '%a'))})
For SQLite you'd probably use strftime
with a %w
format and then convert the numbers to strings by hand. 对于SQLite,您可能会使用
%w
格式的strftime
,然后手动将数字转换为字符串。
Note that using Model.count(:group => ...)
will give you a Hash with holes in it: if there aren't any entries in the table for that day then the Hash won't have a key for it. 请注意,使用
Model.count(:group => ...)
将为您提供一个带孔的哈希:如果当天表中没有任何条目,则哈希将没有密钥。 If you really want seven keys all the time then create a Hash with zeros: 如果您确实一直想要七个键,则创建一个带有零的哈希值:
h = { 'mon' => 0, 'tue' => 0, ... }
and then merge the count
results into it: 然后将
count
结果合并到其中:
h.merge!(Model.count(:group => ...))
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