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按日期列出的一列中不同值的SQL计数

[英]SQL count of different values in one column by date

I have a table that looks like: 我有一张表看起来像:

os         date
--         ---
ios        2014-04-35 21:33:33 
android    2014-04-35 21:33:33
ios        2014-04-35 21:33:33
ios        2014-04-35 21:33:33

I want to get an output to be the counts of ios/android in each month. 我想得到一个输出是每个月的ios / android计数。 So like: 所以喜欢:

Year    Month    ios     android
2015    01       20      100
2015    02       400     20

Something like that. 这样的事情。

This is what I have right now. 这就是我现在所拥有的。 It's not quite right though: 虽然不太对劲:

SELECT year(last_modified_date) as cy
     , month(last_modified_date) as cm
     , (SELECT COUNT(*) FROM device_info WHERE os='ios')
     , (SELECT COUNT(*) FROM device_info WHERE os='android')
FROM device_info
WHERE year(last_modified_date) IN (2014,2015)
GROUP BY year(last_modified_date)
       , month(last_modified_date);

Tru this: 事实是,实际上,事实上:

SELECT YEAR(di.date) AS 'Year', 
       MONTH(di.date) AS 'Month', 
       SUM(di.os = 'ios') AS 'ios', 
       SUM(di.os = 'android') AS 'android'
FROM device_info AS di
GROUP BY YEAR(di.date), MONTH(di.date);

You can use conditional aggregation with case expression: 您可以使用条件聚集case表达式:

select year(`last_modified_date`) as `Year`
     , month(`last_modified_date`) as `Month`
     , sum(`os` = 'ios') as `ios`
     , sum(`os` = 'android') as `android`
from `device_info`
where year(`last_modified_date`) in (2014, 2015)
group by year(`last_modified_date`)
       , month(`last_modified_date`);

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