2011-07-03 83 views
2

我需要编写一个查询,打印日常呼叫统计。 我的电话表看起来像这样:select by month

user date 
1  3/6/2011 
1  15/7/2011 

我的结果应该是这样的:

date    count 

1/6/2011   0 
2/6/2011   0 
3/6/2011   1 
... 
30/6/2011   0 

我怎么能做到这一点?

回答

1

与尝试:

SELECT month.date, count(userid) 
FROM generate_series(DATE '2011-06-01', DATE '2011-06-30', '1 day') month 
LEFT JOIN calls ON (month.date = calls.date) 
GROUP BY month.date 
ORDER BY month.date ASC; 

更普遍的解决方案(这样你就可以指定年份和月份为参数):

SELECT month.date, count(userid) 
FROM generate_series(DATE '2011-06-01', DATE '2011-06-01' + INTERVAL '1 month' - INTERVAL '1 day', '1 day') month 
LEFT JOIN calls c ON (month.date = c.date) 
GROUP BY month.date 
ORDER BY month.date ASC; 
+0

谢谢,这帮了我很多 – Rami

0

它只是你正在寻找的一组?

select date, count(*) from tablename 
group by date 
order by date desc 
+1

这将不会显示缺少日期的值 –

+1

我需要加入我的表与generate_series(开始,停止,步间隔) – Rami