Question

这一定是常见的事情，但我被困住了。

我有一些可以简化的数据：

id    user    unixtime
-----------------------
1     dave    1335312057
2     dave    1335312058
3     steve   1335312128

等

到目前为止，我只需要按天聚合，所以我一直在使用：

SELECT
 UNIX_TIMESTAMP(DATE(FROM_UNIXTIME(unixtime))) AS time,
 count(c.user) AS count
FROM core c
GROUP BY DATE(FROM_UNIXTIME(unixtime))

我在DATE和HOUR尝试使用CONCAT，但是不能让它按预期运行 - 任何想法？

Answer 1

SELECT
 DATE(FROM_UNIXTIME(unixtime)) as date,
 HOUR(FROM_UNIXTIME(unixtime)) AS hour,
 count(c.user) AS count
FROM core c
GROUP BY 1,2

如果您希望将小时作为unix时间戳，请将此查询包装起来以获取它：

SELECT UNIX_TIMESTAMP(DATE_ADD(the_date, INTERVAL the_hour HOUR)), the_count
from (select
    DATE(FROM_UNIXTIME(unixtime)) as the_date,
    HOUR(FROM_UNIXTIME(unixtime)) as the_hour,
    count(c.user) AS the_count
    FROM core c
    GROUP BY 1,2 
) x

注意：在列名上使用the_前缀以避免保留字出现问题

Answer 2

尝试使用

SELECT
     UNIX_TIMESTAMP(DATE_ADD(DATE(FROM_UNIXTIME(unixtime)), INTERVAL HOUR(FROM_UNIXTIME(unixtime)) HOUR)) AS TIME,
     COUNT(c.user) AS COUNT
    FROM core c
    GROUP BY DATE(FROM_UNIXTIME(unixtime)), HOUR(FROM_UNIXTIME(unixtime));

希望这会有所帮助......

基于时间戳按小时聚合数据

2 个答案: