【发布时间】:2019-06-04 19:39:47
【问题描述】:
我有一个看起来像这样的表:
usr_id query_ts
12345 2019/05/13 02:06
123444 2019/05/15 04:06
123444 2019/05/16 05:06
12345 2019/05/16 02:06
12345 2019/05/15 02:06
它包含运行查询时的用户 ID。表中的每个条目代表在给定时间戳运行 1 个查询的 ID。
我正在尝试制作这个:
usr_id day_1 day_2 … day_30
12345 31 13 15
123444 23 41 14
我想显示每个 ID 在过去 30 天内每天运行的查询数,如果当天没有运行查询,则为 0。
这是我提出的查询的一部分,
SELECT
t1.usr_id,
case when t1.count_day_1 is null then 0 else t1.count_day_1 end as day_1,
case when t2.count_day_2 is null then 0 else t2.count_day_2 end as day_2
FROM
(SELECT usr_id, DAY(from_unixtime(unix_timestamp(query_ts ,"yyyy/MM/dd"), "yyyy-MM-dd")) as day_1,
COUNT( DAY(from_unixtime(unix_timestamp(query_ts ,"yyyy/MM/dd"), "yyyy-MM-dd"))) as count_day_1
FROM db.table
WHERE
DAY(from_unixtime(unix_timestamp(query_ts ,"yyyy/MM/dd"), "yyyy-MM-dd")) = 1
AND
from_unixtime(unix_timestamp(query_ts ,"yyyy/MM/dd"), "yyyy-MM-dd")
BETWEEN date_sub(from_unixtime(unix_timestamp()), 30)
AND from_unixtime(unix_timestamp())
GROUP BY usr_id, day_1) t1
LEFT JOIN
(SELECT usr_id, DAY(from_unixtime(unix_timestamp(query_ts ,"yyyy/MM/dd"), "yyyy-MM-dd")) as day_2,
COUNT( DAY(from_unixtime(unix_timestamp(query_ts ,"yyyy/MM/dd"), "yyyy-MM-dd"))) as count_day_2
FROM db.table
WHERE
DAY(from_unixtime(unix_timestamp(query_ts ,"yyyy/MM/dd"), "yyyy-MM-dd")) = 2
AND
from_unixtime(unix_timestamp(query_ts ,"yyyy/MM/dd"), "yyyy-MM-dd")
BETWEEN date_sub(from_unixtime(unix_timestamp()), 30)
AND from_unixtime(unix_timestamp())
GROUP BY usr_id, day_2) t2
ON (t1.usr_id = t2.usr_id)
ORDER BY t1.usr_id;
这很好用,它显示了前 2 天每天运行的查询数,并将 NULL 替换为 0。
问题是要让它工作 30 天,我必须使用 30 个 LEFT JOIN,这会在集群上占用约 400GB+ 的内存。
有更简单的方法吗?
【问题讨论】:
标签: sql hadoop hive query-optimization hiveql