我正在尝试使用Knex在多天内按小时对记录进行分组。因此,例如,9AM将是:
{
hour: 9AM, // for days 12/11, 12/12, 12/13
count: 10 // 10 Ids total over those days for hour 9AM
}
鉴于此快照中的记录,如何在多天内将它们汇总到hour
个桶中?
如果我输出查询结果,您可以看到19:00
和12/12
的{{1}}的两个单独结果。这两天'需要将计数总计为一个12/13
分组:
hour 19:00
我当前的查询:
ROWS [ anonymous {
session_ids: [ 3200 ],
hour: 2016-12-12T14:00:00.000Z,
count: '1' },
anonymous {
session_ids: [ 3201 ],
hour: 2016-12-12T15:00:00.000Z,
count: '1' },
anonymous {
session_ids: [ 3203, 3202 ],
hour: 2016-12-12T19:00:00.000Z,
count: '2' },
anonymous {
session_ids: [ 3204, 3205 ],
hour: 2016-12-13T19:00:00.000Z, // This count should be aggregated into the `19:00` grouping above
count: '2' } ]
答案 0 :(得分:4)
使用EXTRACT
,而不是date_trunc
:
var qry = db.knex
.select(db.knex.raw("array_agg(t2.id) as session_ids, extract('hour' from t2.start_timestamp) as hour"))
.count('*')
.from('sessions as t2')
.groupByRaw("extract('hour' from t2.start_timestamp)")
.orderBy(db.knex.raw("extract('hour' from t2.start_timestamp)"));
date_trunc
将时间戳截断为指定的精度(意味着GROUP BY
不起作用,因为具有相同“小时”字段的两个时间戳的天数可能仍然不同):
SELECT date_trunc('hour', NOW());
┌────────────────────────┐
│ date_trunc │
├────────────────────────┤
│ 2016-12-18 19:00:00+01 │
└────────────────────────┘
(1 row)
而EXTRACT
获取您要求的特定字段:
SELECT extract('hour' from NOW());
┌───────────┐
│ date_part │
├───────────┤
│ 19 │
└───────────┘
(1 row)