跨多天和按小时分组的总和计数

Question

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，上午 9 点将是：

{
  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，我如何将它们聚合到hour多天的存储桶中？

如果我输出查询结果，您可以看到19:00for12/12和 的两个单独结果12/13。这两天的计数需要总结为一hour 19:00组：

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' } ]

我目前的查询：

var qry = db.knex
  .select(db.knex.raw("array_agg(t2.id) as session_ids, date_trunc('hour', t2.start_timestamp) as hour"))
  .count('*')
  .from('sessions as t2')
  .groupByRaw("date_trunc('hour', t2.start_timestamp)")
  .orderBy(db.knex.raw("date_trunc('hour', t2.start_timestamp)"));

Answer 1

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将时间戳截断到指定的精度（这意味着这GROUP BY将不起作用，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)

whileEXTRACT获取您要求的特定字段：

SELECT extract('hour' from NOW());
?????????????
? date_part ?
?????????????
?        19 ?
?????????????
(1 row)