max - 如何在 ArangoDB AQL 查询中获取第一个和最后一个条目

Question

我需要有关 ArangoDB AQL 查询的帮助。我有一个事务详细信息集合( EventTran )，它在其父表( Event )上记录更新详细信息。 EventTran属性包括 timestamp ，以及对父项 _id_event 的引用.我正在尝试制定一个查询以返回仅包含第一个和最后一个的数组(由 timestamp )EventTran指定 id_event 的文档.下面是一个例子:

FOR event IN EventTran
    FILTER event._id_event == "Event/167697"
    SORT event.timestamp DESC
    RETURN event

可能返回:

[
  {
    "_key": "214092",
    "_id": "EventTran/214092",
    "_id_event": "Event/167697",
    "timestamp": 1511202637
  },
  {
    "_key": "213958",
    "_id": "EventTran/213958",
    "_id_event": "Event/167697",
    "timestamp": 1511202542
  },
  {
    "_key": "191809",
    "_id": "EventTran/191809",
    "_id_event": "Event/167697",
    "timestamp": 1511118705
  },
  {
    "_key": "167701",
    "_id": "EventTran/167701",
    "_id_event": "Event/167697",
    "timestamp": 1510965562
  }
]

我想要一个查询，该查询将返回一个仅包含第一个和最后一个项目的数组，即第一个日志条目和最近的日志条目:

[
  {
    "_key": "214092",
    "_id": "EventTran/214092",
    "_id_event": "Event/167697",
    "timestamp": 1511202637
  },
  {
    "_key": "167701",
    "_id": "EventTran/167701",
    "_id_event": "Event/167697",
    "timestamp": 1510965562
  }
]

Answer 1

以下无疑不是最好的解决方案
情况，但它确实避免了排序，这可能是最后一件事
除非集合很小，否则你想做。

想法很简单:确定最小值和最大值，然后收集最小项并选择其中之一，对于最大项也是如此。

LET mnmx = (
  FOR x in EventTran
  FILTER event._id_event == "Event/167697"
  COLLECT AGGREGATE mn = MIN(x.timestamp), mx = MAX(x.timestamp)
  RETURN {mn,mx} )

LET mn = mnmx.mn
LET mx = mnmx.mx

LET least = (
  FOR x in EventTran
  FILTER x.timestamp == mn
  COLLECT y=x INTO minimal
  RETURN minimal[0] )

LET greatest = (
  FOR x in EventTran
  FILTER x.timestamp == mx
  COLLECT y=x INTO maximal
  RETURN maximal[0] )

RETURN {least, greatest}

最后一行是 {"least": least, "greatest": best} 的缩写，least和 greatest是具有最少和最大时间戳的项目。