Question

我想将3个集合的结果合并为单个集合，但我无法合并最终结果的finishedStatus属性。

最终结果集合：

仅包含唯一身份
每个文档都可以保留创建此状态的计算机
合并finishedStatus然后求和finishedStatus.count
（可选）sort finishedStatus [*]。status

鉴于

收藏A

 {"status": ["1","2","3"], "machine": "A", "finishedStatus": [{"status": "4", "count": 10}, {"status": "5", "count": 1}]}
 {"status": ["4","5","6"], "machine": "A", "finishedStatus": [{"status": "3", "count": 5}, {"status": "5", "count": 11}]}

收藏B

{"status": ["1","2","3"], "machine": "B", "finishedStatus": [{"status": "1", "count": 3}, {"status": "5", "count": 14}]}
{"status": ["2","5","2"], "machine": "B", "finishedStatus": [{"status": "5", "count": 5}, {"status": "3", "count": 5}]}

收藏C

{"status": ["2","5","2"], "machine": "C", "finishedStatus": [{"status": "5", "count": 2}, {"status": "3", "count": 5}]}
{"status": ["3","2","1"], "machine": "C", "finishedStatus": [{"status": "2", "count": 6}, {"status": "4", "count": 7}]}

收集结果

{"status": ["1","2","3"], "machine": ["A", "B"] , "finishedStatus": [{"status": "1", "count": 3}, {"status": "4", "count": 10}, {"status": "5", "count": 15}]}
{"status": ["2","5","2"], "machine": ["B", "C"], "finishedStatus": [{"status": "5", "count": 7}, {"status": "3", "count": 10}]}
{"status": ["3","2","1"], "machine": ["C"], "finishedStatus": [{"status": "2", "count": 6}, {"status": "4", "count": 7}]}
{"status": ["4","5","6"], "machine": ["A"], "finishedStatus": [{"status": "3", "count": 5}, {"status": "5", "count": 11}]}

如何编写AQL INSERT / UPDATE / UPSERT以获得最终结果？

这是我的AQL

FOR doc IN A
UPSERT {"status": doc.status}
INSERT {"status": doc.status, "machine": [doc.machine], "finishedStatus": doc.finishedStatus}
UPDATE {
  "machine": APPEND(OLD.machine, doc.machine, true),
  "finishedStatus": <-- I cannot write AQL to update finishedStatus
     How to write AQL to update this property?
}
IN result

谢谢

Answer 1

这是AQL支持的内联if / else格式：

FOR doc IN A
UPSERT {"status": doc.status}
INSERT {"status": doc.status, "machine": [doc.machine], "finishedStatus": doc.finishedStatus}
UPDATE {
  "machine": APPEND(OLD.machine, doc.machine, true),
  "finishedStatus": (
    OLD.finishedStatus[i].status == doc.finishedStatus[i].status ?
    SUM([OLD.finishedStatus[i].count, doc.finishedStatus[i].count]) : 
    APPEND(OLD.finishedStatus, doc.finishedStatus[i])
  )
}
IN result

我无法在没有您的数据集的情况下对此进行测试，但请注意内联if-then-else格式。

( comparator ? is_true : is_false )

(a == 1 ? 'one' : 'not one')

AQL将评估if-then-else，然后将其替换为＆＃39; true＆＃39;或者＆＃39; false＆＃39;你提供的价值。

如何使用UPSERT合并数组？

1 个答案: