AQL

Question

假设我有以下文件：

{"_key": "1", "name": "George Washington"}
{"_key": "2", "name": "George Washington"}
{"_key": "3", "name": "John Adams"}
{"_key": "4", "name": "Thomas Jefferson"}
{"_key": "5", "name": "George Washington"}
{"_key": "6", "name": "Thomas Jefferson"}

我想编写一个 AQL 语句，returns 文档的键按名称分组，但前提是该名称出现不止一次。

所以我想要的输出是：

[["1", "2", "5"], ["4", "6"]]

到目前为止我想出了

FOR doc IN documents
    LET key = doc._key
    COLLECT name = doc.name INTO groups KEEP key
    RETURN (FOR g IN groups RETURN g["key"])

这个returns:

[["1", "2", "5"], ["3"], ["4", "6"]]

如何修改 AQL 命令以仅获取包含两个或更多条目的数组？

Answer 1

已解决：

FOR doc IN documents
    LET key = doc._key
    COLLECT name = doc.name INTO groups KEEP key
    LET groups2 = (FOR group IN groups RETURN group["key"])
    FILTER LENGTH(groups2) >= 2
    RETURN groups2

Answer 2

另一种可能性（由于不涉及子查询，可能更有效）：

FOR doc IN documents
  LET key = doc._key     
  COLLECT name = doc.name INTO groups KEEP key 
  LET keys = groups[*].key 
  FILTER LENGTH(keys) > 1 
  RETURN keys

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