ArangoDB - 有关AQL聚合的帮助-Java 学习之路

我有一个这样的图表：

// [user] -answer-> [question]

for u in user
    filter u._id in ['user/foo', 'user/bar']
    for v, e in 1 outbound u graph 'qaGraph'
        return keep(e, '_from', '_to', 'chosen')

输出：

[
  {
    "_from": "user/foo",
    "_to": "question/A",
    "chosen": 0
  },
  {
    "_from": "user/foo",
    "_to": "question/B",
    "chosen": 0
  },
  {
    "_from": "user/foo",
    "_to": "question/C",
    "chosen": 1
  },
  {
    "_from": "user/bar",
    "_to": "question/A",
    "chosen": 0
  },
  {
    "_from": "user/bar",
    "_to": "question/C",
    "chosen": 0
  }
]

这意味着foo和bar已经回答了两个共同的问题（A＆C），但他们只回答了一个问题（A） .

如何编写AQL以以下格式返回相同的信息？

{
  "questions": 2,
  "match": 1
}

我在这里挣扎但没有成功，所以任何帮助都会受到赞赏 .

谢谢！

编辑：我忘了提及所有问题都是多项选择，只有两种选择：0或1.所以 answer.chosen 代表用户的选择 .

1 回答

我最终得到了这个查询：

let data = (
    for u in user
        filter u._id in ['user/foo', 'user/bar']
        for v, e in 1 outbound u graph 'qaGraph'
            collect question = e._to into answers = e.chosen
            filter count(answers) == 2 // questions answered by both users
            let match = sum(answers) != 1 ? 1 : 0 // flags same answer by both users
            return {question, match}
)
return { 
    questions: count(data), 
    matches: sum(data[*].match) 
}

但是，我想找到一个比这简单的解决方案 .

回复于 2024-05-02T12:24:14+08:00

ArangoDB - 有关AQL聚合的帮助

1 回答

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