3

所以有了这个Person，Company例如：

type Person {
    name: string
    work_for: [Company]
}

type Company {
    name: string
    work_for: [Person]
}

work_for: [uid] @reverse @count .
name: string @index(fulltext, term) @count .

查询要查找：All the companies that had any of "John James Sam" with total count of those names per company构造什么？我努力了：

  q(func: anyofterms(hashed, "Jame John Sam")){
        uid
    @groupby work_for{
      name
      count(uid)
    }
  }
}```

but that gives the person and the companies they worked for plus the count which is not the company and the number of people with those names that worked for it. Thanks in advance
4

1 回答 1

2

检查这是否适合您。（让我们用我的样本来了解这个）

{
  var(func: anyofterms(name, "Alexei Jack Jose Zoe Ivy")) @groupby(works_for) {
   a as count(uid)
  }

  q(func: uid(a), orderdesc: val(a)) {
    name
    total_workers : val(a)
    workHere: ~works_for @groupby(name){
      count(uid)
    }
  }
}

结果

此查询并不完美，因为它利用了 GroupBy 中未实现的功能。但是，您可以使用它或期望 GroupBy 功能得到改进。

{
  "data": {
    "q": [
      {
        "name": "CompanyABC",
        "total_workers": 6,
        "workHere": [
          {
            "@groupby": [
              {
                "name": "Ivy",
                "count": 2
              },
              {
                "name": "Jack",
                "count": 2
              },
              {
                "name": "Zoe",
                "count": 2
              }
            ]
          }
        ]
      },
      {
        "name": "The other company",
        "total_workers": 4,
        "workHere": [
          {
            "@groupby": [
              {
                "name": "Alexei",
                "count": 2
              },
              {
                "name": "Jose",
                "count": 2
              }
            ]
          }
        ]
      }
    ]
  }
}

使用此数据集作为参考：https ://tour.dgraph.io/master/schema/2/

我已经对其进行了编辑，它看​​起来像这样：

{
  set {
    _:company1 <name> "CompanyABC" .
    _:company1 <dgraph.type> "Company" .
    _:company2 <name> "The other company" .
    _:company2 <dgraph.type> "Company" .

    _:company1 <industry> "Machinery" .

    _:company2 <industry> "High Tech" .

    _:jack <works_for> _:company1 .
    _:jack <dgraph.type> "Person" .

    _:ivy <works_for> _:company1 .
    _:ivy <dgraph.type> "Person" .

    _:zoe <works_for> _:company1 .
    _:zoe <dgraph.type> "Person" .

    _:jack <name> "Jack" .
    _:ivy <name> "Ivy" .
    _:zoe <name> "Zoe" .
    _:jose <name> "Jose" .
    _:alexei <name> "Alexei" .

    #duplicated

    _:jack2 <name> "Jack" .
        _:jack2 <works_for> _:company1 .
    _:jack2 <dgraph.type> "Person" .

    _:ivy2 <name> "Ivy" .
        _:ivy2 <works_for> _:company1 .
    _:ivy2 <dgraph.type> "Person" .

    _:zoe2 <name> "Zoe" .
        _:zoe2 <works_for> _:company1 .
    _:zoe2 <dgraph.type> "Person" .

    _:jose2 <name> "Jose" .
        _:jose2 <works_for> _:company2 .
    _:jose2 <dgraph.type> "Person" .

    _:alexei2 <name> "Alexei" .
        _:alexei2 <works_for> _:company2 .
    _:alexei2 <dgraph.type> "Person" .

        #duplicated end

    _:jose <works_for> _:company2 .
    _:jose <dgraph.type> "Person" .
    _:alexei <works_for> _:company2 .
    _:alexei <dgraph.type> "Person" .

    _:ivy <boss_of> _:jack .

    _:alexei <boss_of> _:jose .
  }
}
于 2019-11-03T16:02:09.813 回答