Question

我有一个简单的文档细分，其中一个属性名称。每个细分可以包含细分类型的 0..N儿童，他们的孩子也可以有细分儿童0..N，所以基本树结构。
从子段到父级创建边缘。

我需要的是在一个命令中使用所有子孙和孙子等获取NodeJS根目录，所以我可以检索这样的内容：

    [
      {
        "@rid": "#1:1",
        "name": "Segment 1",
        "segments": [
          {
            "@rid": "#1:2",
            "name": "Segment 1.1",
            "segments": [
              {
                "@rid": "#1:3",
                "name": "Segment 1.1.1"
              },
              {
                "@rid": "#1:4",
                "name": "Segment 1.1.2"
              }
            ]
          },
          {
            "@rid": "#1:5",
            "name": "Segment 1.2",
            "segments": [
              {
                "@rid": "#1:6",
                "name": "Segment 1.2.1"
              },
              {
                "@rid": "#1:7",
                "name": "Segment 1.2.2"
              } 
            ]
          }
        ]
      },
      {
        "@rid": "#1:8",
        "name": "Segment 2",
        "segments": [
          {
            "@rid": "#1:9",
            "name": "Segment 2.1",
            "segments": [
              {
                "@rid": "#1:10",
                "name": "Segment 2.1.1"
              },
              {
                "@rid": "#1:11",
                "name": "Segment 2.1.2"
              }
            ]
          },
          {
            "@rid": "#1:12",
            "name": "Segment 2.2",
            "segments": [
              {
                "@rid": "#1:13",
                "name": "Segment 2.2.1"
              },
              {
                "@rid": "#1:14",
                "name": "Segment 2.2.2"
              } 
            ]
          }
        ]
      }
    ]

Answer 1

试试这个：

var OrientDB = require('orientjs');
var db = OrientDB({
    host: 'localhost',
    port: 2424,
});
var db = db.use({
    name: 'your db name',
    username: 'your username',
    password: 'your password'
});
db.query('traverse * from Segment')
.then(function (response) {
console.log(response);
});
db.close();

希望有所帮助

此致

Answer 2

function parseSegment(segment) {
    var parsedSegment = {
        id : segment['@rid'].toString(),
        name: segment.name
    };

    if (segment.in_) {
        var children = segment.in_.all().map(edge => parseSegment(edge.out));

        if (children.length) {
            parsedSegment.children = children;
        }
    }

    return parsedSegment;
}

db.select()
    .from('Segment')
    .where('out().size() = 0')
    .fetch({'in_':'-1'})
    .transform(parseSegment)
    .all()
    .then(segments => {
        console.log(JSON.stringify(segments, null, 2));
    })
    .finally(() => {db.close();});

OrientDB：如何在NodeJS中的树结构模型中获取所有文档及其子项（包含所有属性）

2 个答案: