node.js - 如何获取仅包含数据的node-neo4j返回 Node

Question

我正在使用node-neo4j连接到neo4j图形数据库。我注意到，每当我尝试获取所有 Node (例如用户)时，返回的 json 结果都会包含太多我不需要的信息。

这是返回所有用户 Node 的代码:

User.getAll = function (callback) {
    var query = [
       'MATCH (user:User)',
       'RETURN user',
    ].join('\n');

    db.query(query, null, function (err, results) {
        if (err) return callback(err);
        var users = results.map(function (result) {
           return new User(result['user']);
        });
        callback(null, users);
    });
 };

它给了我这些 json 响应；

[
{
    "_node": {
        "_nodeNeo4j": {
            "version": "1.1.0",
            "constructor": "Node"
        },
        "_data": {
            "extensions": {},
            "outgoing_relationships": "http://localhost:7474/db/data/node/13/relationships/out",
            "labels": "http://localhost:7474/db/data/node/13/labels",
            "all_typed_relationships": "http://localhost:7474/db/data/node/13/relationships/all/{-list|&|types}",
            "traverse": "http://localhost:7474/db/data/node/13/traverse/{returnType}",
            "self": "http://localhost:7474/db/data/node/13",
            "property": "http://localhost:7474/db/data/node/13/properties/{key}",
            "properties": "http://localhost:7474/db/data/node/13/properties",
            "outgoing_typed_relationships": "http://localhost:7474/db/data/node/13/relationships/out/{-list|&|types}",
            "incoming_relationships": "http://localhost:7474/db/data/node/13/relationships/in",
            "create_relationship": "http://localhost:7474/db/data/node/13/relationships",
            "paged_traverse": "http://localhost:7474/db/data/node/13/paged/traverse/{returnType}{?pageSize,leaseTime}",
            "all_relationships": "http://localhost:7474/db/data/node/13/relationships/all",
            "incoming_typed_relationships": "http://localhost:7474/db/data/node/13/relationships/in/{-list|&|types}",
            "data": {
                "uid": "53c7a820-f0b4-11e3-af63-28373723792e",
                "name": "user1"
            }
        }
    }
},

有没有办法只返回密码结果的“数据”部分？或者我应该在将结果返回给客户端之前去除node.js服务器中不需要的部分？

谢谢!

Answer 1

通过更改返回的响应，我设法仅返回数据字段

User.getAll = function (callback) {
var query = [
    'MATCH (user:User)',
    'RETURN user',
].join('\n');

db.query(query, null, function (err, results) {
       if (err) return callback(err);
       var users = results.map(function (result) {
          return new User(result['user']['data']);
       });
       callback(null, users);
    });
};

我更改了return new User(result['user']['data'])，它成功返回了响应的“数据”部分。