我有这个在工作室中运行的查询,对于给定的工作人员,它会返回那些推荐它们的其他工作者。
[工人:V] - > [RecommendedBy:E] - > [工人:V]
我构建SQL以从推荐的E和V返回一些选定的数据。以下作品
SELECT out('RecommendedBy').firstName as recommendedByFirstName,
out('RecommendedBy').lastName as recommendedByLastName,
out('RecommendedBy').@rid as recommendedByRID,
outE('RecommendedBy').recommendationHeadline as headline
FROM Worker WHERE userName = 'paulw';
如何使用@ this.toJSON将对上述adhoc查询的响应转换为单个JSON对象?我可以得到类似下面的东西Ok:
SELECT @this.toJson('rid,version,fetchPlan:in_RecommendedBy:1') FROM Worker WHERE userName = 'paulw';
但不适用于第一个SQL。任何帮助表示感谢!
答案 0 :(得分:0)
我不确定我理解你需要什么,这有用吗?
SELECT @this.toJson() FROM (
SELECT out('RecommendedBy').firstName as recommendedByFirstName,
out('RecommendedBy').lastName as recommendedByLastName,
out('RecommendedBy').@rid as recommendedByRID,
outE('RecommendedBy').recommendationHeadline as headline
FROM Worker WHERE userName = 'paulw')