Question

所以我尝试从单个密码查询中返回2组：

鉴于父节点在索引中，那么

父节点的子节目
无子节点的父节点。

我能够得到一个类似的查询，它返回一组中的索引中的父项，以及另一组中的子项，如下所示

START Parents = node:app_fulltext('name:"City"'),
 MATCH Parents-[?:ChildOf]-Apps 
 RETURN collect(Apps.Title), collect(Parents.Name);

==> +--------------------------------------------------------------+
==> | collect(Apps.Title) | collect(Parents.Name)                  |
==> +--------------------------------------------------------------+
==> | ["Empty City 3D"]   | ["Empty City 3D","Empty City 3D Fake"] |
==> +--------------------------------------------------------------+

这接近我想要的。但是，我想从Parents.Name中过滤掉Apps.Title集合中已有Child的项目。

这是我想要的结果集。 “Empty City 3D Fake”没有任何子节点，因此在Parents.Name中返回。

==> +--------------------------------------------------------------+
==> | collect(Apps.Title) | collect(Parents.Name)                  |
==> +--------------------------------------------------------------+
==> | ["Empty City 3D"]   | ["Empty City 3D Fake"]                 |

当我刚刚添加了Where r is null子句时，它什么都没有返回，所以我最终必须添加两个相同的开始父集（Parents和Parents2）才能执行此操作。然而，这看起来非常狡猾，所以我希望有更好的方法。

START Parents = node:app_fulltext('name:"City"'), 
      Parents2 = node:app_fulltext('name:"City"')
      MATCH Parents-[r?:ChildOf]-Children, Parents2-[:ChildOf]-Apps 
      Where r is null 
      return collect(Apps.Title), collect(Parents.Name);


==> +--------------------------------------------------------------+
==> | collect(Apps.Title) | collect(Parents.Name)                  |
==> +--------------------------------------------------------------+
==> | ["Empty City 3D"]   | ["Empty City 3D Fake"]                 |
==> +--------------------------------------------------------------+

Answer 1

你可以尝试一下 -

START Parents = node:app_fulltext('name:"City"'),
 MATCH Parents-[?:ChildOf]-Apps 
 WITH collect(Apps.Title) as myapps, collect(Parents.Name) as myparents
 RETURN myapps, filter(x in parents : not x in myapps) as myfilteredparents

如何使用空过滤器在一个cypher查询中返回2个集合

1 个答案: