将valueMap与match()一起使用 [英] Using valueMap with match()

问题描述

我正在调用JanusGraph远程，默认情况下它将返回ReferenceVertex.为了同样检索属性，我使用了valueMap()，它适用于简单的查询.

I am calling JanusGraph remote, which returns ReferenceVertex by default. In order to retrieve properties as well, I use valueMap(), which works well for simple queries.

但是，在我的用例中，我需要建立一个联接，该联接基于ReferenceVertex可以很好地工作，如下所示:

However, in my use case I need to build a join, which works well based on ReferenceVertex as follows:

// select t1.table2_ID, t2.table2_ID from table1 as t1 inner join table2 as t2 on t1.table2_ID = t2.table2_ID

GraphTraversal<?, ?> t1 = __.start().as("table1").out("relatesTo").hasLabel("table2").as("table2");
GraphTraversal<?, ?> t2 = g.V().hasLabel("table1").match(t1).select("table1", "table2");
List<?> l = t2.toList();

将valueMap添加到遍历中以检索属性时，它将失败.我想包括以下特定属性:

When adding valueMap to the traversal to retrieve the properties it fails. I want to include specific properties as follows:

// select t1.column1, t2.column2 from table1 as t1 inner join table2 as t2 on p.table2_ID = c.table2_ID

GraphTraversal<?, ?> t1 = __.start().as("table1").out("relatesTo").hasLabel("table2").valueMap(true, "column2").as("table2");
GraphTraversal<?, ?> t2 = g.V().hasLabel("table1").valueMap(true, "column1").match(t1).select("table1", "table2");
List<?> l = t2.toList();

-> java.util.HashMap无法转换为org.apache.tinkerpop.gremlin.structure.Element

我是否建立了错误的遍历，或者这是Tinkerpop中的限制/错误?

Did I build the wrong traversal, or is this a limitation / bug in Tinkerpop?

谢谢

推荐答案

您可以调制select()以将valueMap()应用于返回的每一列.这是使用TinkerPop附带的现代玩具图的示例:

You can just modulate the select() to apply valueMap() to each column you return. Here's an example using the modern toy graph that ships with TinkerPop:

gremlin> g.V().as('a').out().as('b').select('a','b').by(valueMap())
==>[a:[name:[marko],age:[29]],b:[name:[lop],lang:[java]]]
==>[a:[name:[marko],age:[29]],b:[name:[vadas],age:[27]]]
==>[a:[name:[marko],age:[29]],b:[name:[josh],age:[32]]]
==>[a:[name:[josh],age:[32]],b:[name:[ripple],lang:[java]]]
==>[a:[name:[josh],age:[32]],b:[name:[lop],lang:[java]]]
==>[a:[name:[peter],age:[35]],b:[name:[lop],lang:[java]]]

因此，在您的情况下，您只需这样做:

so in your case you would just do:

GraphTraversal<?, ?> t1 = __.start().as("table1").out("relatesTo").hasLabel("table2").as("table2");
GraphTraversal<?, ?> t2 = g.V().hasLabel("table1").match(t1).select("table1", "table2").by(__.valueMap());
List<?> l = t2.toList();

也就是说，除非您出于这个问题的目的将代码缩写了一下，否则我不确定在这种情况下是否需要match().似乎您可以将其简化为:

That said, unless you've abbreviated your code a bit for purpose of this question, I'm not sure I see the need for match() in this case. Seems like you could just simplify that to just:

List<?> l = g.V().hasLabel("table1").
              out("relatesTo").hasLabel("table2").as("table2").
              select("table1", "table2").
                by(__.valueMap()).toList();

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