将普通的JSON转换为GraphSON格式

Question

将普通的JSON转换为GraphSON格式

javascriptjsongremlingremlin-servergraphson

5

我有两个问题：

我在哪里可以找到保证能被Gremlin控制台成功加载的GraphSON文件的基本格式？我正在尝试将一个具有10-20个字段的JSON转换为另一个可由Gremlin查询的文件，但实际上我找不到有关图形格式保留字段或如何处理ID等信息的相关信息。我导出了他们提供的现代图表，但它甚至不是一个有效的JSON（多个JSON根元素），而是一个JSON列表[1]。我还看到像outE、inE这样的字段...这些字段是我手动创建的吗？
如果我能创建JSON，我在哪里告诉服务器在启动时将其作为基础图形加载？在配置文件中还是脚本中？

谢谢！ Adrian

[1] https://pastebin.com/drwXhg5k

{"id":1,"label":"person","outE":{"created":[{"id":9,"inV":3,"properties":{"weight":0.4}}],"knows":[{"id":7,"inV":2,"properties":{"weight":0.5}},{"id":8,"inV":4,"properties":{"weight":1.0}}]},"properties":{"name":[{"id":0,"value":"marko"}],"age":[{"id":1,"value":29}]}}
{"id":2,"label":"person","inE":{"knows":[{"id":7,"outV":1,"properties":{"weight":0.5}}]},"properties":{"name":[{"id":2,"value":"vadas"}],"age":[{"id":3,"value":27}]}}
{"id":3,"label":"software","inE":{"created":[{"id":9,"outV":1,"properties":{"weight":0.4}},{"id":11,"outV":4,"properties":{"weight":0.4}},{"id":12,"outV":6,"properties":{"weight":0.2}}]},"properties":{"name":[{"id":4,"value":"lop"}],"lang":[{"id":5,"value":"java"}]}}
{"id":4,"label":"person","inE":{"knows":[{"id":8,"outV":1,"properties":{"weight":1.0}}]},"outE":{"created":[{"id":10,"inV":5,"properties":{"weight":1.0}},{"id":11,"inV":3,"properties":{"weight":0.4}}]},"properties":{"name":[{"id":6,"value":"josh"}],"age":[{"id":7,"value":32}]}}
{"id":5,"label":"software","inE":{"created":[{"id":10,"outV":4,"properties":{"weight":1.0}}]},"properties":{"name":[{"id":8,"value":"ripple"}],"lang":[{"id":9,"value":"java"}]}}
{"id":6,"label":"person","outE":{"created":[{"id":12,"inV":3,"properties":{"weight":0.2}}]},"properties":{"name":[{"id":10,"value":"peter"}],"age":[{"id":11,"value":35}]}}

- Adrian Pop

请提供您想要转换的原始JSON文件，即使晚些时候我可能会有不同的方法可用。 - Wolfgang Fahl

1个回答

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- stephen mallette · Accepted Answer

在哪里可以找到保证能够被Gremlin控制台成功加载的GraphSON文件的基本格式？目前有多个版本的GraphSON。您可以在Apache TinkerPop的IO文档IO Documentation中找到参考资料。当您写道“成功加载到Gremlin控制台”时，我假设您是指使用此处描述的GraphSONReader方法。如果是这样的话，那么您展示的格式是一种可用的形式。正如您所看到的，它不是有效的JSON格式，但您可以使用设置为true的wrapAdjacencyList选项构建读取器/编写器，并且它将生成有效的JSON。以下是一个例子：

gremlin> graph = TinkerFactory.createModern();
==>tinkergraph[vertices:6 edges:6]
gremlin> writer =  graph.io(IoCore.graphson()).writer().wrapAdjacencyList(true).create()
==>org.apache.tinkerpop.gremlin.structure.io.graphson.GraphSONWriter@24a298a6
gremlin> os = new FileOutputStream('wrapped-adjacency-list.json')
==>java.io.FileOutputStream@6d3c232f
gremlin> writer.writeGraph(os, graph)
gremlin> os.close()
gremlin> newGraph = TinkerGraph.open()
==>tinkergraph[vertices:0 edges:0]
gremlin> ins = new FileInputStream('wrapped-adjacency-list.json')
==>java.io.FileInputStream@7435a578
gremlin> reader = graph.io(IoCore.graphson()).reader().unwrapAdjacencyList(true).create()
==>org.apache.tinkerpop.gremlin.structure.io.graphson.GraphSONReader@63da207f
gremlin> reader.readGraph(ins, newGraph)
gremlin> newGraph
==>tinkergraph[vertices:6 edges:6]

默认情况下不会得到有效的JSON，这是因为GraphSON文件的标准格式需要支持Hadoop和其他分布式处理引擎的可拆分性。因此，它会产生每个顶点的一行-一个StarGraph格式。

如果我能创建JSON，那么我在哪里告诉服务器在启动时将其作为基础图形加载？在配置文件中还是脚本中？

脚本可以工作。在TinkerGraph上，gremlin.tinkergraph.graphLocation 和gremlin.tinkergraph.graphFormatconfiguration options也可以使用。

但总的来说，如果你已经有JSON并且没有加载数百万个图元素，最简单的方法可能就是解析它并使用标准的g.addV()和g.addE()方法构建图形。

gremlin> import groovy.json.*
==>org.apache.tinkerpop.gremlin.structure.*,...
gremlin> graph = TinkerGraph.open()
==>tinkergraph[vertices:0 edges:0]
gremlin> g = graph.traversal()
==>graphtraversalsource[tinkergraph[vertices:0 edges:0], standard]
gremlin> jsonSlurper = new JsonSlurper()
==>groovy.json.JsonSlurper@53e3a87a
gremlin> object = jsonSlurper.parseText('[{ "name": "John Doe" }, { "name" : "Jane Doe" }]')
==>[name:John Doe]
==>[name:Jane Doe]
gremlin> object.each {g.addV('name',it.name).iterate() }
==>[name:John Doe]
==>[name:Jane Doe]
gremlin> g.V().valueMap()
==>[name:[John Doe]]
==>[name:[Jane Doe]]

尝试将其转换为GraphSON相对于上述方法来说过于复杂。