Question

我有两个问题：

在哪里我可以找到GraphSON文件的基本格式，保证由gremlin控制台成功加载？我正在尝试将JSON（大约10-20个字段）转换为可以由gremlin查询的另一个文件，但我实际上找不到有关graphson格式保留的字段的任何相关信息或我应该如何处理ID我导出了他们提供的现代图表，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 ...这样的字段是我手动拥有的这些字段创建？
如果我能够创建JSON，当我启动时，我在哪里告诉服务器将其作为基本图加载？在配置文件或脚本中？

谢谢！阿德里安

{"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}]}}

Answer 1

在哪里我可以找到GraphSON文件的基本格式，保证由gremlin控制台成功加载？

此时有多个版本的GraphSON。您可以在Apache TinkerPop IO Documentation中获取引用。当你写作时，＆＃34;由gremlin控制台成功加载＆＃34;我假设你的意思是GraphSONReader所描述的方法here。因此，您在上面显示的格式是您可以使用的一种形式。尽管您可以在wrapAdjacencyList选项设置为true的情况下构建读取器/写入器，但它不是有效的JSON，它将生成有效的JSON。这是一个例子：

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，您没有获得有效JSON的原因是因为需要为Hadoop和其他分布式处理引擎拆分GraphSON文件的标准格式。因此，它为每个顶点生成一行 - StarGraph格式。

如果我能够创建JSON，当我启动它时，我在哪里告诉服务器将其作为基本图加载？在配置文件或脚本中？

脚本可行。就像TinkerGraph上的gremlin.tinkergraph.graphLocation和gremlin.tinkergraph.graphFormat configuration options一样。

最终，如果你有现有的JSON并且你没有加载数千万个图元素，那么解析它并使用标准的g.addV()和g.addE()方法可能是最简单的。构建图表：

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]]

与上述方法相比，尝试将其转换为GraphSON过于复杂。

普通JSON到GraphSON格式

1 个答案: