普通 JSON 到 GraphSON 格式
Normal JSON to GraphSON format
我有两个问题：
我在哪里可以真正找到 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 （大约有 10-20 个字段）转换为另一个可以由 gremlin 查询的文件，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如何应该处理 ID 等。我导出了他们提供的现代图形，它甚至不是有效的 JSON（多个 JSON 根元素），而是 JSONs [1] 的列表我也看到像 outE、inE 这样的字段……这些字段是我必须手动创建的吗？
如果我能够创建 JSON，我应该在哪里告诉服务器在启动时将其加载为基础图形？在配置文件中还是在脚本中？
谢谢！
阿德里安
[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}]}}
Where I can actually find the basic format for a GraphSON file, that is guaranteed to be successfully loaded by the gremlin console?
此时GraphSON有多个版本。您可以在 Apache TinkerPop IO Documentation. When you write, "successfully loaded by the gremlin console" I assume that you mean with the GraphSONReader methods described here 中获得参考。当然，您上面显示的格式是您可以使用的一种形式。如您所见，它是无效的 JSON，尽管您可以在将 wrapAdjacencyList 选项设置为 true 的情况下构建 reader/writer，并且它将生成有效的 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 GraphSON 文件的标准格式需要为 Hadoop 和其他分布式处理引擎拆分。因此它为每个顶点生成一条线 - StarGraph 格式。
If I am able to create the JSON, where do I tell the server to load it as the base graph when I start it? In the config file or in the script?
一个脚本就可以了。 TinkerGraph 上的 gremlin.tinkergraph.graphLocation 和 gremlin.tinkergraph.graphFormat configuration options 也是如此。
最终，如果您有现有的 JSON 并且您没有加载数千万个图形元素，那么解析它并使用标准 g.addV() 和 [=19 可能是最简单的=] 构建图表的方法：
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。
我有两个问题：
我在哪里可以真正找到 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 （大约有 10-20 个字段）转换为另一个可以由 gremlin 查询的文件，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如何应该处理 ID 等。我导出了他们提供的现代图形，它甚至不是有效的 JSON（多个 JSON 根元素），而是 JSONs [1] 的列表我也看到像 outE、inE 这样的字段……这些字段是我必须手动创建的吗？
如果我能够创建 JSON，我应该在哪里告诉服务器在启动时将其加载为基础图形？在配置文件中还是在脚本中？
谢谢！ 阿德里安
[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}]}}
Where I can actually find the basic format for a GraphSON file, that is guaranteed to be successfully loaded by the gremlin console?
此时GraphSON有多个版本。您可以在 Apache TinkerPop IO Documentation. When you write, "successfully loaded by the gremlin console" I assume that you mean with the GraphSONReader methods described here 中获得参考。当然，您上面显示的格式是您可以使用的一种形式。如您所见，它是无效的 JSON，尽管您可以在将 wrapAdjacencyList 选项设置为 true 的情况下构建 reader/writer，并且它将生成有效的 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 GraphSON 文件的标准格式需要为 Hadoop 和其他分布式处理引擎拆分。因此它为每个顶点生成一条线 - StarGraph 格式。
If I am able to create the JSON, where do I tell the server to load it as the base graph when I start it? In the config file or in the script?
一个脚本就可以了。 TinkerGraph 上的 gremlin.tinkergraph.graphLocation 和 gremlin.tinkergraph.graphFormat configuration options 也是如此。
最终，如果您有现有的 JSON 并且您没有加载数千万个图形元素，那么解析它并使用标准 g.addV() 和 [=19 可能是最简单的=] 构建图表的方法：
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。