在 TitanDB 中使用 Gremlin 进行排序答案

【问题标题】：Sorting in TitanDB with Gremlin在 TitanDB 中使用 Gremlin 进行排序
【发布时间】：2017-04-26 11:08:22
【问题描述】：

每个顶点和边都有以下属性：

Request [req_id: long, submit_time:long, priority:int, status:char]  
Test_B [tb_id: long, tb_name: string]  
TestSuite [suite_id: long, Suite_Status: char]  
Edge-A [sequence_num: int]  
Edge-B [ sequence_num: int, suite_status: char]

顶点和边之间的关系：

Request----EdgeA----Test-B  
Request----EdgeB----TestSuite  
TestSuite----EdgeC----Test-B

以下查询按升序排序（从请求中提交时间），并将显示 TestSuite 详细信息：

g.V().order().by("submit_time",incr).outE("Edge-B").
  has("suite_status","Q").inV().valueMap()

这是一个示例图：

rj1=graph.addVertex(label,"request","req_id",1138,"status","S","priority",3,"submit_time",1330647978000)
rj2=graph.addVertex(label,"request","req_id",1133,"status","S","priority",2,"submit_time",1330662012000)
rj3=graph.addVertex(label,"request","req_id",1136,"status","S","priority",1,"submit_time",1330652298000)
rj4=graph.addVertex(label,"request","req_id",1135,"status","S","priority",1,"submit_time",1330666768000)

ts1=graph.addVertex(label,"test_suite","test_suite_id",8713)
ts2=graph.addVertex(label,"test_suite","test_suite_id",8718)
ts3=graph.addVertex(label,"test_suite","test_suite_id",8688)
ts4=graph.addVertex(label,"test_suite","test_suite_id",8729)

tb1=graph.addVertex(label,"test_b","tb_id",574)
tb2=graph.addVertex(label,"test_b","tb_id",573)
tb3=graph.addVertex(label,"test_b","tb_id",577)


e1=rj1.addEdge("edgeB",ts1,"sequence",1,"job_suite_status","Q")
e2=rj1.addEdge("edgeB",ts2,"sequence",2,"job_suite_status","Q")
e1=rj2.addEdge("edgeB",ts1,"sequence",1,"job_suite_status","Q")
e2=rj2.addEdge("edgeB",ts3,"sequence",2,"job_suite_status","Q")
e1=rj3.addEdge("edgeB",ts4,"sequence",1,"job_suite_status","Q")
e2=rj3.addEdge("edgeB",ts2,"sequence",2,"job_suite_status","Q")
e1=rj4.addEdge("edgeB",ts3,"sequence",1,"job_suite_status","Q")
e2=rj4.addEdge("edgeB",ts4,"sequence",2,"job_suite_status","Q")

e1=rj1.addEdge("edgeA",tb1,"sequence",1)
e2=rj1.addEdge("edgeA",tb2,"sequence",2)
e3=rj2.addEdge("edgeA",tb1,"sequence",1)
e4=rj2.addEdge("edgeA",tb1,"sequence",2)
e5=rj3.addEdge("edgeA",tb2,"sequence",1)
e6=rj3.addEdge("edgeA",tb3,"sequence",2)
e7=rj4.addEdge("edgeA",tb3,"sequence",2)
e8=rj4.addEdge("edgeA",tb1,"sequence",1)

这里，rj1、rj2、rj4 连接到 tb1(edgeA)：因此对于每个 test_b（例如：tb_id=574），根据 submit_time 对 request(rj1,rj2,rj4) 进行排序，并取 test_suite_id 和序列。

最后我需要打印test_suite_id、sequence和对应的req_id。请帮我解决这个问题。谢谢。

【问题讨论】：

能否提供一些节点的示例和预期结果？
@cattani 我已经用示例数据更新了这个问题。你能帮我解决这个问题吗？
遍历似乎不适用于您的示例数据。

标签： java cassandra titan gremlin tinkerpop3

【解决方案1】：

它有助于使用as() 标记步骤，然后使用select() 准备结果。

g.V().has("tb_id", 574).
  in("edgeA").dedup().as("r").
  order().by("submit_time", Order.incr).as("st").
  outE("edgeB").has("job_suite_status", "Q").as("s").
  inV().as("t").
  select("t", "s", "r", "st").
    by("test_suite_id").by("sequence").by("req_id").by("submit_time")

从 tb_id 574 的测试开始，然后遍历它的请求（标签为r），按提交时间排序请求（标签为st），遍历序列（标签为s），最后遍历测试套件（标签为t）。

使用select() 准备结果输出，使用by() 选择属性：测试套件、序列、请求和提交时间。

【讨论】：