java - 使用 Gremlin 在 TitanDB 中排序

Question

每个顶点和边都具有以下属性:

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(请求中的提交时间)，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() 

这是一个示例图表:

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对请求(rj1，rj2，rj4)进行排序，并获取test_suite_id和序列。

最后我需要打印test_suite_id、序列和相应的req_id。请帮我解决这个问题。谢谢。

Answer 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，遍历到其请求(标签为r)，按提交时间排序(标签为st)，遍历到序列(标签为s)，最后遍历到测试套件(标签为t)。

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