问题描述
我的情况是随机选择参加比赛的参与者，例如2名。并为比赛创建一个优势（用于更新分数，即，获胜者200分，与会人员100分）。
对于获奖者来说，它运行良好。但是，对于其余的顶点（即参与者），winGift edge正在从参与者到竞争者创建多次。
如果在12名参与者中产生2名获胜者，则获胜者应获得200分，其余参与者应获得100分。并且对于每个获胜者和参与者，都应添加新游戏顶点以及属性gameID（应随机生成）。而且对于具有gameID ...的游戏，我正在获得唯一性违反了约束错误。。我如何创建具有不同gameId的新游戏顶点作为赢家和参与者的属性。 如果我在不添加游戏顶点的情况下尝试查询，那么正在为参与者创建多个wonGift边缘
这是查询：
g.withSideEffect('curTime',1599469059825)
.V().hasLabel('Competition').has('competitionId','competition-OY-KzEt_H')
.has('active',true).as('c')
.project('participantCount','winner').
by(select('c').inE('participatedIn').has('batch',1).has('isEligible',true).count()).
by(select('c').inE('participatedIn').has('batch',true).as('pe')
.choose(
has('picked',true),outV().valueMap().fold(),sample(2).property('earlybirdPicked',true).outV().as('winner')
.sideEffect(select('pe').hasNot('picked').property('picked',false))
.sideEffect(
choose(
select('c').has('competitionType','online'),select('winner')
.addE('wonGift').property('typeOfGift','points')
.property('pointsscored',200)
.property('on',1599469059825).to('c')
.select('winner')
.addV('Game')
.property('gameId',`game-${Math.random()}`)
.property('pointsscored',200)
.addE('participates')
.to('winner')
.select('c')
.inE('participatedIn')
.has('batch',1)
.has('isEligible',true)
.has('picked',false)
.outV().as('participant')
.addE('wonGift').property('typeOfGift',100)
.property('on',1599469059825).to('c')
.select('participant')
.addV('Game')
.property('gameId',100)
.addE('participates')
.to('participant')
)
).select('winner').valueMap().fold()
)
)
g.addV('Person').as('1').
property(single,'callingName','Sharath').
property(single,'personId','zefasdafas').
addV('Person').as('2').
property(single,'Raj').
property(single,'sVzqUTNVkar').
addV('Person').as('3').
property(single,'Hima').
property(single,'ajlsdfj').
addV('Competition').as('4').
property(single,'competitionId','ag-competition-sVzVkar').
property(single,'active',true).
property(single,'competitionType','online').
addV('Person').as('5').
property(single,'Ram').
property(single,'personID','asdfasewr').
addV('Person').as('6').
property(single,'shiva').
property(single,'98iejalsd').
addV('Person').as('7').
property(single,'Andhrew').
property(single,'aui;lkj').
addV('Person').as('8').
property(single,'Naveen').
property(single,'naedloli').
addV('Person').as('9').
property(single,'Euler').
property(single,'aek,kndal').
addV('Person').as('10').
property(single,'Rahul').
property(single,'rahoil;lj').
addV('Person').as('11').
property(single,'Vijay').
property(single,'vijIoUik').
addV('Person').as('12').
property(single,'Ravali').
property(single,'raikjlka').
addV('Person').as('13').
property(single,'Sirisha').
property(single,'siriosjdkl').
addE('participatedIn').from('2').to('4').
property('picked',false).
property('isEligible',true).
property('batch',1).addE('participatedIn').
from('1').to('4').property('picked',1).addE('participatedIn').
from('3').to('4').property('picked',1).addE('participatedIn').
from('5').to('4').property('picked',1).addE('participatedIn').
from('6').to('4').property('picked',1).addE('participatedIn').
from('7').to('4').property('picked',1).addE('participatedIn').
from('8').to('4').property('picked',1).addE('participatedIn').
from('9').to('4').property('picked',1).addE('participatedIn').
from('10').to('4').property('picked',1).addE('participatedIn').
from('11').to('4').property('picked',1).addE('participatedIn').
from('12').to('4').property('picked',1).addE('participatedIn').
from('13').to('4').property('picked',1)
预先感谢
解决方法
我可以看到出了什么问题，但是我不知道您要如何解决。这行：
sample(2).property('earlybirdPicked',true).outV().as('winner')
随机选择两个边，然后为每个边choose('c').inE('participatedIn')进行选择，这当然意味着您将那些“ participatedIn”边重复两次，然后两次addE()。如果您更改为sample(3)，则应该使参与者获得100分，甚至可以创建更多次。
一种解决方法是只更新sideEffect()中的“ 200分赢家”，然后再处理“ 100分赢家”。
gremlin> g.withSideEffect('curTime',1599469059825).
......1> V().hasLabel('Competition').
......2> has('competitionId','ag-competition-sVzVkar').
......3> has('active',true).as('c').
......4> project('participantCount','winner').
......5> by(inE('participatedIn').has('batch',1).
......6> has('isEligible',true).count()).
......7> by(inE('participatedIn').has('batch',1).
......8> has('isEligible',true).
......9> choose(has('picked',true),.....10> outV(),.....11> sideEffect(hasNot('picked').property('picked',false)).
.....12> sample(2).property('earlybirdPicked',true).outV().
.....13> sideEffect(choose(select('c').has('competitionType','online'),.....14> addE('wonGift').
.....15> property('typeOfGift','points').
.....16> property('pointsScored',200).
.....17> property('on',1599469059825).to('c')))).
.....18> valueMap().fold()).
.....19> sideEffect(select('c').
.....20> inE('participatedIn').has('batch',1).has('isEligible',true).has('picked',false).
.....21> outV().as('participant').
.....22> addE('wonGift').
.....23> property('typeOfGift','points').
.....24> property('pointsScored',100).
.....25> property('on',1599469059825).to('c'))
==>[participantCount:12,winner:[[callingName:[Sirisha],personID:[siriosjdkl]],[callingName:[Ravali],personID:[raikjlka]]]]
gremlin> g.E().hasLabel('wonGift').elementMap()
==>[id:64,label:wonGift,IN:[id:9,label:Competition],OUT:[id:22,label:Person],pointsScored:100,typeOfGift:points,on:1599469059825]
==>[id:65,OUT:[id:25,on:1599469059825]
==>[id:52,OUT:[id:37,pointsScored:200,on:1599469059825]
==>[id:53,OUT:[id:34,on:1599469059825]
==>[id:54,OUT:[id:28,on:1599469059825]
==>[id:55,OUT:[id:31,on:1599469059825]
==>[id:56,on:1599469059825]
==>[id:57,on:1599469059825]
==>[id:58,OUT:[id:3,on:1599469059825]
==>[id:59,OUT:[id:0,on:1599469059825]
==>[id:60,OUT:[id:6,on:1599469059825]
==>[id:61,OUT:[id:13,on:1599469059825]
==>[id:62,OUT:[id:16,on:1599469059825]
==>[id:63,OUT:[id:19,on:1599469059825]
我认为这可以为您提供正确的边数。我还摆脱了一堆步骤标签和回溯，我认为这使遍历变得更容易阅读。我认为还有更多优化的余地，因为这种方法使您遍历“ participatedIn”边的总数达到了三倍。我想可以将它重组为一次，但这可能会牺牲可读性。您在这里也有一些相当具体的业务逻辑，我试图避免混乱，所以我不想对整体查询结构进行过多修改。