Kotlin Corda：在单个事务中添加多个输出状态

在我们的用例中，我们必须向一个事务添加两个具有不同契约的输出状态。我试图在

with items

中将多个状态添加到事务中。我从RPC调用此流，但它在签名事务时引发异常，如

java.lang.IllegalArgumentException:List有多个元素。

这就是我添加多个输出状态的方式：-

val outputstate1 = IouState1(dataModel1, me, otherParty)
val outputstate2 = IouState2(dataModel2, me, otherParty)
val cmd1 = Command(IouContract1.Commands.Create(), state1.participants.map { it.owningKey })
val cmd2 = Command(IouContract2.Commands.Initiated(), state2.participants.map { it.owningKey })
val txBuilder = TransactionBuilder(TransactionType.General, notary).withItems(state1, state2, cmd1, cmd2)
progressTracker.currentStep = VERIFYING_TRANSACTION

// Verifying the transaction.
txBuilder.toWireTransaction().toLedgerTransaction(serviceHub).verify()

progressTracker.currentStep = SIGNING_TRANSACTION
// Signing the transaction.
val partSignedTx = serviceHub.signInitialTransaction(txBuilder)

progressTracker.currentStep = GATHERING_SIGS
// Send the state to the counterparty, and receive it back with their signature.
val fullySignedTx = subFlow(CollectSignaturesFlow(partSignedTx, GATHERING_SIGS.childProgressTracker()))

// Finalising the transaction.
progressTracker.currentStep = FINALISING_TRANSACTION
// Notarise and record the transaction in both parties' vaults.
return subFlow(FinalityFlow(fullySignedTx, FINALISING_TRANSACTION.childProgressTracker())).single()

堆栈跟踪：-

net.corda.core.flows.FlowException: java.lang.IllegalArgumentException: List has more than one element.
    at net.corda.core.flows.SignTransactionFlow.call(CollectSignaturesFlow.kt:202) ~[corda-core-0.14.0.jar:?]
    at net.corda.core.flows.SignTransactionFlow.call(CollectSignaturesFlow.kt:175) ~[corda-core-0.14.0.jar:?]
    at net.corda.core.flows.FlowLogic.subFlow(FlowLogic.kt:166) ~[corda-core-0.14.0.jar:?]
    at com.dwlabcorda.salescontract.LOCReceiverFlow.call(LOCContractFlow.kt:107) ~[tradefinancecorda-0.1.jar:?]
    at com.dwlabcorda.salescontract.LOCReceiverFlow.call(LOCContractFlow.kt:95) ~[tradefinancecorda-0.1.jar:?]
    at net.corda.node.services.statemachine.FlowStateMachineImpl.run(FlowStateMachineImpl.kt:101) [corda-node-0.14.0.jar:?]
    at net.corda.node.services.statemachine.FlowStateMachineImpl.run(FlowStateMachineImpl.kt:33) [corda-node-0.14.0.jar:?]
    at co.paralleluniverse.fibers.Fiber.run1(Fiber.java:1067) [quasar-core-0.7.6-jdk8.jar:0.7.6]
    at co.paralleluniverse.fibers.Fiber.exec(Fiber.java:767) [quasar-core-0.7.6-jdk8.jar:0.7.6]
    at co.paralleluniverse.fibers.RunnableFiberTask.doExec(RunnableFiberTask.java:100) [quasar-core-0.7.6-jdk8.jar:0.7.6]
    at co.paralleluniverse.fibers.RunnableFiberTask.run(RunnableFiberTask.java:91) [quasar-core-0.7.6-jdk8.jar:0.7.6]
    at java.util.concurrent.Executors$RunnableAdapter.call(Unknown Source) [?:1.8.0_144]
    at java.util.concurrent.FutureTask.run(Unknown Source) [?:1.8.0_144]
    at java.util.concurrent.ScheduledThreadPoolExecutor$ScheduledFutureTask.access$201(Unknown Source) [?:1.8.0_144]
    at java.util.concurrent.ScheduledThreadPoolExecutor$ScheduledFutureTask.run(Unknown Source) [?:1.8.0_144]
    at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source) [?:1.8.0_144]
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source) [?:1.8.0_144]
    at net.corda.node.utilities.AffinityExecutor$ServiceAffinityExecutor$1$thread$1.run(AffinityExecutor.kt:69) [corda-node-0.14.0.jar:?]
Caused by: java.lang.IllegalArgumentException: List has more than one element.
    at kotlin.collections.CollectionsKt___CollectionsKt.single(_Collections.kt:471) ~[kotlin-stdlib-1.1.1.jar:1.1.1]
    at com.dwlabcorda.salescontract.LOCReceiverFlow$call$signTransactionFlow$1.checkTransaction(LOCContractFlow.kt:100) ~[tradefinancecorda-0.1.jar:?]
    at net.corda.core.flows.SignTransactionFlow.call(CollectSignaturesFlow.kt:199) ~[corda-core-0.14.0.jar:?]
    ... 17 more

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可能您在某个契约中使用了.single（）强制转换，该契约现在引发异常

这应该没问题。我怀疑你的问题在别处。你能发布流的其余部分吗？更新了整个流的代码..谢谢-你能添加完整的堆栈跟踪吗？因为我没有看到上面有什么不好的地方，我猜问题可能在合同代码中（CollectSignaturesFlow将调用该代码）。正如Stefano所说，检查您是否正在对合同中的输出调用

single

。是的，我使用了val out=tx.outputsOfType（）.single（）“发送方和接收方不能是同一个实体。”使用（out.party1！=out.party2）…如何更改它..get（0）和get（1）抛出异常..您不能使用.single（）然后创建两个IoState类型的输出，首先您必须选择是否可以输出多个IoState或仅输出一个IoState

get（0）

不应引发异常。您看到的bakriOnFire出现了什么异常？@bakriOnFire您必须对单个输入执行检查，然后转换为单个输入。如果你不明白，你可以pm我