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Posted to dev@rocketmq.apache.org by GitBox <gi...@apache.org> on 2021/11/16 11:11:24 UTC

[GitHub] [rocketmq] SignUp-StealWheel opened a new issue #3494: 如何捕获在本地事务执行方法中抛出的异常？

SignUp-StealWheel opened a new issue #3494:
URL: https://github.com/apache/rocketmq/issues/3494


   如下图所示，本地事务执行过程中直接抛出一个RuntimeException。
   ![image](https://user-images.githubusercontent.com/31411237/141974025-db008b8a-55a2-4071-813b-aeb2ce61c30f.png)
   当异常发生后调用方无法捕获到抛出的异常，只接收到了TransactionSendResult，此时TransactionSendResult中的SendStatus为SEND_OK并且LocalTransactionState为UNKNOW，在TransactionSendResult类中没有找到任何和原始异常相关的信息，我应该如何捕获原始异常呢？
   如下图所示逻辑，我的目的是根据原始异常信息来判断给前端返回什么错误信息（例如：service逻辑中的某些参数限制条件不满足抛出的错误），但我只能做到如下图所示的逻辑，在不引入其他组件的前提下有什么好的实现方法吗？
   ![image](https://user-images.githubusercontent.com/31411237/141974714-3f4cff81-c848-4b7a-a189-3e7c6d729dea.png)
   
   


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[GitHub] [rocketmq] Git-Yang commented on issue #3494: 如何捕获在本地事务执行方法中抛出的异常？

Posted by GitBox <gi...@apache.org>.
Git-Yang commented on issue #3494:
URL: https://github.com/apache/rocketmq/issues/3494#issuecomment-970201220


   Maybe you can catch the exception in executeLocalTransaction and put the information you want to get into the custom parameter arg.


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[GitHub] [rocketmq] SignUp-StealWheel commented on issue #3494: 如何捕获在本地事务执行方法中抛出的异常？

Posted by GitBox <gi...@apache.org>.
SignUp-StealWheel commented on issue #3494:
URL: https://github.com/apache/rocketmq/issues/3494#issuecomment-970278666


   > Maybe you can catch the exception in executeLocalTransaction and put the information you want to get into the custom parameter arg.
   
   @Git-Yang Thanks for your answer! I was just thinking how to catch an Exception but i forgot to use the custom parameter arg.


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[GitHub] [rocketmq] odbozhou closed issue #3494: 如何捕获在本地事务执行方法中抛出的异常？

Posted by GitBox <gi...@apache.org>.
odbozhou closed issue #3494:
URL: https://github.com/apache/rocketmq/issues/3494


   


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