嵌套事务异常Transaction rolled back because it has been marked as rollback-only导致的原因和处理方法
参考:https://fangshixiang.blog.csdn.net/article/details/80445912?utm_medium=distribute.pc_relevant.none-task-blog-searchFromBaidu-2.not_use_machine_learn_pai&depth_1-utm_source=distribute.pc_relevant.none-task-blog-searchFromBaidu-2.not_use_machine_learn_pai
根本原因:手动捕捉了事务中的异常导致事务失效,提交已经标记必须回滚的事务
例如:
@Transactional public void testA(){ mapper.insert(...) try{ testB(); }catch(Exception e){ e.printStackTrace(); } } @Transactional public void testB(){ int a = 1/0; }
testA和testB都需要事务,事务传播属性默认为REQUIRED,调用testA时,创建新事务,调用testB时已经有事务了,则加入testA的事务,当发生异常时,拦截器拦截到testB异常,将共同的事务标记为rollback-only
testA方法中由于使用了try catch,导致事务失效,截器无法捕捉到异常,testA正常提交,提交时发现事务已经被标记成rollback-only了,所以会抛出Transaction rolled back because it has been marked as rollback-only
解决办法:
1. 不要在事务控制的方法中手动try catch 捕捉异常
2. 在try catch中捕捉到异常后继续抛出,让拦截器拦截到
2.在catch中手动回滚 TransactionAspectSupport.currentTransactionStatus().setRollbackOnly(); //手动设置回滚
@Transactional public void testA(){ mapper.insert(...) try{ testB(); }catch(Exception e){ e.printStackTrace();
// 以下方法二选一 //1. 设置手动回滚 TransactionAspectSupport.currentTransactionStatus() .setRollbackOnly();
//2. 继续往外抛出异常
throw new RuntimeException(e.toString); } } @Transactional public void testB(){ int a = 1/0; }