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大事务存在的一些问题
并发数据不一致public Boolean transactionCommit(String userName) {
//查询用户
SysUser sysUser = userMapper.selectUserByUserName(userName,null);
// 堆代码 duidaima.com
transactionTemplate.execute(new TransactionCallbackWithoutResult() {
@Override
protected void doInTransactionWithoutResult(TransactionStatus transactionStatus) {
try {
if (null != sysUser) {
//用户信息状态更新 status更新为1
userMapper.updateStatus(userName);
}
} catch (Exception e){
//回滚
transactionStatus.setRollbackOnly();
}
}
});
//再次查询
SysUser sysUser1 = userMapper.selectUserByUserName(userName,"1");
/log/.info("状态为1的用户信息"+JSON./toJSONString/(sysUser1));
return true;
}
编程式事务的灵活点在于可以控制事务执行方法,运用transactionTemplate类进行事务操作,查询操作可以写在外面,这样查询获取数据的操作就不会进入mysql事务表。List<List<ReceivableFeeSaveDTO>> partition = Lists.partition(receivableFeeSaveDTOList, 50);
CompletableFuture<Object> task1 =CompletableFuture.supplyAsync(() -> {
System.out.println("单号check线程" + Thread.currentThread().getId());
//单号check接口 校验失败抛出异常
return "账单实体信息";
}, executor);
CompletableFuture<Object> task2 = CompletableFuture.supplyAsync(() -> {
System.out.println("收款单生成线程" + Thread.currentThread().getId());
try {
//收款单生成
return “账单编号”;
Thread.sleep(3000);
System.out.println("任务2结束:");
} catch (InterruptedException e) {
e.printStackTrace();
}
}, executor);
//task1、task2 执行完执行task3 ,需要感知task1和task2的执行结果
CompletableFuture<Boolean> future = task1.thenCombineAsync(task2, (t1, t2) -> {
System.out.println("账单金额回写线程" + Thread.currentThread().getId());
// t1 、t2返回判断
//回写返回结果
return ture;
}, executor);