仔细研究了下,发现sql server里面的explicit transaction(显示事务)仍是有点复杂的。如下是有些总结:html
Commit transaction 会提交全部嵌套的transaction修改。可是若是嵌套的transaction里面有rollback tran to save point, 那么save point以后的部分会revert掉。sql
delete from dbo.numbertable begin tran out1 insert into dbo.numbertable values(1) insert into dbo.numbertable values(2) begin tran inn1 insert into dbo.numbertable values(3) insert into dbo.numbertable values(4) save tran inn1SavePoint insert into dbo.numbertable values(5) rollback tran inn1SavePoint commit tran inn1 commit tran out1
@@TRANCOUNT能够用来记录当前session transaction的个数,对于嵌套的transaction来说,每次begin transaction都让它加一,每次commit tran都会让它减一。因此在语句里面能够经过select @@TRANCOUNT 来检查当前是否在一个transaction里面。若是当前@@TRANCOUNT为0,那调用commit仍是rollback都会出现语句错误。在嵌套的transaction里面,rollback是很特殊的,它会直接把@@TRANCOUNT设置为0。session
begin tran begin tran begin tran print @@trancount rollback tran print @@trancount
对于嵌套的transaction来说,rollback的写法是很特殊。若是嵌套,rollback transaction后面是不能带transaction的name的,要带也只能是最外面的transaction的name。Rollback会抛弃全部嵌套transaction在rollback语句以前的修改。不过Rollback以后的更新依然会提交就是了,缘由在于:rollback以后,@@trancount为0,那么rollback以后的语句就不属于explicit transaction, 属于autocmmit transaction了,自动提交。spa
delete from dbo.numbertable begin tran t1 insert into dbo.numbertable values(1) begin tran t2 insert into dbo.numbertable values(2) rollback tran print 'after rollback in innert transaction, the transaction count is: '+cast(@@trancount, varchar(5)) insert into dbo.numbertable values(3) --commit tran select * from dbo.numbertable
存储过程里面也能够begin transaction,若是调用存储过程的地方也begin transaction,那么这种状况也属于嵌套transaction,若是在存储过程里面rollback,获得的结果和上面同样。可是有一点特殊的地方在与,执行存储过程结束的时候会比较开始执行存储过程的@@trancount和结束时候@@trancount的值,若是不同,Sqlserver会给出一个消息像“Transaction count after EXECUTE indicates that a COMMIT or ROLLBACK TRANSACTION statement is missing. Previous count = 1, current count = 0.”这个给出的消息并不会影响其后的执行。code
CREATE PROCEDURE [dbo].[AddNumber] AS BEGIN begin tran insert into dbo.numbertable values(1) insert into dbo.numbertable values(2) insert into dbo.numbertable values(3) rollback tran END delete from dbo.numbertable begin tran out1 exec dbo.addnumber print @@trancount insert into dbo.numbertable values(3) select * from dbo.numbertable
若是在存储过程里面作rollback了,那到外面再作commit或者rollback都是没有效果的而且会报错,由于嵌套transaction内部的transaction一旦调用了rollback,@@trancount就为0了,在外面commit或rollback就会直接报错。好比以下sp,我想像在最外面rollback,那就出错了,由于sp里面语句rollback了。最后表里面始终会插入值3。server
delete from dbo.numbertable begin tran out1 exec dbo.addnumber print @@trancount insert into dbo.numbertable values(3) rollback tran out1 select * from dbo.numbertable
因此对于嵌套的transaction来说,若是内部transaction一旦rollback,就会给外部的transaction留下一个大坑。为了解决这个为题,有两种解决方案:htm
1.在外部的transaction里面检查@@trancount,若是这个值跟你代码begin tran的时候一致,那说明内部transaction没有rollback,那能够继续commit或者rollback。blog
delete from dbo.numbertable begin tran t1 insert into dbo.numbertable values(1) begin transaction t2 insert into dbo.numbertable values(2) rollback tran if @@trancount = 1 begin insert into dbo.numbertable values(3) commit tran end
2.在全部的内部transaction里面,只能commit,不能rollback。若是必须rollback,那怎么办?save point就能够派上用场了。好比sp改为这样子:事务
ALTER PROCEDURE [dbo].[AddNumber] AS BEGIN begin tran save tran pp insert into dbo.numbertable values(1) insert into dbo.numbertable values(2) insert into dbo.numbertable values(3) rollback tran pp commit tran END begin tran out1 exec dbo.addnumber print @@trancount insert into dbo.numbertable values(3) commit tran out1
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