blocks|key|3129739|text|SQL事务，就像所有数据库平台上的事务一样，将数据隔离以涵盖整个ACID缩写(原子、一致、隔离和持久)。所以答案是肯定的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3129740|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

SQL transactions, like transactions on all database platforms, put the data in isolation to cover the entire ACID acronym (atomic, consistent, isolated and durable). So the answer is yes.

blocks|key|1514313|text|原子意味着事务中的操作要么发生，要么不发生。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1514314|如果1,000条语句中有一条语句失败，则事务内的操作将提交none。事务中的语句样本越小--比如100个--意味着可以提交导致错误的100个块(比如501个)(前400个；500个块不提交，600%2B块提交)。|1514315|1514316|，但这是否也意味着，当我完成操作时，将保证这些行是连续的？|blockquote|1514317|1514318|1514319|你必须提供更多关于你在事务中所做的事情的上下文，才能是“顺序的”。|1514320|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|Q|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|R|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|S|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|T|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|U|8|@]|9|@]|A|$]]]|M|$]]

Atomic means the operation(s) within the transaction with either occur, or they don't.

If one of the 1,000 statements fails, none of the operations within the transaction will commit. The smaller the sample of statements within a transaction -- say 100 -- means that the blocks of 100 leading up to the error (say at the 501st) can be committed (the first 400; the 500 block won't, and the 600+ blocks will).

<blockquote>
 But does it also mean that those rows would be guaranteed to be sequential when I'm done?
</blockquote>

You'll have to provide more context about what you're doing in a transaction to be "sequential".

blocks|key|3129790|text|事务保证原子性。这就是问题所在。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3129791|你的问题是，在你做完插入之后，它们只是“连续的”，直到下一个东西出现并接触到一个新的记录。|3129792|如果流程中的另一个步骤要求它们仍然是连续的，那么该步骤也需要在您的原始事务中。|3129793|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

A transaction guarantees atomicity. That is the point. 

You problem is that after you do the insert, they are only "Sequential" until the next thing comes along and touches one of the new records.

If another step in you process requires them to still be sequential then that step, too, needs to be within your original transaction.

blocks|key|3129837|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3129838|，但这是否也意味着，当我完成操作时，将保证这些行是连续的？|blockquote|3129839|3129840|3129841|不是的。这与事务无关，因为您所要求的根本不存在:关系表没有顺序，要求“连续行”是错误的问题。您可以将问题重新表述为“更新的1000行是否包含从1到1000的整个序列，没有间隙”？很可能是这样的，但事实是，根据您进行更新的方式，可能会有差距。这些间隙不会出现，因为更新后的行在提交之前被修改，而是因为更新将是无操作(不会更新任何行)，这是读-修改-写回类型的更新的常见问题(由于并发操作，行在读和写回之间“消失”)。|3129842|为了更准确地回答您的问题，您的代码是否正确，您必须发布执行更新的确切代码，以及确切的表结构，包括所有索引。|3129843|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|L|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|M|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|N|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|O|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|P|7|@]|8|@]|9|$]]|$1|H|3|I|4|5|6|Q|7|@]|8|@]|9|$]]|$1|J|3|-4|4|5|6|R|7|@]|8|@]|9|$]]]|K|$]]

<blockquote>
 But does it also mean that those rows would be guaranteed to be sequential when I'm done?
</blockquote>

No. This has nothing to do with transactions, because what you're asking for simply doesn't exists: relational tables have no order an asking for 'sequential rows' is the wrong question to ask. You can rephrase the question as 'will the 1000 updated rows contain the entire sequence from 1 to 1000, w/o gaps' ? Most likely yes, but the truth of the matter is that there could be gaps depending on the way you do the updates. Those gaps would not appear because updated rows are modified after the update before commit, but because the update will be a no-op (will not update any row) which is a common problem of read-modify-write back type of updates ( the row 'vanishes' between the read and the write-back due to concurrent operations).

To answer your question more precisely whether your code is correct or not you have to post the exact code you're doing the update with, as well as the exact table structure, including all indexes.

blocks|key|3414667|text|这两点是不相关的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3414668|顺序|3414669|如果您插入1到1000之间的值，它将是连续的，并带有WHERE和ORDER+BY，以限制您在某些列中只能插入这1000行。除非有副本，否则需要一个唯一的约束|3414670|如果你依赖于一个身份，就不能保证：Do+Inserted+Records+Always+Receive+Contiguous+Identity+Values。|offset|length|3414671|原子性|3414672|所有事务都是原子的：|3414673|3414674|Is+neccessary+to+encapsulate+a+single+merge+statement+(with+insert,+delete+and+update)+in+a+transaction?|unordered-list-item|3414675|SQL+Server+and+connection+loss+in+the+middle+of+a+transaction|3414676|Does+it+delete+partially+if+execute+a+delete+statement+without+transaction?|3414677|3414678|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/q/4219201/27535|1|https://stackoverflow.com/questions/5117288/is-neccessary-to-encapsulate-a-single-merge-statement-with-insert-delete-and-up/5117330#5117330|2|https://stackoverflow.com/questions/4978372/sql-server-and-connection-loss-in-the-middle-of-a-transaction/4978511#4978511|3|https://stackoverflow.com/questions/6502588/does-it-delete-partially-if-execute-a-delete-statement-without-transaction/6502695#6502695^0|0|0|0|H|1P|0|0|0|0|0|0|2W|1|0|0|1P|2|0|0|23|3|0|0^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1B|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|1C|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|1D|8|@]|9|@$H|1E|I|1F|1|1G]]|A|$]]|$1|J|3|K|5|6|7|1H|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1I|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]|$1|O|3|P|5|Q|7|1K|8|@]|9|@$H|1L|I|1M|1|1N]]|A|$]]|$1|R|3|S|5|Q|7|1O|8|@]|9|@$H|1P|I|1Q|1|1R]]|A|$]]|$1|T|3|U|5|Q|7|1S|8|@]|9|@$H|1T|I|1U|1|1V]]|A|$]]|$1|V|3|-4|5|6|7|1W|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1X|8|@]|9|@]|A|$]]]|X|$Y|$5|Z|10|11|A|$12|13]]|14|$5|Z|10|11|A|$12|15]]|16|$5|Z|10|11|A|$12|17]]|18|$5|Z|10|11|A|$12|19]]]]

The 2 points are unrelated

<h2>Sequential</h2>

If you insert values 1 to 1000, it will be sequential with an WHERE and ORDER BY to limit you to these 1000 rows in some column. Unless there are duplicates, so you'd need a unique constraint

If you rely on an IDENTITY, it isn't guaranteed: <a href="https://stackoverflow.com/q/4219201/27535">Do Inserted Records Always Receive Contiguous Identity Values</a>.

<h2>Atomicity</h2>

All transactions are atomic:

<ul>
<li><a href="https://stackoverflow.com/questions/5117288/is-neccessary-to-encapsulate-a-single-merge-statement-with-insert-delete-and-up/5117330#5117330">Is neccessary to encapsulate a single merge statement (with insert, delete and update) in a transaction?</a></li>
<li><a href="https://stackoverflow.com/questions/4978372/sql-server-and-connection-loss-in-the-middle-of-a-transaction/4978511#4978511">SQL Server and connection loss in the middle of a transaction</a></li>
<li><a href="https://stackoverflow.com/questions/6502588/does-it-delete-partially-if-execute-a-delete-statement-without-transaction/6502695#6502695">Does it delete partially if execute a delete statement without transaction?</a></li>
</ul>

I'm trying to better understand a nuance of SQL Server transactions.

Say I have a query that updates 1,000 existing rows, updating one of the columns to have the values 1 through 1,000. It's possible to execute this query and, when completed, those rows would not be numbered sequentially. This is because it's possible for another query to modify one of those rows before my query finishes.

On the other hand, if I wrap those updates in a transaction, that guarantees that if any one update fails, I can fail all updates. But does it also mean that those rows would be guaranteed to be sequential when I'm done?

In other words, are transactions always atomic?

Are Transactions Always Atomic?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在尝试更好地了解SQL Server事务的细微差别。假设我有一个更新1,000个现有行的查询，将其中一列更新为从1到1,000的值。可以执行此查询，当查询完成时，这些行将不会按顺序编号。这是因为在我的查询完成之前，另一个查询可能会修改其中的一行。另一方面，如果我将这些更新包装在一个事务中，这就保证了如果任何一个更新...

问事务总是原子的吗？
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问事务总是原子的吗？EN