blocks|key|1135756|text|您可以根据文件内容本身生成散列。这样做有两个很好的理由：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1135757|1135758|允许您永远不会将同一文件存储两次-例如，如果您有两个内容相同的音乐文件副本，您可以检查您是否已经存储了该文件，然后只存储一次。|ordered-list-item|1135759|您将元数据(文件名就是元数据)与blob分开。因此，您将拥有一个按文件内容的散列进行索引的存储系统，然后将文件元数据与该散列查找代码相关联。|1135760|1135761|根据散列的大小，找到两个计算相同散列但内容并不相同的文件的风险很低，您可以通过将文件分块散列来有效地缓解这种风险(这可能会导致一些有趣的存储优化场景:P)。|1135762|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|N|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|P|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|L|$]]

You could produce a hash based on the file contents itself. There are two good reasons to do this:

<ol>
<li>Allows you to never store the same file twice - for example, if you have two copies of a music file which are identical in content you could check to see if you have already stored that file, and just store it once.</li>
<li>You separate meta-data (file name is just meta data) from the blob. So you would have a storage system which is indexed by the hash of the file contents, and you then associate the file meta-data with that hash lookup code.</li>
</ol>

The risk of finding two files that compute the same hash that aren't indeed the same contents, depending on the size of the hash would be low, and you can effectively mitigate that by perhaps hashing the file in chunks (which could then lead to some interesting storage optimisation scenarios :P).

blocks|key|539170|text|GUIDs是单向的。你基本上保证不会得到任何重复(如果你有一个合适的随机生成器)。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|539171|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Globally_Unique_Identifier^0|0|5|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

<a href="http://en.wikipedia.org/wiki/Globally_Unique_Identifier" rel="nofollow noreferrer">GUIDs</a> are one way. You're basically guaranteed to not get any repeats (if you have a proper random generator).

blocks|key|3759603|text|您还可以附加自纪元以来的时间。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3759604|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|$]]

You could also append with the time since epoch.

blocks|key|539341|text|已经提到了最好的解决方案。我只想补充一些想法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|539342|最简单的解决方案是在每个新文件上都有一个计数器并递增。只要只有一个线程创建新文件，这种方法就能很好地工作。如果多个线程、进程甚至系统添加了新文件，事情就会变得有点复杂。您必须使用锁定或任何类似的同步方法来协调新ids的创建。您还可以为每个进程分配id范围，以减少同步工作，或者通过唯一的进程id扩展文件id。|539343|更好的解决方案可能是在此场景中使用GUID，而不必关心进程之间的同步。|539344|最后，您可以将一些随机数据添加到每个标识符，以使它们更难猜测这是否是必需的。|539345|此外，coommon将文件存储在目录结构中，其中文件的位置取决于其名称。abcdef1234.xyz文件可能存储为/ab/cd/ef/1234.xyz。这避免了包含大量文件的目录。我真的不知道为什么这样做-可能是文件系统的限制，性能问题-但这是相当常见的。如果文件直接存储在数据库中，我不知道类似的事情是否常见。|539346|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|N|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|O|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

The best solution have already been mentioned. I just want to add some thoughts.

The simplest solution is to have a counter and increment on every new file. This works quite well as long as only one thread creates new files. If multiple threads, processes or even systems add new files, things get a bit more complicated. You must coordinate the creation of new ids with locking or any similar synchronisation method. You could also assign id ranges to every proceses to reduce the synchronisation work, or extend the file id by a unique process id.

A better solution might be to use GUIDs in this scenario and do not have to care about synchronisation between processes.

Finally, you can at some random data to every identifier to make them harder to guess if this is a requirement.

Also coommon is storing files in a directory structure where the location of a file depends on its name. File abcdef1234.xyz might be stored as /ab/cd/ef/1234.xyz. This avoids directories with a huge number of files. I am not really aware why this is done - may be file system limitations, performance issues - but it is quite common. I do not know if similar things are common if the files are stored directly in the database.

blocks|key|1135707|text|最好的方法是简单地使用计数器。第一个文件是1，下一个是2，另一个是3，依此类推...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1135708|但是，似乎你想要随机的。要快速完成此操作，您可以确保随机数大于上一个创建的文件。你可以缓存最后一个文件，然后用它的姓氏来偏移你的随机数。|1135709|file+=+last_file+%2B+random(1+through+10)|code-block|syntax|javascript|1135710|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The best way is to simply use a counter. The first file is 1, the next is 2, another is 3, and so on...

But, it seems you want random. To quickly do this, you could make sure that your random number is greater than the last file created. You can cache the last file and then just offset your random number with its last name. 

<pre><code>file = last_file + random(1 through 10)
</code></pre>

I'm writing a "file sharing hosting" and I want to rename all the files when uploading to a unique name and somehow keep track of the names on the database. Since I don't want two or more files having same name (which is surely impossible), I'm looking for an algorithm which based on key or something generates random names for me.

Moreover, I don't want to generate a name and search the database to see if the file already exists. I want to make sure 100% or 99% that the generated filename has never been created earlier by my application.

Any idea how I can write such application?

Generate File Names Automatically without collision

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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我正在写一个“文件共享主机”，我想要重命名所有的文件时，上传到一个唯一的名称，并以某种方式保持跟踪数据库中的名称。由于我不希望两个或更多的文件具有相同的名称(这肯定是不可能的)，我正在寻找一种基于键或其他东西为我生成随机名称的算法。此外，我不想生成一个名称并搜索数据库来查看该文件是否已经存在。我想要确保100%或99%生成的文件名从未被我的应用程序早先创建过。你知道我该怎么写这样的应用程序吗？

问自动生成文件名而不发生冲突
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