blocks|key|892548|text|是。从安全角度来看，MD5已经完全崩溃，但意外碰撞的可能性仍然很小。只需确保这些文件不是由您不信任的人创建的，并且这些人可能具有恶意。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|892549|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|$]]

Yes. MD5 has been completely broken from a security perspective, but the probability of an accidental collision is still vanishingly small. Just be sure that the files aren't being created by someone you don't trust and who might have malicious intent.

blocks|key|892737|text|如果你没有对手，MD5就足够好了。然而，有人可以(故意)创建两个不同的文件，它们散列为相同的值(这称为冲突)，这可能是问题，也可能不是问题，这取决于您的实际情况。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|892738|由于了解已知的MD5弱点是否适用于给定的上下文是一个微妙的问题，因此建议不要使用MD5。使用防冲突哈希函数(SHA-256或SHA-512)是安全的答案。此外，使用MD5是不好的公关(如果您使用MD5，请准备好为自己辩护；而没有人会质疑您使用的是SHA-256)。|892739|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

MD5 will be good enough if you have no adversary. However, someone can (purposely) create two distinct files which hash to the same value (that's called a collision), and this may or may not be a problem, depending on your exact situation.

Since knowing whether known MD5 weaknesses apply to a given context is a subtle matter, it is recommended not to use MD5. Using a collision-resistant hash function (SHA-256 or SHA-512) is the safe answer. Also, using MD5 is bad public relations (if you use MD5, be prepared to have to justify yourselves; whereas nobody will question your using SHA-256).

blocks|key|892821|text|md5可能会产生冲突。从理论上讲，尽管可能性很小，但一百万个文件可以产生相同的散列。不要在存储值之前测试您的运气和检查md5冲突。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|892822|我个人喜欢创建随机字符串的md5，这减少了散列大文件的开销。当发现冲突时，我使用附加的循环计数器进行迭代和重新散列。|892823|你可以在pigeonhole+principle上阅读。|offset|length|892824|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Pigeonhole_principle^0|0|0|4|K|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@$F|S|G|T|1|U]]|A|$]]|$1|H|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|I|$J|$5|K|L|M|A|$N|O]]]]

An md5 can produce collisions. Theoretically, although highly unlikely, a million files in a row can produce the same hash. Don't test your luck and check for md5 collisions before storing the value.

I personally like to create md5 of random strings, which reduces the overhead of hashing large files. When collisions are found, I iterate and re-hash with the appended loop counter.

You may read on the <a href="http://en.wikipedia.org/wiki/Pigeonhole_principle" rel="noreferrer">pigeonhole principle</a>.

blocks|key|1107862|text|我不建议你这么做。如果应用程序可以在多用户系统上工作，可能会有用户，有两个文件具有相同的md5散列(他可能是工程师和玩这样的文件，或者只是好奇-他们很容易从http://www2.mat.dtu.dk/people/S.Thomsen/wangmd5/samples.html下载，我自己在写这个答案的过程中下载了两个样本)。另一件事是，一些应用程序可能会出于某种原因存储这样的副本(我不确定是否有这样的应用程序，但存在这种可能性)。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1107863|如果你是唯一地识别你的程序生成的文件，我会说使用MD5是可以的。否则，我会推荐任何其他还没有冲突的散列函数。|1107864|entityMap|0|LINK|mutability|MUTABLE|url|http://www2.mat.dtu.dk/people/S.Thomsen/wangmd5/samples.html^0|26|1O|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]]|C|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@]|C|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]]]

I wouldn't recommend it. If the application would work on multi-user system, there might be user, that would have two files with the same md5 hash (he might be engineer and play with such files, or be just curious - they are easily downloadable from <a href="http://www2.mat.dtu.dk/people/S.Thomsen/wangmd5/samples.html">http://www2.mat.dtu.dk/people/S.Thomsen/wangmd5/samples.html</a> , I myself during writing this answer downloaded two samples). Another thing is, that some applications might store such duplicates for whatever reason (I'm not sure, if there are any such applications but the possibility exists). 

If you are uniquely identifying files generated by your program I would say it is ok to use MD5. Otherwise, I would recommend any other hash function where no collisions are known yet.

blocks|key|1107609|text|就我个人而言，我认为人们使用其他对象的原始校验和(选择你的方法)来充当唯一标识符的方式太多了，而他们真正想做的是拥有唯一标识符。为了这种用途而对对象进行指纹识别并不是我们的意图，而且可能比使用uuid或类似的完整性机制需要更多的考虑。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1107610|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|$]]

Personally i think people use raw checksums (pick your method) of other objects to act as unique identifiers way too much when they really want to do is have unique identifiers. Fingerprinting an object for this use wasn't the intent and is likely to require more thinking than using a uuid or similar integrity mechanism.

blocks|key|1107535|text|MD5已被破坏，您可以使用SHA1+(在大多数语言中实现)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1107536|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|$]]

MD5 has been broken, you could use SHA1 instead (implemented in most languages)

blocks|key|892974|text|当散列短时(<几个K+?)字符串(或文件)一个可以创建两个md5散列密钥，一个用于实际的字符串，另一个用于与短的非对称字符串连接的字符串的反向。示例:+md5+(+reverse+(+string+%7C%7C+'1010‘)+)。添加额外的字符串可以确保即使是由一系列相同的位组成的文件也会生成两个不同的密钥。请理解，即使在此方案下，对于不相同的字符串，理论上也存在两个哈希键相同的可能性，但概率似乎非常小-大约是单个md5冲突概率的平方，并且当文件数量增加时，节省的时间可以相当可观。也可以考虑创建第二个字符串的更复杂的方案，但我不确定这些方案是否会大幅提高几率。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|892975|要检查冲突，可以对db中所有bit_vectors的md5散列键的唯一性运行此测试：|892976|select+md5+(+bit_vector+)，count(*)，bit_and+(+bit_vector+)+from+db+with+bit_vector|892977|+group+by+md5(+bit\_vector+),+bit\_vector+++++having++bit\_and+(+bit\_vector+)+<>+bit\_vector|code-block|syntax|javascript|892978|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

When hashing short (&lt; a few K ?) strings (or files) one can create two md5 hash keys, one for the actual string and a second one for the reverse of the string concatenated with a short asymmetric string. Example : md5 ( reverse ( string || '1010' ) ). Adding the extra string ensures that even files consisting of a series of identical bits generate two different keys. Please understand that even under this scheme there is a theoretical chance of the two hash keys being identical for non-identical strings, but the probability seems exceedingly small - something in the order of the square of the single md5 collision probability, and the time saving can be considerable when the number of files is growing. More elaborate schemes for creating the second string could be considered as well, but I am not sure that these would substantially improve the odds.

To check for collisions one can run this test for the uniqueness of the md5 hash keys for all bit_vectors in a db:

select md5 ( bit_vector ), count(*), bit_and ( bit_vector)
 from db with bit_vector 
 group by md5( bit_vector ), bit_vector
 having bit_and ( bit_vector ) &lt;> bit_vector

blocks|key|1108004|text|我喜欢将MD5视为存储大量文件数据时的概率指标。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1108005|如果哈希值相等，那么我知道我必须逐字节比较文件，但这可能只会因为错误的原因而发生几次，否则(哈希值不相等)我可以确定我们正在谈论的是两个不同的文件。|1108006|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

I like to think of MD5 as an indicator of probability when storing a large amount of file data.

If the hashes are equal I then know I have to compare the files byte by byte, but that might only happen a few times for a false reason, otherwise (hashes are not equal) I can be certain we're talking about two different files.

Is MD5 hashing a file still considered a good enough method to uniquely identify it given all the breaking of MD5 algorithm and security issues etc? Security is not my primary concern here, but uniquely identifying each file is. 

Any thoughts?

Is MD5 still good enough to uniquely identify files?

哈希函数

考虑到所有MD5算法和安全问题的破坏，MD5散列文件仍然被认为是一种足够好的唯一标识它的方法吗？在这里，安全性不是我最关心的问题，但唯一地标识每个文件才是。有什么想法吗？

问MD5是否仍然足够好，可以唯一地标识文件？
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问MD5是否仍然足够好，可以唯一地标识文件？
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