blocks|key|1971708|text|盐|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1971709|要回答这个问题，重要的是要知道盐是什么。|1971710|Salts是针对预先计算的表的攻击而设计的。例如彩虹桌。彩虹表是巨大的表，所有可能的密码变化达到一定的长度。(使用巧妙的内存/时间权衡。)|1971711|如果攻击者只想破解单个密码，他们不会提供任何优势。|1971712|如果满足以下条件，则上述语句不成立|1971713|1971714|数据库不使用盐。那么就可以使用普通的彩虹桌了。|unordered-list-item|1971715|盐太短了。如果盐太短，就和密码长了效果一样。|1971716|这是一种和盐一样的普通盐。可以肯定的是，已经有一些彩虹表中包含了这种盐。|1971717|1971718|使用彩虹表的攻击者通常想要破解尽可能多的账户。|1971719|你的哪种方法最安全？|1971720|除了第三个方法之外，所有的方法都是不安全的。这是因为使用任何其他方法都允许攻击者计算整个数据库的彩虹表。|offset|length|style|BOLD|1971721|因为盐依赖于密码。也不要让它依赖于用户名，这仍然允许攻击者为100个最常见的用户名创建一个彩虹表。|1971722|请牢记|1971723|1971724|仅使用加密安全的随机生成器来计算盐。|1971725|请勿使用MD5。它已经被认为是损坏的，请至少使用SHA1。|1971726|1971727|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|H|4|1F|1|0|0|1D|0|0|3|0|0|0|I|0|0|T|0|0^^$0|@$1|2|3|4|5|6|7|1E|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1F|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|1G|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|1H|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|1I|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]|$1|K|3|L|5|M|7|1K|8|@]|9|@]|A|$]]|$1|N|3|O|5|M|7|1L|8|@]|9|@]|A|$]]|$1|P|3|Q|5|M|7|1M|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|1N|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1O|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1P|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1Q|8|@$Y|1R|Z|1S|10|11]|$Y|1T|Z|1U|10|11]]|9|@]|A|$]]|$1|12|3|13|5|6|7|1V|8|@$Y|1W|Z|1X|10|11]]|9|@]|A|$]]|$1|14|3|15|5|6|7|1Y|8|@$Y|1Z|Z|20|10|11]]|9|@]|A|$]]|$1|16|3|-4|5|6|7|21|8|@]|9|@]|A|$]]|$1|17|3|18|5|M|7|22|8|@$Y|23|Z|24|10|11]]|9|@]|A|$]]|$1|19|3|1A|5|M|7|25|8|@$Y|26|Z|27|10|11]]|9|@]|A|$]]|$1|1B|3|-4|5|6|7|28|8|@]|9|@]|A|$]]|$1|1C|3|-4|5|6|7|29|8|@]|9|@]|A|$]]]|1D|$]]

<h2>Salts</h2>

To answer this question it's important to know for what salts are.

Salts are designed against attacks with pre-calculated tables. For example rainbow tables. Rainbow tables are huge tables with all possible password variations up to a certain length. (Using a clever memory/time tradeoff.)

If the attacker only wants to crack a single password, they don't offer an advantage.

The statement above is not true if

<ul>
<li>The database doesn't use salts. Then a common rainbow table can be used.</li>
<li>The salt is too short. If the salt is too short, it has the same effect as just having a longer password.</li>
<li>It's a common salt like salt. There are for sure already some rainbow tables with that salt included.</li>
</ul>

Attackers using rainbow tables usually want to crack as many accounts as possible.

<h2>Which of your methods is most secure?</h2>

All of your methods except the third are insecure. This is because using any of the other methods allows the attacker to calculate a rainbow-table for your whole database. 

Because the salt is dependent on the password. Don't make it dependent on the username either, this would still allow an attacker to create a rainbow table for the 100 most common usernames.

<h2>Keep in mind</h2>

<ul>
<li>Use only a cryptographically secure random generator to calculate your salts.</li>
<li>Don't use MD5. It's already considered broken, use at least SHA1.</li>
</ul>

blocks|key|810100|text|严格地说，您只有一种加盐方法，用来计算散列。前三行是生成盐的不同方式。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|810101|因此，这里有一个盐来阻止预计算查找表发现密码。它应该是某个人存储的固定值，优选地，对于被散列的纯文本是唯一的。|810102|最安全的方法是使用加密安全的随机数生成器来生成盐，然后将其与密码一起存储。|810103|如果您创建了一个salt，它是密码的MD5，那么它必须与散列和salted密码值一起存储，这意味着您有一个容易受到预计算查找表攻击的非salt散列，除非您计划每次都计算它，这会对性能造成很小的影响。通过对MD5散列进行SHA，可以减少出现纯文本值的可能性，因为MD5散列值的数量是有限的，因为它们的长度是固定的。这意味着彩虹表查找可能比真正随机的盐更有可能成功。|810104|所以请使用随机的盐。|810105|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|$]]

Well strictly speaking you only have one salting method, where you calculate the hash. The first three lines are different ways of generating a salt.

So a salt is there to stop precomputed lookup tables from discovering passwords. It should be a fixed value stored someone that is, preferably, unique to the plain text being hashed.

The most secure would be to use a cryptographically secure random number generator to produce a salt which is then stored along side the password.

If you created a salt which was an MD5 of the password then it would have to stored alongside the hashed and salted password value, which means you have an unsalted hash which is vulnerable to precomputed lookup tables, unless you plan to calculate it every time which is a small performance hit. By taking a SHA hash of an MD5 hash you're reducing the possibility of the plain text values, as there's a finite number of MD5 hash values as they are fixed length. This would mean that a rainbow table lookup might have a greater chance of success than a truly random salt.

So use the random salt please.

blocks|key|1974164|text|前两种方法毫无价值。加盐的全部意义在于，相同的密码并不总是产生相同的加密/散列字符串。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1974165|如果您使“盐”仅依赖于密码，则相同的密码将始终导致相同的散列。因此，基本上结果是相同的，就好像你使用了一个略有不同的散列函数，没有任何盐。|1974166|使用第三种方法时，两个具有相同密码的用户通常会得到不同的salt，并且对于两个用户来说，密码的散列版本看起来会不同。从散列很难看出它们都有相同的密码。|1974167|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|$]]

The first two methods are worthless. The whole point of salting is that the same password does not always result in the same encrypted/hashed string.

If you make the "salt" dependent on just the password, the same password will always result in the same hash. So basically the result is the same as if you'd use a slightly different hash function without any salt.

With the third method two users with the same password will usually get a different salt and the hashed version of the password will look different for both users. It will be hard to tell by the hashes that they both have the same password.

blocks|key|810202|text|思考彩虹表的一个有用的方法是，它们可以为任何只有一个输入的单向(或“陷阱门”)函数构建。也就是说，如果您对所有密码使用相同的函数F+:+hash+=F(密码)。F可以是MD5，也可以是SHA1，或者其他什么。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|810203|现在让我们来看一下盐。您使用一个加盐函数G，hash+=+G(salt，password)。如果您的数据库中的所有密码都使用相同的salt，您可以构造一个函数G，其中G(password)+=+F("your+salt"，password)，因此只有一个输入函数，因此您可以构建一个彩虹表。|810204|如果盐依赖于密码怎么办？假设盐=I(密码)，我们可以构建J(密码)=G(I(密码)，密码)，这是一个单输入函数，所以可以构建彩虹表。|810205|因此，每个密码都需要有自己的salt。这意味着在攻击者破解所有密码的时间内，他们只能破解一个密码。|810206|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

A useful way to think of rainbow tables is that they can be built for any one-way (or "trapdoor") function that only has one input. That is, if you use the same function, F, for all your passwords: hash = F(password). F could be MD5 or SHA1 or whatever.

Now lets look at salts. You use a salted function G, hash = G(salt, password). If all passwords in your database use the same salt, you can construct a function G, where G(password) = F("your salt", password), so there is a single input function, and thus you can build a rainbow table.

What if the salt depends on the password? Say the salt = I(password), we can build J(password) = G(I(password), password), a single input function, so rainbow tables can be built.

So, each password needs to have its own salt. This means that in the time it would take an attacker to crack all of your passwords, they can only crack one.

What are the advantages / disadvantages of those 3 methods to create a salt?

<pre><code>$salt = md5($password);

$salt = sha1(md5($password));

$salt = generate_random_number();
</code></pre>

Computing hash:

<pre><code>$hash = sha1($salt + $password);
</code></pre>

Salting Algorithm Strength

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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这3种创建盐的方法的优缺点是什么？$salt = md5($password);$salt = sha1(md5($password));$salt = generate_random_number();计算哈希：$hash = sha1($salt + $password);

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