自动标记用户代理字符串以进行统计？
我们在我们的网站中跟踪用户代理字符串。我想对它们进行一些统计，看看我们有多少 IE6 用户（这样我们就知道我们要针对什么进行开发），以及我们有多少移动用户。
所以我们有这样的日志整体：
Mozilla/4.0 (compatible; MSIE 7.0; Windows NT 5.1; FunWebProducts)
Mozilla/4.0 (compatible; MSIE 7.0; Windows NT 5.1; FunWebProducts; .NET CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; .NET CLR 2.0.50727)
理想情况下，看到所有“有意义”的字符串会非常整洁，这意味着字符串可能长于一定的长度。例如，我可能想查看有多少条目包含 FunWebProducts、.NET CLR 或 .NET CLR 1.0.3705 - - 但我不想想知道有多少个有分号。所以我不一定要寻找唯一的字符串，而是所有字符串，甚至子集。因此，我想查看所有 Mozilla 的计数，知道这包括 Mozilla/5.0 和 Mozilla/4.0 的计数。如果有一个嵌套显示，从最短的字符串开始，一直向下，那就太好了。也许是这样的：
4,2093 Mozilla
1,093 Mozilla/5.0
468 Mozilla/5.0 (Windows;
47 Mozilla/5.0 (Windows; U
2,398 Mozilla/4.0
这听起来像是计算机科学作业。这叫什么？那里有这样的东西吗，还是我自己写？
We keep track of user agent strings in our website. I want to do some statistics on them, to see how many IE6 users we have ( so we know what we have to develop against), and also how many mobile users we have.
So we have log entires like this:
Mozilla/4.0 (compatible; MSIE 7.0; Windows NT 5.1; FunWebProducts)
Mozilla/4.0 (compatible; MSIE 7.0; Windows NT 5.1; FunWebProducts; .NET CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; .NET CLR 2.0.50727)
And ideally, it would be pretty neat to see all the 'meaningful' strings, which would just mean probably strings longer than a certain length. For instance, I might like to see how many entries have FunWebProducts in it, or .NET CLR, or .NET CLR 1.0.3705 -- but I don't want to see how many have a semi-colon. So I'm not necessarily looking for unique strings, but all strings, even sub-sets. So, I would want to see the count of all Mozilla, knowing that this includes the counts for Mozilla/5.0 and Mozilla/4.0. It would be nice if there were a nested display for this, starting with the shortest strings, and working its way down. Something perhaps like
4,2093 Mozilla
1,093 Mozilla/5.0
468 Mozilla/5.0 (Windows;
47 Mozilla/5.0 (Windows; U
2,398 Mozilla/4.0
This sounds like a computer science homework. What would this be called? Does something like this exist out there, or do I write my own?
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您正在查看最长公共子字符串问题，或者，根据上面的具体示例，最长公共子字符串前缀问题，可以通过 trie 来解决。
但是，从上面的示例来看，您可能甚至不需要对此保持高效。相反，简单地：
对某些标点子集上的字符串进行标记，例如
[ ;/]对
对于每个前缀，获取其匹配的记录的计数并保存
You are looking at a longest common substring problem, or, given your specific example above, a longest common prefix problem, which can be approached with a trie.
However, going from your example above, you probably don't even need to be efficient about this. Instead, simply:
Tokenize strings on some punctuation subset, like
[ ;/]Save each unique prefix of however many tokens, replacing the original delimiters
For each prefix, get a count of which records it matches and save that
如果您将其分解为主要名称（左括号之前的部分），然后将每个部分以分号分隔存储为子记录，您可以进行任何您想要的分析。例如，将其存储在关系数据库中：
然后记录对浏览器和部件的引用，您可以进行任何类型的分析。
If you break it up into the major name (part before the opening paren), and then store each part separated by semicolon as a child record, you could do whatever analysis you want. For example, store it in a relational database:
Then log references to browser and parts and you can do any type of analysis you want.
使用正则表达式将用户代理字符串解析为其相关组成部分怎么样？用户代理字符串的基本规范是“
[name]/[version]”或“[name][version]”'。有了这些信息，我们可以使用像([^\(\)\/\\;\n]+)([ ]((?=\d*\.+\d*|\d*_ +\d*)[\d\.Xx_]+)|[/]([^\(\)\/; \n]+))获取匹配集，其中第一个匹配项是[name]，集合中的第二个匹配项是[version]。当然，您必须从集合中的第二个匹配中删除空格和/，或者修改正则表达式以使用lookbehind（几种正则表达式风格不支持，所以我没有将其包含在这里）。获得所有这些元组后，您可以根据需要对它们进行操作和计数。
What about using a regex to parse the user agent string into its relevant component parts? The basic spec for a user agent string is '
[name]/[version]' or '[name][version]'. With that information we can use a regex like([^\(\)\/\\;\n]+)([ ]((?=\d*\.+\d*|\d*_+\d*)[\d\.Xx_]+)|[/]([^\(\)\/; \n]+))to get match sets where the first match in a set is the[name]and the second match in a set is the[version]. Of course, you'll have to strip the spaces and/from the second match in the set, or modify the regex to use lookbehind (which several regex flavors don't support, so I didn't include it here).After you get all these tuples you can manipulate and count them however you want.