java - 如何在正则表达式中定义切换案例？

Question

如何制作执行以下操作的单个正则表达式：

“一般来说，做一个通用的正则表达式。但如果特定字符相互跟随，请以不同的方式检查这些特定字符后面的字符串”？

one == two && three | four

一般的正则表达式是：[&|=\s]+.

拆分时会导致：one, two, three, four.

但是，如果我想在每次有=字符时应用不同的正则表达式，并且希望后面的表达式=只在|字符处停止呢？这样我就可以得到结果：one, two && three, four.

我怎么能这样做？

Answer 1

这是一种可能性：

(?=[&|=\s])[&|\s]*(?:=[^|]*?[|][&|=\s]+)?

或者在带有解释的自由间距模式下：

(?=[&|=\s])  # make sure there is at least a single separator character ahead;
             # this is important, otherwise you would get empty matches, and
             # depending on your implementation split every non-separator
             # character apart.
[&|\s]*      # same as before but leave out =
(?:          # start a (non-capturing) group; it will be optional and treat the
             # special =...| case
  =          # match a literal =
  [^|]*?     # match 0 or more non-| characters (as few as possible)
  [|]        # match a literal | ... this is the same as \|, but this is more
             # readable IMHO
  [&|=\s]+   # and all following separator characters
)?           # make the whole thing optional

试试看。

编辑：

我刚刚意识到，这吞噬了中心部分，但你也想归还它。在这种情况下，您最好使用匹配而不是拆分（使用find）。这种模式应该可以解决问题：

=[&|=\s]*([^|]+?)[&|=\s]*[|]|([^&|=\s]+)

现在，第一个或第二个捕获组将包含所需的结果。这是一个解释：

#this consists of two alternatives, the first one is the special case
  =            # match a literal =
  [&|=\s]*     # consume more separator characters
  ([^|]+?)     # match 1 or more non-| characters (as few as possible) and
               # capture (group 1)
  [&|=\s]*     # consume more separator characters
  [|]          # match a literal |
|              # OR
  ([^&|=\s]+)  # match 1 or more non-separator characters (as many as possible)
               # and capture (group 2)

试试看。