Java 如何在正则表达式中定义开关情况？

如何生成一个执行以下操作的正则表达式：

“一般来说，使用通用正则表达式。但是如果特定字符紧跟在后面，请以不同的方式检查这些特定字符后面的字符串”

一般正则表达式是：

[&|=\s]+

拆分时会产生以下结果：

1、2、3、4

但是，如果每次出现

=

字符时都要应用不同的正则表达式，并且希望

=

后面的表达式仅在

字符处停止，该怎么办？ 这样我就得到了结果：

1，2&&3，4

---

我该怎么做呢？

这里有一种可能性：

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

或在自由间距模式下进行解释：

(?=[&|=\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

#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)

编辑：

我刚刚意识到，这会吞下中间部分，但你也要把它还给我。在这种情况下，您最好使用匹配而不是拆分（使用

find

）。这种模式应该可以做到：

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

现在，第一个或第二个捕获组将包含所需的结果。下面是一个解释：

(?=[&|=\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

#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)

---

你的问题不是很清楚（至少对我来说是这样）。你是这样想的：

[=one |][124;[=two |=three |][124;[=four |]