java - 一个正则表达式来匹配一个子字符串，该子字符串后面没有某个其他子字符串

Question

我需要一个匹配 blahfooblah 的正则表达式但不是 blahfoobarblah

我希望它只匹配 foo 和 foo 周围的所有内容，只要它后面没有 bar。

我试过用这个:foo.*(?<!bar)相当接近，但匹配 blahfoobarblah .背后的负面形象需要匹配任何东西，而不仅仅是条形图。

我使用的特定语言是 Clojure，它在后台使用 Java 正则表达式。

编辑:更具体地说，我还需要它通过 blahfooblahfoobarblah但不是 blahfoobarblahblah .

Answer 1

试试:

/(?!.*bar)(?=.*foo)^(\w+)$/

测试:

blahfooblah            # pass
blahfooblahbarfail     # fail
somethingfoo           # pass
shouldbarfooshouldfail # fail
barfoofail             # fail

正则表达式解释

NODE                     EXPLANATION
--------------------------------------------------------------------------------
  (?!                      look ahead to see if there is not:
--------------------------------------------------------------------------------
    .*                       any character except \n (0 or more times
                             (matching the most amount possible))
--------------------------------------------------------------------------------
    bar                      'bar'
--------------------------------------------------------------------------------
  )                        end of look-ahead
--------------------------------------------------------------------------------
  (?=                      look ahead to see if there is:
--------------------------------------------------------------------------------
    .*                       any character except \n (0 or more times
                             (matching the most amount possible))
--------------------------------------------------------------------------------
    foo                      'foo'
--------------------------------------------------------------------------------
  )                        end of look-ahead
--------------------------------------------------------------------------------
  ^                        the beginning of the string
--------------------------------------------------------------------------------
  (                        group and capture to \1:
--------------------------------------------------------------------------------
    \w+                      word characters (a-z, A-Z, 0-9, _) (1 or
                             more times (matching the most amount
                             possible))
--------------------------------------------------------------------------------
  )                        end of \1
--------------------------------------------------------------------------------
  $                        before an optional \n, and the end of the
                           string

其他正则表达式

如果只想在foo之后直接排除bar，可以使用

/(?!.*foobar)(?=.*foo)^(\w+)$/

---

编辑

您更新了您的问题以使其具体化。

/(?=.*foo(?!bar))^(\w+)$/

新测试

fooshouldbarpass               # pass
butnotfoobarfail               # fail
fooshouldpassevenwithfoobar    # pass
nofuuhere                      # fail

新的解释

(?=.*foo(?!bar)) 确保找到 foo 但不直接跟随 bar