一个正则表达式匹配一个子字符串，后面不跟一个特定的其他子字符串 [英] A regex to match a substring that isn't followed by a certain other substring

问题描述

我需要一个正则表达式匹配 blahfooblah ，但不匹配 blahfoobarblah

我希望它只匹配foo和foo周围的所有东西，只要它没有跟随bar。

我试过使用这个： foo。*（？<！bar）这是相当接近的，但它匹配 blahfoobarblah 。

我使用的具体语言是Clojure，它使用了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 。

解决方案

试用：

  /（？！。* bar）（？=。* foo）^（\w +）$ / 
  

测试：

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

正则表达式解释

 节点说明
 -------------------------- -------------------------------------------------- ---- 
（？！向前看看是否有没有：
 -------------------------- -------------------------------------------------- ---- 
。*除了\\\
之外的任何字符（0或更多次
（匹配最多的可能））
 ------------- -------------------------------------------------- ----------------- 
 bar'bar'
 --------------------- -------------------------------------------------- --------- 
）结束前瞻
 --------------------------- -------------------------------------------------- --- 
（？=向前看看是否有：
 ---------------------------- -------------------------------------------------- -  
。*除了\\\
之外的任何字符（0或更多次
（匹配最多的可能））
 --------------- -------------------------------------------------- --------------- 
 foo'foo'
 ----------- -------------------------------------------------- ------- 
）结束前瞻
 ----------------------------- -------------------------------------------------- -  
 ^字符串的开头
 ----------------------------------- --------------------------------------------- 
（组并捕获到\1：
 -------------------------------------- ------------------------------------------ 
 \w + word字符（az，AZ，0-9，_）（1或
多次（匹配最多的金额
可能））
 ------------- -------------------------------------------------- ----------------- 
）\1结尾
 ------------------- -------------------------------------------------- ----------- 
 $在可选的\\\
之前，结束
字符串
  

其他正则表达式

如果您只想排除 bar 它直接在 foo 之后，可以使用

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

---

编辑

您对问题进行了更新，使其具体。

  /（？=。* foo（？！bar））^（\w +）$ / 
   
 
新测试
 
 
 
  fooshouldbarpass＃pass 
 butnotfoobarfail＃fail 
 fooshouldpassevenwithfoobar＃pass 
 nofuuhere＃fail 
  
 
 
 
新说明
 
 
 
 （？=。* foo（？！bar））确保找到 foo 但不会直接跟随 bar  
 
I need a regex that will match blahfooblah but not blahfoobarblah

I want it to match only foo and everything around foo, as long as it isn't followed by bar.

I tried using this: foo.*(?<!bar) which is fairly close, but it matches blahfoobarblah. The negative look behind needs to match anything and not just bar.

The specific language I'm using is Clojure which uses Java regexes under the hood.

EDIT: More specifically, I also need it to pass blahfooblahfoobarblah but not blahfoobarblahblah.
解决方案
Try:

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


Tests:

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


Regular expression explanation

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


Other regex

If you only want to exclude bar when it is directly after foo, you can use
/(?!.*foobar)(?=.*foo)^(\w+)$/




Edit

You made an update to your question to make it specific.
/(?=.*foo(?!bar))^(\w+)$/


New tests

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


New explanation

(?=.*foo(?!bar)) ensures a foo is found but is not followed directly bar

                        
这篇关于一个正则表达式匹配一个子字符串，后面不跟一个特定的其他子字符串的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！