javascript - 匹配所有内容但不匹配带引号的字符串

Question

我想匹配所有内容，但不匹配带引号的字符串。

我可以用这个匹配所有引用的字符串:/(("([^"\\]|\\.)*")|('([^'\\]|\\.)* '))/所以我尝试用这个匹配所有但没有引号的字符串: /[^(("([^"\\]|\\.)*")|('([^'\\]|\\.)*'))]/ 但它不起作用。

我只想使用正则表达式，因为我想替换它并想在它返回后得到引用的文本。

string.replace(regex, function(a, b, c) {
   // return after a lot of operations
});

带引号的字符串对我来说是这样的"bad string" 或这个'cool string'

所以如果我输入:

he\'re is "watever o\"k" efre 'dder\'4rdr'?

它应该输出这样的匹配:

["he\'re is ", " efre ", "?"]

而且我不想替换它们。

我知道我的问题很难，但并非不可能!没有什么是不可能的。

谢谢

Answer 1

编辑:重写以涵盖更多边缘情况。

这个可以做，但是有点复杂。

result = subject.match(/(?:(?=(?:(?:\\.|"(?:\\.|[^"\\])*"|[^\\'"])*'(?:\\.|"(?:\\.|[^"'\\])*"|[^\\'])*')*(?:\\.|"(?:\\.|[^"\\])*"|[^\\'])*$)(?=(?:(?:\\.|'(?:\\.|[^'\\])*'|[^\\'"])*"(?:\\.|'(?:\\.|[^'"\\])*'|[^\\"])*")*(?:\\.|'(?:\\.|[^'\\])*'|[^\\"])*$)(?:\\.|[^\\'"]))+/g);

会回来

, he said. 
, she replied. 
, he reminded her. 
, 

从此字符串中(为清楚起见，添加了换行符并删除了引号):

"Hello", he said. "What's up, \"doc\"?", she replied. 
'I need a 12" crash cymbal', he reminded her. 
"2\" by 4 inches", 'Back\"\'slashes \\ are OK!'

解释:(有点令人难以置信)

分解正则表达式:

(?:
 (?=      # Assert even number of (relevant) single quotes, looking ahead:
  (?:
   (?:\\.|"(?:\\.|[^"\\])*"|[^\\'"])*
   '
   (?:\\.|"(?:\\.|[^"'\\])*"|[^\\'])*
   '
  )*
  (?:\\.|"(?:\\.|[^"\\])*"|[^\\'])*
  $
 )
 (?=      # Assert even number of (relevant) double quotes, looking ahead:
  (?:
   (?:\\.|'(?:\\.|[^'\\])*'|[^\\'"])*
   "
   (?:\\.|'(?:\\.|[^'"\\])*'|[^\\"])*
   "
  )*
  (?:\\.|'(?:\\.|[^'\\])*'|[^\\"])*
  $
 )
 (?:\\.|[^\\'"]) # Match text between quoted sections
)+

首先，你可以看到有两个相似的部分。这两个先行断言确保前面的字符串中有偶数个单引号/双引号，忽略转义引号和相反类型的引号。我将用单引号部分显示它:

(?=                   # Assert that the following can be matched:
 (?:                  # Match this group:
  (?:                 #  Match either:
   \\.                #  an escaped character
  |                   #  or
   "(?:\\.|[^"\\])*"  #  a double-quoted string
  |                   #  or
   [^\\'"]            #  any character except backslashes or quotes
  )*                  # any number of times.
  '                   # Then match a single quote
  (?:\\.|"(?:\\.|[^"'\\])*"|[^\\'])*'   # Repeat once to ensure even number,
                      # (but don't allow single quotes within nested double-quoted strings)
 )*                   # Repeat any number of times including zero
 (?:\\.|"(?:\\.|[^"\\])*"|[^\\'])*      # Then match the same until...
 $                    # ... end of string.
)                     # End of lookahead assertion.

双引号部分作用相同。

然后，在字符串中这两个断言成功的每个位置，正则表达式的下一部分实际上会尝试匹配某些内容:

(?:      # Match either
 \\.     # an escaped character
|        # or
 [^\\'"] # any character except backslash, single or double quote
)        # End of non-capturing group

整个过程重复一次或多次，尽可能多次。 /g 修饰符确保我们获得字符串中的所有匹配项。

See it in action here on RegExr .