Regex needs to break the string into "."

Question

I need regex in javascript. I have a line:

'*window.some1.some\.2.(a.b + ")" ? cc\.c : d.n [a.b, cc\.c]).some\.3.(this.o.p ? ".mike." [ff\.]).some5'

I want to split this string into periods to get an array:

[
    '*window',
    'some1',
    'some\.2',   //ignore the . because it escaped
    '(a.b ? cc\.c : d.n [a.b, cc\.c])',  //ignore everything inside ()
    'some\.3',
    '(this.o.p ? ".mike." [ff\.])',
    'some5'
]

What will this regex do?

Answer 1

var string = '*window.some1.some\\.2.(a.b + ")" ? cc\\.c : d.n [a.b, cc\\.c]).some\\.3.(this.o.p ? ".mike." [ff\\.]).some5';
var pattern = /(?:\((?:(['"])\)\1|[^)]+?)+\)+|\\\.|[^.]+?)+/g;
var result = string.match(pattern);
result = Array.apply(null, result); //Convert RegExp match to an Array

Fiddle: http://jsfiddle.net/66Zfh/3/
RegExp explanation. Match a sequential character set satisfying:

/             Start of RegExp literal
(?:            Create a group without reference (example: say, group A)
   \(          `(` character
   (?:         Create a group without reference (example: say, group B)
      (['"])     ONE `'` OR `"`, group 1, referable through `\1` (inside RE)
      \)         `)` character
      \1         The character as matched at group 1, either `'` or `"`
     |          OR
      [^)]+?     Any non-`)` character, at least once (see below)
   )+          End of group (B). Let this group occur at least once
  |           OR
   \\\.        `\.` (escaped backslash and dot, because they're special chars)
  |           OR
   [^.]+?      Any non-`.` character, at least once (see below)
)+            End of group (A). Let this group occur at least once
/g           "End of RegExp, global flag"
        /*Summary: Match everything which is not satisfying the split-by-dot
                 condition as specified by the OP*/

There is a difference between +and +?. One plus is trying to match as many characters as possible, but +?matches only those characters needed to get RegExp matching. Example 123 using \d+? > 1 and \d+ > 123.

String.match - /g. match g , .

g , . :

Index 0: <Whole match>
Index 1: <Group 1>

Answer 2

:

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

. 1 , .

:

var myregexp = /(?:(\(.*?[^'"]\)|.*?[^\\])(?:\.|$))/g;
var match = myregexp.exec(subject);
while (match != null) {
    for (var i = 0; i < match.length; i++) {
        // matched text: match[i]
    }
    match = myregexp.exec(subject);
}

:

// (?:(\(.*?[^'"]\)|.*?[^\\])(?:\.|$))
// 
// Match the regular expression below «(?:(\(.*?[^'"]\)|.*?[^\\])(?:\.|$))»
//    Match the regular expression below and capture its match into backreference number 1 «(\(.*?[^'"]\)|.*?[^\\])»
//       Match either the regular expression below (attempting the next alternative only if this one fails) «\(.*?[^'"]\)»
//          Match the character "(" literally «\(»
//          Match any single character that is not a line break character «.*?»
//             Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
//          Match a single character NOT present in the list "'"" «[^'"]»
//          Match the character ")" literally «\)»
//       Or match regular expression number 2 below (the entire group fails if this one fails to match) «.*?[^\\]»
//          Match any single character that is not a line break character «.*?»
//             Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
//          Match any character that is NOT a "A \ character" «[^\\]»
//    Match the regular expression below «(?:\.|$)»
//       Match either the regular expression below (attempting the next alternative only if this one fails) «\.»
//          Match the character "." literally «\.»
//       Or match regular expression number 2 below (the entire group fails if this one fails to match) «$»
//          Assert position at the end of the string (or before the line break at the end of the string, if any) «$»

Answer 3

, , Javascript.

. , Regex's:

• , " " - /(?:(\\.)|([\(\[\{])|([\)\]\}])|(\.))/g
• string.replace(pattern, function (...)), .
• .
• , .

, , , string.replace, , !

:
, , .
, Regex .
:

• Regex
• ,
• , \

.

Answer 4

.

var s = '*window.some1.some\.2.(a.b + ")" ? cc\.c : d.n [a.b, cc\.c]).some\.3.(this.o.p ? ".mike." [ff\.]).some5';

console.log(s.match(/(?:\([^\)]+\)|.*?\.)/g));

:

  ["*window.", "some1.", "some.", "2.", "(a.b + ")", "" ? cc.", "c : d.", "n [a.", "b, cc.", "c]).", "some.", "3.", "(this.o.p ? ".mike." [ff.])", "."]

Answer 5

, , , @FailedDev , .:)

, . .

((\(.*?((?<!")\)(?!")))|((\\\.)|([^.]))+)

Unfortunately, this will not work in your case, since I am using a negative lookbehind, which, it seems to me, is not supported by the javascript regex engine. However, it should work in accordance with other engines, which can be confirmed here: http://gskinner.com/RegExr/ . Replace with $ 1 \ n.