How to ignore elements escaped in a rule in regular expression?

Question

I’m wanting to do with regex (regular expression), for example (if it’s javascript):

var str = '[abc\[0123\]] [efg\[987\]h] [olá \[mundo\]!] [foo [baz]]';
str.match(/\[(.*?)\]/g);

Exit: ["[abc[0123]", "[efg[987]h", "[olá [mundo]!", "[foo [baz]"]

Or

var str = '{abc\{0123\}} {efg\{987\}h} {olá \{mundo\}!} {foo {baz}}';
str.match(/\{(.*?)\}/g);

Exit: ["{abc{0123}", "{efg{987}", "{olá {mundo}", "{foo {baz}"]

But I would like the first to ignore places not escaped as [foo [baz]] and just take the [baz] and the escapees, like this:

 ["[abc[0123]]", "[efg[987]h]", "[olá [mundo]!]", "[baz]"]

And in the second it returns:

 {"{abc{0123}}", "{efg{987}h}", "{olá {mundo}!}", "{baz}"]

My intention initially is for study, but I also intend to use in things like a structure that is similar to CSS selectors, so for example:

  input[name=\[0\]], input[name=foo\[baz\]\[bar\]]

It would return this:

  [0], [1]

Or a map of Urls I intend to create:

  /{nome}/{foo\{bar}/{baz\{foo\}}/

And return this:

 {nome}, {foo{bar}, {baz{foo}}

What I want is to ignore the escaped characters, how can I do this? Can provide an example in any language, the most important is Regex

Answer 1

You need to make the content to be married consume both the backslash and the subsequent character as if it were one thing:

\\.|.

That is, it houses a backslash followed by anything (2 characters), and only if the first is not a backslash it matches a single character.

As for the last example (where you only want the innermost bracket), you can achieve this in this particular case (but not in general, because balancing brackets/brackets/keys does not constitute a regular language) requiring that the married content not contain a clasp opener, unless escaped:

\\.|[^\[]

The full regex would therefore be so:

\[((?:\\.|[^\[])*?)\]
\{((?:\\.|[^{])*?)\}

Example:

var str = '[abc\\[0123\\]] [efg\\[987\\]h] [olá \\[mundo\\]!] [foo [baz]]';
var regex = /\[((?:\\.|[^\[])*?)\]/g;
   
document.getElementById("saida").innerHTML += "<pre>" + str.match(regex) + "</pre><br/>"

var str = '{abc\\{0123\\}} {efg\\{987\\}h} {olá \\{mundo\\}!} {foo {baz}}';
var regex = /\{((?:\\.|[^{])*?)\}/g;

document.getElementById("saida").innerHTML += "<pre>" + str.match(regex) + "</pre><br/>"

<div id="saida"></div>

Notes:

1. In the example, I needed to use two \ in the string because otherwise the backslash would not be considered an escape character.

2. The output includes the bars; if you want to remove them, you would need to process the output of the match using maybe a replace:

   str.match(regex).replace(/\\([\[\]{}])/g, "$1");
   

3. The ?: was placed so that the parenthesis does not become a capture group. If you are not using groups, it may be omitted.