Javascript Regular Expression excluding a word

Question

i am familiar with javascript regular expression to not match a word but it does not help me much. when given a string (with any type of characters), i wish to parse it around two tokens, "//" and "\\". i did the following:

var patt = /.*[^"//"]/gm;
patt.exec(str);

but it seems to match any occurrences of the characters between the quotes, i.e. "/" and "//". how may i achieve it?

What are you trying to do? Give us an example for `str` and the expected result. — greut, Jan 15 '12 at 13:20
What exactely do you want? Can you include some **consistent** examples in your question? Such as a couple of Input-output pairs. — Rob W, Jan 15 '12 at 13:20
Could you please give an example of what you want to achieve? It's not fully clear what you want to do here (at least to me). — Aleksandrs Ulme, Jan 15 '12 at 13:21
@RobW i wish to parse a given string between specified tokens, which are "//" and "\\". for instance var str = "hey//you//how\\you\\are//doing"; — MrIzik, Jan 15 '12 at 13:31
@MrIzik: Your question says the closing delimiter is one backslash, your comment says two. Which is it? — T.J. Crowder, Jan 15 '12 at 13:32
@T.J.Crowder It's `\\ ` and `//`. Click on `edit` question, markdown turned two backslashes in one. — Rob W, Jan 15 '12 at 13:33

score 3 · Answer 1 · edited Jun 20 '20 at 09:12

When you start a character class with ^ (as you have in [^"//"]), it means "any character except the ones listed." So [^"//"] means "match one of any character except " and / (it ignores the fact you've listed each of them twice).

If you're trying to match the text between two slashes (//) and one backslash (\) (as per your question; you later made a comment suggesting it's two backslashes, I address that later), then:

var match = str.match(/\/\/(.*?)\\/);    // Match between // and \

Note that we have to escape the slashes because the slash is the regular expression delimiter; and we have to escape backslashes because the backslash is the escape character.

The above means "match two slashes followed by zero or more of any character followed by a backslash." The ? after * makes * non-greedy (so it will consume as few characters as it can to satisfy the expression). The () create a capture group, which in the match object will receive the characters that matched in that position.

Example:

test("foo");
test("foo //bar");
test("foo //bar\\");
test("foo //bar\\ baz");
test("bar\\ baz");
test("//bar\\ baz");
test("foo //bar\\ baz \\ more \\ more");

function test(str) {
    var m = str.match(/\/\/(.*?)\\/),
        cap = (m && m[1]) || "<em>nothing</em>";
    display("Str: <code>" + str + "</code>: Captured <code>" + cap + "</code>");
}

Output:

Str: foo: Captured nothing

Str: foo //bar: Captured nothing

Str: foo //bar\: Captured bar

Str: foo //bar\ baz: Captured bar

Str: bar\ baz: Captured nothing

Str: //bar\ baz: Captured bar

Str: foo //bar\ baz \ more \ more: Captured bar

Live copy

Or for two backslashes:

var match = str.match(/\/\/(.*?)\\\\/);  // Match between // and \\

Live copy (the output is the same, just with two backslashes)

Some reading on JavaScript regular expressions:

The spec (that's a handy linkable HTML rendering of it; the canonical one is a PDF here)
Mozilla's excellent RegExp page

thank you for your answer though i did not understand how to use it, here is an example of what i am trying to parse -- "hey[[you[[how[[you]]are[[doing"; -- and the delimiters are "[[" and "]]". — MrIzik, Jan 15 '12 at 13:38
@MrIzik: What do you want to do with that example? Specifically, with the nesting? You may need a greedy capture and a loop rather than the non-greedy version, although those delimiters are unbalanced. Or do you want just to get `["hey", "you", "how", "you", "are", "doing"]` (which would be completely different from the question actually asked). — T.J. Crowder, Jan 15 '12 at 13:44

score 1 · Answer 2 · answered Jan 15 '12 at 14:02

1

How about simply

var tokens = str.split(/\/\/|\\\\/);

answered Jan 15 '12 at 14:02

Borodin

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Javascript Regular Expression excluding a word

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