Develop Reference

Regex match apostrophe inside, but not around words, inside a character set

I'm counting how many times different words appear in a text using Regular Expressions in JavaScript. My problem is when I have quoted words: 'word' should be counted simply as word (without the quotes, otherwise they'll behave as two different words), while it's should be counted as a whole word.
(?<=\w)(')(?=\w)
This regex can identify apostrophes inside, but not around words. Problem is, I can't use it inside a character set such as [\w]+.
(?<=\w)(')(?=\w)|[\w]+
Will count it's a 'miracle' of nature as 7 words, instead of 5 (it, ', s becoming 3 different words). Also, the third word should be selected simply as miracle, and not as 'miracle'.
To make things even more complicated, I need to capture diacritics too, so I'm using [A-Za-zÀ-ÖØ-öø-ÿ] instead of \w.
How can I accomplish that?

1) You can simply use /[^\s]+/g regex
const str = `it's a 'miracle' of nature`;
const result = str.match(/[^\s]+/g);
console.log(result.length);
console.log(result);
2) If you are calculating total number of words in a string then you can also use split as:
const str = `it's a 'miracle' of nature`;
const result = str.split(/\s+/);
console.log(result.length);
console.log(result);
3) If you want a word without quote at the starting and at the end then you can do as:
const str = `it's a 'miracle' of nature`;
const result = str.match(/[^\s]+/g).map((s) => {
s = s[0] === "'" ? s.slice(1) : s;
s = s[s.length - 1] === "'" ? s.slice(0, -1) : s;
return s;
});
console.log(result.length);
console.log(result);

You might use an alternation with 2 capture groups, and then check for the values of those groups.
(?<!\S)'(\S+)'(?!\S)|(\S+)
(?<!\S)' Negative lookbehind, assert a whitespace boundary to the left and match '
(\S+) Capture group 1, match 1+ non whitespace chars
'(?!\S) Match ' and assert a whitespace boundary to the right
| Or
(\S+) Capture group 2, match 1+ non whitespace chars
See a regex demo.
const regex = /(?<!\S)'(\S+)'(?!\S)|(\S+)/g;
const s = "it's a 'miracle' of nature";
Array.from(s.matchAll(regex), m => {
if (m[1]) console.log(m[1])
if (m[2]) console.log(m[2])
});

Regex replace all character except last 5 character and whitespace with plus sign

I wanted to replace all characters except its last 5 character and the whitespace with +
var str = "HFGR56 GGKDJ JGGHG JGJGIR"
var returnstr = str.replace(/\d+(?=\d{4})/, '+');
the result should be "++++++ ++++ +++++ JGJGIR" but in the above code I don't know how to exclude whitespace

You need to match each character individually, and you need to allow a match only if more than six characters of that type follow.
I'm assuming that you want to replace alphanumeric characters. Those can be matched by \w. All other characters will be matched by \W.
This gives us:
returnstr = str.replace(/\w(?=(?:\W*\w){6})/g, "+");
Test it live on regex101.com.

The pattern \d+(?=\d{4}) does not match in the example string as is matches 1+ digits asserting what is on the right are 4 digits.
Another option is to match the space and 5+ word characters till the end of the string or match a single word character in group 1 using an alternation.
In the callback of replace, return a + if you have matched group 1, else return the match.
\w{5,}$|(\w)
Regex demo
let pattern = / \w{5,}$|(\w)/g;
let str = "HFGR56 GGKDJ JGGHG JGJGIR"
.replace(pattern, (m, g1) => g1 ? '+' : m);
console.log(str);

Another way is to replace a group at a time where the number of +
replaced is based on the length of the characters matched:
var target = "HFGR56 GGKDJ JGGHG JGJGIR";
var target = target.replace(
/(\S+)(?!$|\S)/g,
function( m, g1 )
{
var len = parseInt( g1.length ) + 1;
//return "+".repeat( len ); // Non-IE (quick)
return Array( len ).join("+"); // IE (slow)
} );
console.log ( target );

You can use negative lookahead with string end anchor.
\w(?!\w{0,5}$)
Match any word character which is not followed by 0 to 5 characters and end of string.
var str = "HFGR56 GGKDJ JGGHG JGJGIR"
var returnstr = str.replace(/\w(?!\w{0,5}$)/g, '+');
console.log(returnstr)

Regex - ignoring text between quotes / HTML(5) attribute filtering

So I have this Regular expression, which basically has to filter the given string to a HTML(5) format list of attributes. It currently isn't doing my fulfilling, but that's about to change! (I hope so)
I'm trying to achieve that whenever an occurrence is found, it selects the text until the next occurrence OR the end of the string, as the second match. So if you'd take a look at the current regular expression:
/([a-zA-Z]+|[a-zA-Z]+-[a-zA-Z0-9]+)=["']/g
A string like this: hey="hey world" hey-heyhhhhh3123="Hello world" data-goed="hey"
Would be filtered / matched out like this:
MATCH 1. [0-3] `hey`
MATCH 2. [16-32] `hey-heyhhhhh3123`
MATCH 3. [47-56] `data-goed`
This has to be seen as the attribute-name(s), and now.. we just have to fetch the attribute's value(s). So the mentioned string has to have an outcome like this:
MATCH 1.
1 [0-3] `hey`
2 [6-14] `hey world`
MATCH 2.
1 [16-32] `hey-heyhhhhh3123`
2 [35-45] `Hello world`
MATCH 3.
1 [47-56] `data-goed`
2 [59-61] `hey`
Could anyone try and help me to get my fulfilling? It would be appericiated a lot!

You can use
/([^\s=]+)=(?:"([^"\\]*(?:\\.[^"\\]*)*)"|(\S+))/g
See regex demo
Pattern details:
([^\s=]+) - Group 1 capturing 1 or more characters other than whitespace and = symbol
= - an equal sign
(?:"([^"\\]*(?:\\.[^"\\]*)*)"|(\S+)) - a non-capturing group of 2 alternatives (one more '([^'\\]*(?:\\.[^'\\]*)*)' alternative can be added to account for single quoted string literals)
"([^"\\]*(?:\\.[^"\\]*)*)" - a double quoted string literal pattern:
" - a double quote
([^"\\]*(?:\\.[^"\\]*)*) - Group 2 capturing 0+ characters other than \ and ", followed with 0+ sequences of any escaped symbol followed with 0+ characters other than \ and "
" - a closing dlouble quote
| - or
(\S+) - Group 3 capturing one or more non-whitespace characters
JS demo (no single quoted support):
var re = /([^\s=]+)=(?:"([^"\\]*(?:\\.[^"\\]*)*)"|(\S+))/g;
var str = 'hey="hey world" hey-heyhhhhh3123="Hello \\"world\\"" data-goed="hey" more=here';
var res = [];
while ((m = re.exec(str)) !== null) {
if (m[3]) {
res.push([m[1], m[3]]);
} else {
res.push([m[1], m[2]]);
}
}
console.log(res);
JS demo (with single quoted literal support)
var re = /([^\s=]+)=(?:"([^"\\]*(?:\\.[^"\\]*)*)"|'([^'\\]*(?:\\.[^'\\]*)*)'|(\S+))/g;
var str = 'pseudoprefix-before=\'hey1"\' data-hey="hey\'hey" more=data and="more \\"here\\""';
var res = [];
while ((m = re.exec(str)) !== null) {
if (m[2]) {
res.push([m[1], m[2]])
} else if (m[3]) {
res.push([m[1], m[3]])
} else if (m[4]) {
res.push([m[1], m[4]])
}
}
console.log(res);

Regular Expression Url Address

My URL String:
https://stackoverflow.com/questions
My regex : [\w.]+
Result: ["http","stackoverflow.com","questions"]
How I ignore .* in stackoverflow.com only one regex.
Result I want: ["http","stackoverflow","questions"]

You can use this regex that captures into Group 1 all alphanumeric/underscore chunks that are not preceded with .:
/(?:^|[^.])\b(\w+)\b/g
See the regex demo.
Breakdown:
(?:^|[^.]) - matches (but does not store in a group buffer) the beginning of a string or any character but a literal dot
\b - leading word boundary
(\w+) - Group 1 capturing one or more word characters
\b - trailing word boundary
Sample code:
var re = /(?:^|[^.])\b(\w+)\b/g;
var str = 'http://stackoverflow.com/questions';
var res = [];
while ((m = re.exec(str)) !== null) {
res.push(m[1]);
}
document.body.innerHTML = "<pre>" + JSON.stringify(res, 0, 4) + "</pre>";
Another solution based on the assumption the word character should not be followed with /:
\b\w+\b(?!\/)
See another regex demo
var re = /\b\w+\b(?!\/)/g;
var str = 'http://stackoverflow.com/questions';
var res = str.match(re);
document.body.innerHTML = "<pre>" + JSON.stringify(res, 0, 4) + "</pre>";
Note that both solutions require a word boundary to work properly, just a negated character class (#1) or a lookahead (#2) won't work by themselves (partial matches will be rejected thanks to \b).

regex to match all words but AND, OR and NOT

In my javascript app I have this random string:
büert AND NOT 3454jhadf üasdfsdf OR technüology AND (bar OR bas)
and i would like to match all words special chars and numbers besides the words AND, OR and NOT.
I tried is this
/(?!AND|OR|NOT)\b[\u00C0-\u017F\w\d]+/gi
which results in
["büert", "3454jhadf", "asdfsdf", "technüology", "bar", "bas"]
but this one does not match the ü or any other letter outside the a-z alphabet at the beginning or at the end of a word because of the \b word boundary.
removing the \b oddly ends up matching part or the words i would like to exclude:
/(?!AND|OR|NOT)[\u00C0-\u017F\w\d]+/gi
result is
["büert", "ND", "OT", "3454jhadf", "üasdfsdf", "R", "technüology", "ND", "bar", "R", "bas"]
what is the correct way to match all words no matter what type of characters they contain besides the ones i want exclude?

The issue here has its roots in the fact that \b (and \w, and other shorthand classes) are not Unicode-aware in JavaScript.
Now, there are 2 ways to achieve what you want.
1. SPLIT WITH PATTERN(S) YOU WANT TO DISCARD
var re = /\s*\b(?:AND|OR|NOT)\b\s*|[()]/;
var s = "büert AND NOT 3454jhadf üasdfsdf OR technüology AND (bar OR bas)";
var res = s.split(re).filter(Boolean);
document.body.innerHTML += JSON.stringify(res, 0, 4);
// = > [ "büert", "3454jhadf üasdfsdf", "technüology", "bar", "bas" ]
Note the use of a non-capturing group (?:...) so as not to include the unwanted words into the resulting array. Also, you need to add all punctuation and other unwanted characters to the character class.
2. MATCH USING CUSTOM BOUNDARIES
You can use groupings with anchors/reverse negated character class in a regex like this:
(^|[^\u00C0-\u017F\w])(?!(?:AND|OR|NOT)(?=[^\u00C0-\u017F\w]|$))([\u00C0-\u017F\w]+)(?=[^\u00C0-\u017F\w]|$)
The capure group 2 will hold the values you need.
See regex demo
JS code demo:
var re = /(^|[^\u00C0-\u017F\w])(?!(?:AND|OR|NOT)(?=[^\u00C0-\u017F\w]|$))([\u00C0-\u017F\w]+)(?=[^\u00C0-\u017F\w]|$)/gi;
var str = 'büert AND NOT 3454jhadf üasdfsdf OR technüology AND (bar OR bas)';
var m;
var arr = [];
while ((m = re.exec(str)) !== null) {
arr.push(m[2]);
}
document.body.innerHTML += JSON.stringify(arr);
or with a block to build the regex dynamically:
var bndry = "[^\\u00C0-\\u017F\\w]";
var re = RegExp("(^|" + bndry + ")" + // starting boundary
"(?!(?:AND|OR|NOT)(?=" + bndry + "|$))" + // restriction
"([\\u00C0-\\u017F\\w]+)" + // match and capture our string
"(?=" + bndry + "|$)" // set trailing boundary
, "g");
var str = 'büert AND NOT 3454jhadf üasdfsdf OR technüology AND (bar OR bas)';
var m, arr = [];
while ((m = re.exec(str)) !== null) {
arr.push(m[2]);
}
document.body.innerHTML += JSON.stringify(arr);
Explanation:
(^|[^\u00C0-\u017F\w]) - our custom boundary (match a string start with ^ or any character outside the [\u00C0-\u017F\w] range)
(?!(?:AND|OR|NOT)(?=[^\u00C0-\u017F\w]|$)) - a restriction on the match: the match is failed if there are AND or OR or NOT followed by string end or characters other than those in the \u00C0-\u017F range or non-word character
([\u00C0-\u017F\w]+) - match word characters ([a-zA-Z0-9_]) or those from the \u00C0-\u017F range
(?=[^\u00C0-\u017F\w]|$) - the trailing boundary, either string end ($) or characters other than those in the \u00C0-\u017F range or non-word character.