Develop Reference

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)

Replace all “?” by “&” except first

I’d would to replace all “?” by “&” except the first one by javascript. I found some regular expressions but they didn’t work.
I have something like:
home/?a=1
home/?a=1?b=2
home/?a=1?b=2?c=3
And I would like:
home/?a=1
home/?a=1&b=2
home/?a=1&b=2&c=3
Someone know how to I can do it?
Thanks!

I don't think it's possible with regex but you can split the string and then join it back together, manually replacing the first occurance:
var split = 'home/?a=1?b=2'.split('?'); // [ 'home/', 'a=1', 'b=2' ]
var replaced = split[0] + '?' + split.slice(1).join('&') // 'home/?a=1&b=2'
console.log(replaced);

You could match from the start of the string not a question mark using a negated character class [^?]+ followed by matching a question mark and capture that in the first capturing group. In the second capturing group capture the rest of the string.
Use replace and pass a function as the second parameter where you return the first capturing group followed by the second capturing group where all the question marks are replaced by &
let strings = [
"home/?a=1",
"home/?a=1?b=2",
"home/?a=1?b=2?c=3"
];
strings.forEach((str) => {
let result = str.replace(/(^[^?]+\?)(.*)/, function(match, group1, group2) {
return group1 + group2.replace(/\?/g, '&')
});
console.log(result);
});

You can split it by "?" and then rewrap the array:
var string = "home/?a=1?b=2";
var str = string.split('?');
var new = str[0] + '?'; // text before first '?' and first '?'
for( var x = 1; x < str.length; x++ ) {
new = new + str[x];
if( x != ( str.length - 1 ) ) new = new + '&'; //to avoid place a '&' after the string
}

You can use /([^\/])\?/ as pattern in regex that match any ? character that isn't after / character.
var str = str.replace(/([^\/])\?/g, "$1&");
var str = "home/?a=1\nhome/?a=1?b=2\nhome/?a=1?b=2?c=3\n".replace(/([^\/])\?/g, "$1&");
console.log(str);

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.

Regular expression to parse jQuery-selector-like string

text = '#container a.filter(.top).filter(.bottom).filter(.middle)';
regex = /(.*?)\.filter\((.*?)\)/;
matches = text.match(regex);
log(matches);
// matches[1] is '#container a'
//matchss[2] is '.top'
I expect to capture
matches[1] is '#container a'
matches[2] is '.top'
matches[3] is '.bottom'
matches[4] is '.middle'
One solution would be to split the string into #container a and rest. Then take rest and execute recursive exec to get item inside ().
Update: I am posting a solution that does work. However I am looking for a better solution. Don't really like the idea of splitting the string and then processing
Here is a solution that works.
matches = [];
var text = '#container a.filter(.top).filter(.bottom).filter(.middle)';
var regex = /(.*?)\.filter\((.*?)\)/;
var match = regex.exec(text);
firstPart = text.substring(match.index,match[1].length);
rest = text.substring(matchLength, text.length);
matches.push(firstPart);
regex = /\.filter\((.*?)\)/g;
while ((match = regex.exec(rest)) != null) {
matches.push(match[1]);
}
log(matches);
Looking for a better solution.

This will match the single example you posted:
<html>
<body>
<script type="text/javascript">
text = '#container a.filter(.top).filter(.bottom).filter(.middle)';
matches = text.match(/^[^.]*|\.[^.)]*(?=\))/g);
document.write(matches);
</script>
</body>
</html>
which produces:
#container a,.top,.bottom,.middle
EDIT
Here's a short explanation:
^ # match the beginning of the input
[^.]* # match any character other than '.' and repeat it zero or more times
#
| # OR
#
\. # match the character '.'
[^.)]* # match any character other than '.' and ')' and repeat it zero or more times
(?= # start positive look ahead
\) # match the character ')'
) # end positive look ahead
EDIT part II
The regex looks for two types of character sequences:
one ore more characters starting from the start of the string up to the first ., the regex: ^[^.]*
or it matches a character sequence starting with a . followed by zero or more characters other than . and ), \.[^.)]*, but must have a ) ahead of it: (?=\)). This last requirement causes .filter not to match.

You have to iterate, I think.
var head, filters = [];
text.replace(/^([^.]*)(\..*)$/, function(_, h, rem) {
head = h;
rem.replace(/\.filter\(([^)]*)\)/g, function(_, f) {
filters.push(f);
});
});
console.log("head: " + head + " filters: " + filters);
The ability to use functions as the second argument to String.replace is one of my favorite things about Javascript :-)

You need to do several matches repeatedly, starting where the last match ends (see while example at https://developer.mozilla.org/en/Core_JavaScript_1.5_Reference/Global_Objects/RegExp/exec):
If your regular expression uses the "g" flag, you can use the exec method multiple times to find successive matches in the same string. When you do so, the search starts at the substring of str specified by the regular expression's lastIndex property. For example, assume you have this script:
var myRe = /ab*/g;
var str = "abbcdefabh";
var myArray;
while ((myArray = myRe.exec(str)) != null)
{
var msg = "Found " + myArray[0] + ". ";
msg += "Next match starts at " + myRe.lastIndex;
print(msg);
}
This script displays the following text:
Found abb. Next match starts at 3
Found ab. Next match starts at 9
However, this case would be better solved using a custom-built parser. Regular expressions are not an effective solution to this problem, if you ask me.

var text = '#container a.filter(.top).filter(.bottom).filter(.middle)';
var result = text.split('.filter');
console.log(result[0]);
console.log(result[1]);
console.log(result[2]);
console.log(result[3]);

text.split() with regex does the trick.
var text = '#container a.filter(.top).filter(.bottom).filter(.middle)';
var parts = text.split(/(\.[^.()]+)/);
var matches = [parts[0]];
for (var i = 3; i < parts.length; i += 4) {
matches.push(parts[i]);
}
console.log(matches);