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Lookbehind alternative with both lookbehind and lookahead

I'm looking for a regex to split user supplied strings on the : character but not when the user has escaped the colon \: or it's part of a url, e.g. https://stackoverflow...
In javascript the majority of browsers don't yet support lookbehinds. Is it possible to apply some other approach for the lookbehind part?
In clojure/ Clojurescript on Chrome (which does support lookbehinds) this regex does the trick:
#"(?<!\):(?!//)"
but not in Safari (for example).

The main problem is that currently browsers aren't supporting the lookbehind, which is required to find and negate the prefix \ so we don't include \:.
One workaround (not very pretty but it works) is to first substitute the \: with some "symbol" you know will not occur naturally in your text, do your split, and the substitute back any \:.
For example, this method will return an empty element "" if you have "::" in your string:
let regex = /:(?!\/\/)/
//original string literal \: has to be expressed as \\:
let str = "http://example.com::hello:dolly:12\\:00\\:PM";
//substitute out any \:
str = str.replace(/\\:/g,"<colon>"); //http://example.com::hello:dolly:12<colon>00<colon>PM
//now we split 'normally' without lookbehind
let arr = str.split(regex); //[ 'http://example.com', '', 'hello', 'dolly', '12\\:00\\:PM' ]
//substitute back \:
arr = arr.map(element => element.replace(/<colon>/g, "\\:")); //[ 'http://example.com', '', 'hello', 'dolly', '12\\:00\\:PM' ]
console.log(arr);
If you're just after non-empty elements you can just do an arr.filter(Boolean) on it, or just use #Skeeve's matching solution as it's more elegant for this purpose.

An alternative could be to not search for the separator but to search for the elements:
var str="this:is\\:a:test:https://stackoverflow:80:test::test";
var elements= str.match(/((?:[^\\:]|\\:|:\/\/)+)/g);
// elements= [ "this", "is\\:a", "test", "https://stackoverflow", "80", "test", "test" ]
The elements may not be empty (Observe the"+" in the regexp) and how the empty element between the last 2 "test" is missing
You forgot that an URL can contain multiple colons. What about `http://me:password#myhost.com:8080/path?value=d:f'
Besides these I think it should work for you.
I think you can only overcome the disadvantages with a more or less sophisticated loop using regexp-exec.
P.S. I know the grouping isn't required here, but if you want to use it in regexp-exec, you'll need it.
Disadvantages:
P.P.S. Fixed the typo #chatnoir found

You might also make use of replace and pass a function as the second parameter.
You could use a pattern to match what you don't want and capture in a group what you want to keep. Then you can replace the part that you want to keep with a marker just as in the approach of #chatnoir and afterwards split on that marker.
:\/\/\S+|\\:|(:)
Explanation
:\/\/\S+ Match :// followed by 1+ times a non whitespace char
| Or
\\: Match \:
| Or
(:) Capture a : in group 1
Regex demo
let pattern = /:\/\/\S+|\\:|(:)/g;
let str = "string\\: or https://www.example.com:8000 or split:me or te\\:st or \\:test or notsplit\\:me:splitted or \\: or ftp://example.com :";
str = str.replace(pattern, function(match, group1) {
return group1 === undefined ? match : "<split>"
});
console.log(str.split("<split>").filter(Boolean));

Remove specific words except last in JavaScript?

I have a sentence that I would like to have only the last 'and' remaining, and remove the others.
"Lions, and tigers, and bears, and elephants", and I would like to turn this into:
"Lions, tigers, bears, and elephants".
I have tried using a regex pattern like str = str.replace(/and([^and]*)$/, '$1'); which obviously didn't work. Thanks.

Use this regex:
and (?=.*and)
and matches any and followed by a space. Space is matched so it is removed on replacement, to prevent having 2 spaces
(?=.*and) is a lookahead, meaning it will only match if followed by .*and, if followed by and
Use this code:
str = str.replace(/and (?=.*and)/g, '');

You can use a positive look-ahead (?=...), to see if there is another and ahead of the current match. You also need to make the regex global with g.
function removeAllButLastAnd(str) {
return str.replace(/and\s?(?=.*and)/g, '');
}
console.log(removeAllButLastAnd("Lions, and tigers, and bears, and elephants"));

var multipleAnd = "Lions, and tigers, and bears, and elephants";
var lastAndIndex = multipleAnd.lastIndexOf(' and');
var onlyLastAnd = multipleAnd.substring(0, lastAndIndex).replace(/ and/gi, '')+multipleAnd.substring(lastAndIndex);
console.log(onlyLastAnd);

Replace all besides the Regex group?

I was given a task to do which requires a long time to do.
The image say it all :
This is what I have : (x100 times):
And I need to extract this value only
How can I capture the value ?
I have made it with this regex :
DbCommand.*?\("(.*?)"\);
As you can see it does work :
And after the replace function (replace to $1) I do get the pure value :
but the problem is that I need only the pure values and not the rest of the unmatched group :
Question : In other words :
How can I get the purified result like :
Eservices_Claims_Get_Pending_Claims_List
Eservices_Claims_Get_Pending_Claims_Step1
Here is my code at Online regexer
Is there any way of replacing "all besides the matched group" ?
p.s. I know there are other ways of doing it but I prefer a regex solution ( which will also help me to understand regex better)

Unfortunately, JavaScript doesn't understand lookbehind. If it did, you could change your regular expression to match .*? preceded (lookbehind) by DbCommand.*?\(" and followed (lookahead) by "\);.
With that solution denied, i believe the cleanest solution is to perform two matches:
// you probably want to build the regexps dynamically
var regexG = /DbCommand.*?\("(.*?)"\);/g;
var regex = /DbCommand.*?\("(.*?)"\);/;
var matches = str.match(regexG).map(function(item){ return item.match(regex)[1] });
console.log(matches);
// ["Eservices_Claims_Get_Pending_Claims_List", "Eservices_Claims_Get_Pending_Claims_Step1"]
DEMO: http://jsbin.com/aqaBobOP/2/edit

You should be able to do a global replace of:
public static DataTable.*?{.*?DbCommand.*?\("(.*?)"\);.*?}
All I've done is changed it to match the whole block including the function definition using a bunch of .*?s.
Note: Make sure your regex settings are such that the dot (.) matches all characters, including newlines.
In fact if you want to close up all whitespace, you can slap a \s* on the front and replace with $1\n:
\s*public static DataTable.*?{.*?DbCommand.*?\("(.*?)"\);.*?}
Using your test case: http://regexr.com?37ibi

You can use this (without the ignore case and multiline option, with a global search):
pattern: (?:[^D]+|\BD|D(?!bCommand ))+|DbCommand [^"]+"([^"]+)
replace: $1\n

Try simply replacing the whole document replacing using this expression:
^(?: |\t)*(?:(?!DbCommand).)*$
You will then only be left with the lines that begin with the string DbCommand
You can then remove the spaces in between by replacing:
\r?\n\s* with \n globally.
Here is an example of this working: http://regexr.com?37ic4

JS XRegExp Replace all non characters

My objective is to replace all characters which are not dash (-) or not number or not letters in any language in a string.All of the #!()[], and all other signs to be replaced with empty string. All occurences of - should not be replaced also.
I have used for this the XRegExp plugin but it seems I cannot find the magic solution :)
I have tryed like this :
var txt = "Ad СТИНГ (ALI) - Englishmen In New York";
var regex = new XRegExp('\\p{^N}\\p{^L}',"g");
var b = XRegExp.replace(txt, regex, "")
but the result is : AСТИН(AL EnglishmeINeYork ... which is kind of weird
If I try to add also the condition for not removing the '-' character leads to make the RegEx invalid.

\\p{^N}\\p{^L} means a non-number followed by a non-letter.
Try [^\\p{N}\\p{L}-] that means a non-number, non-letter, non-dash.
A jsfiddle where to do some tests... The third XRegExp is the one you asked.

\p{^N}\p{^L}
is a non-number followed by a non-letter. You probably meant to say a character that is neither a letter nor a number:
[^\p{N}\p{L}]

// all non letters/numbers in a string => /[^a-zA-z0-9]/g
I dont know XRegExp.
but in js Regexp you can replace it by
b.replace(/[^a-zA-z0-9]/g,'')

java script Regular Expressions patterns problem

My problem start with like-
var str='0|31|2|03|.....|4|2007'
str=str.replace(/[^|]\d*[^|]/,'5');
so the output becomes like:"0|5|2|03|....|4|2007" so it replaces 31->5
But this doesn't work for replacing other segments when i change code like this:
str=str.replace(/[^|]{2}\d*[^|]/,'6');
doesn't change 2->6.
What actually i am missing here.Any help?

I think a regular expression is a bad solution for that problem. I'd rather do something like this:
var str = '0|31|2|03|4|2007';
var segments = str.split("|");
segments[1] = "35";
segments[2] = "123";
Can't think of a good way to solve this with a regexp.

Here is a specific regex solution which replaces the number following the first | pipe symbol with the number 5:
var re = /^((?:\d+\|){1})\d+/;
return text.replace(re, '$15');
If you want to replace the digits following the third |, simply change the {1} portion of the regex to {3}
Here is a generalized function that will replace any given number slot (zero-based index), with a specified new number:
function replaceNthNumber(text, n, newnum) {
var re = new RegExp("^((?:\\d+\\|){"+ n +'})\\d+');
return text.replace(re, '$1'+ newnum);
}

Firstly, you don't have to escape | in the character set, because it doesn't have any special meaning in character sets.
Secondly, you don't put quantifiers in character sets.
And finally, to create a global matching expression, you have to use the g flag.

[^\|] means anything but a '|', so in your case it only matches a digit. So it will only match anything with 2 or more digits.
Second you should put the {2} outside of the []-brackets
I'm not sure what you want to achieve here.