Develop Reference

Why would the replace with regex not work even though the regex does?

There may be a very simple answer to this, probably because of my familiarity (or possibly lack thereof) of the replace method and how it works with regex.
Let's say I have the following string: abcdefHellowxyz
I just want to strip the first six characters and the last four, to return Hello, using regex... Yes, I know there may be other ways, but I'm trying to explore the boundaries of what these methods are capable of doing...
Anyway, I've tinkered on http://regex101.com and got the following Regex worked out:
/^(.{6}).+(.{4})$/
Which seems to pass the string well and shows that abcdef is captured as group 1, and wxyz captured as group 2. But when I try to run the following:
"abcdefHellowxyz".replace(/^(.{6}).+(.{4})$/,"")
to replace those captured groups with "" I receive an empty string as my final output... Am I doing something wrong with this syntax? And if so, how does one correct it, keeping my original stance on wanting to use Regex in this manner...
Thanks so much everyone in advance...

The code below works well as you wish
"abcdefHellowxyz".replace(/^.{6}(.+).{4}$/,"$1")
I think that only use ()to capture the text you want, and in the second parameter of replace(), you can use $1 $2 ... to represent the group1 group2.
Also you can pass a function to the second parameter of replace,and transform the captured text to whatever you want in this function.
For more detail, as #Akxe recommend , you can find document on https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replace.

You are replacing any substring that matches /^(.{6}).+(.{4})$/, with this line of code:
"abcdefHellowxyz".replace(/^(.{6}).+(.{4})$/,"")
The regex matches the whole string "abcdefHellowxyz"; thus, the whole string is replaced. Instead, if you are strictly stripping by the lengths of the extraneous substrings, you could simply use substring or substr.
Edit
The answer you're probably looking for is capturing the middle token, instead of the outer ones:
var str = "abcdefHellowxyz";
var matches = str.match(/^.{6}(.+).{4}$/);
str = matches[1]; // index 0 is entire match
console.log(str);

Regex appears to ignore multiple piped characters

Apologies for the awkward question title, I have the following JavaScript:
var wordRe = new RegExp('\\b(?:(?![<^>"])fox|hello(?![<\/">]))\\b', 'g'); // Words regex
console.log('<span>hello</span> <hello>fox</hello> fox link hello my name is fox'.replace(wordRe, 'foo'));
What I'm trying to do is replace any word that isn't nested in a HTML tag, or part of a HTML tag itself. I.e I want to only match "plain" text. The expression seems to be ignoring the rule for the first piped match "fox", and replacing it when it shouldn't be.
Can anyone point out why this is? I think I might have organised the expression incorrectly (at least the negative lookahead).
Here is the JSFiddle.
I'd also like to add that I am aware of the implications of using regex with HTML :)

For your regex work, you want lookbehind. However, as of this writing, this feature is not supported in Javascript.
Here is a workaround:
Instead of matching what we want, we will match what we don't want and remove it from our input string. Later, we can perform the replace on the cleaned input string.
var nonWordRe = new RegExp('<([^>]+).*?>[^<]+?</\\1>', 'g');
var test = '<span>hello</span> <hello>fox</hello> fox link hello my name is fox';
var cleanedTest = test.replace(nonWordRe, '');
var final = cleanedTest.replace(/fox|hello/, 'foo'); // once trimmed final=='foo my name is foo'
NOTA:
I have build this workaround based on your sample. But here are some points that may need to be explored if you face them:
you may need to remove self closing tags (<([^>]+).*?/\>) from the test string
you may need to trim the final string (final)
you may need a descent html parser if tags can contain other tags as HTML allow this.
Javascript doesn't, again as of this writing, recursive patterns.
Demo
http://jsfiddle.net/yXd82/2/

Regex to match all instances not inside quotes

From this q/a, I deduced that matching all instances of a given regex not inside quotes, is impossible. That is, it can't match escaped quotes (ex: "this whole \"match\" should be taken"). If there is a way to do it that I don't know about, that would solve my problem.
If not, however, I'd like to know if there is any efficient alternative that could be used in JavaScript. I've thought about it a bit, but can't come with any elegant solutions that would work in most, if not all, cases.
Specifically, I just need the alternative to work with .split() and .replace() methods, but if it could be more generalized, that would be the best.
For Example:
An input string of: +bar+baz"not+or\"+or+\"this+"foo+bar+
replacing + with #, not inside quotes, would return: #bar#baz"not+or\"+or+\"this+"foo#bar#

Actually, you can match all instances of a regex not inside quotes for any string, where each opening quote is closed again. Say, as in you example above, you want to match \+.
The key observation here is, that a word is outside quotes if there are an even number of quotes following it. This can be modeled as a look-ahead assertion:
\+(?=([^"]*"[^"]*")*[^"]*$)
Now, you'd like to not count escaped quotes. This gets a little more complicated. Instead of [^"]* , which advanced to the next quote, you need to consider backslashes as well and use [^"\\]*. After you arrive at either a backslash or a quote, you need to ignore the next character if you encounter a backslash, or else advance to the next unescaped quote. That looks like (\\.|"([^"\\]*\\.)*[^"\\]*"). Combined, you arrive at
\+(?=([^"\\]*(\\.|"([^"\\]*\\.)*[^"\\]*"))*[^"]*$)
I admit it is a little cryptic. =)

Azmisov, resurrecting this question because you said you were looking for any efficient alternative that could be used in JavaScript and any elegant solutions that would work in most, if not all, cases.
There happens to be a simple, general solution that wasn't mentioned.
Compared with alternatives, the regex for this solution is amazingly simple:
"[^"]+"|(\+)
The idea is that we match but ignore anything within quotes to neutralize that content (on the left side of the alternation). On the right side, we capture all the + that were not neutralized into Group 1, and the replace function examines Group 1. Here is full working code:
<script>
var subject = '+bar+baz"not+these+"foo+bar+';
var regex = /"[^"]+"|(\+)/g;
replaced = subject.replace(regex, function(m, group1) {
if (!group1) return m;
else return "#";
});
document.write(replaced);
Online demo
You can use the same principle to match or split. See the question and article in the reference, which will also point you code samples.
Hope this gives you a different idea of a very general way to do this. :)
What about Empty Strings?
The above is a general answer to showcase the technique. It can be tweaked depending on your exact needs. If you worry that your text might contain empty strings, just change the quantifier inside the string-capture expression from + to *:
"[^"]*"|(\+)
See demo.
What about Escaped Quotes?
Again, the above is a general answer to showcase the technique. Not only can the "ignore this match" regex can be refined to your needs, you can add multiple expressions to ignore. For instance, if you want to make sure escaped quotes are adequately ignored, you can start by adding an alternation \\"| in front of the other two in order to match (and ignore) straggling escaped double quotes.
Next, within the section "[^"]*" that captures the content of double-quoted strings, you can add an alternation to ensure escaped double quotes are matched before their " has a chance to turn into a closing sentinel, turning it into "(?:\\"|[^"])*"
The resulting expression has three branches:
\\" to match and ignore
"(?:\\"|[^"])*" to match and ignore
(\+) to match, capture and handle
Note that in other regex flavors, we could do this job more easily with lookbehind, but JS doesn't support it.
The full regex becomes:
\\"|"(?:\\"|[^"])*"|(\+)
See regex demo and full script.
Reference
How to match pattern except in situations s1, s2, s3
How to match a pattern unless...

You can do it in three steps.
Use a regex global replace to extract all string body contents into a side-table.
Do your comma translation
Use a regex global replace to swap the string bodies back
Code below
// Step 1
var sideTable = [];
myString = myString.replace(
/"(?:[^"\\]|\\.)*"/g,
function (_) {
var index = sideTable.length;
sideTable[index] = _;
return '"' + index + '"';
});
// Step 2, replace commas with newlines
myString = myString.replace(/,/g, "\n");
// Step 3, swap the string bodies back
myString = myString.replace(/"(\d+)"/g,
function (_, index) {
return sideTable[index];
});
If you run that after setting
myString = '{:a "ab,cd, efg", :b "ab,def, egf,", :c "Conjecture"}';
you should get
{:a "ab,cd, efg"
:b "ab,def, egf,"
:c "Conjecture"}
It works, because after step 1,
myString = '{:a "0", :b "1", :c "2"}'
sideTable = ["ab,cd, efg", "ab,def, egf,", "Conjecture"];
so the only commas in myString are outside strings. Step 2, then turns commas into newlines:
myString = '{:a "0"\n :b "1"\n :c "2"}'
Finally we replace the strings that only contain numbers with their original content.

Although the answer by zx81 seems to be the best performing and clean one, it needes these fixes to correctly catch the escaped quotes:
var subject = '+bar+baz"not+or\\"+or+\\"this+"foo+bar+';
and
var regex = /"(?:[^"\\]|\\.)*"|(\+)/g;
Also the already mentioned "group1 === undefined" or "!group1".
Especially 2. seems important to actually take everything asked in the original question into account.
It should be mentioned though that this method implicitly requires the string to not have escaped quotes outside of unescaped quote pairs.

Building a Hashtag in Javascript without matching Anchor Names, BBCode or Escaped Characters

I would like to convert any instances of a hashtag in a String into a linked URL:
#hashtag -> should have "#hashtag" linked.
This is a #hashtag -> should have "#hashtag" linked.
This is a [url=http://www.mysite.com/#name]named anchor[/url] -> should not be linked.
This isn't a pretty way to use quotes -> should not be linked.
Here is my current code:
String.prototype.parseHashtag = function() {
return this.replace(/[^&][#]+[A-Za-z0-9-_]+(?!])/, function(t) {
var tag = t.replace("#","")
return t.link("http://www.mysite.com/tag/"+tag);
});
};
Currently, this appears to fix escaped characters (by excluding matches with the amperstand), handles named anchors, but it doesn't link the #hashtag if it's the first thing in the message, and it seems to grab include the 1-2 characters prior to the "#" in the link.
Halp!

How about the following:
/(^|[^&])#([A-Za-z0-9_-]+)(?![A-Za-z0-9_\]-])/g
matches the hashtags in your example. Since JavaScript doesn't support lookbehind, it tries to either match the start of the string or any character except & before the hashtag. It captures the latter so it can later be replaced. It also captures the name of the hashtag.
So, for example:
subject.replace(/(^|[^&])#([A-Za-z0-9_-]+)(?![A-Za-z0-9_\]-])/g, "$1http://www.mysite.com/tag/$2");
will transform
#hashtag
This is a #hashtag and this one #too.
This is a [url=http://www.mysite.com/#name]named anchor[/url]
This isn't a pretty way to use quotes
into
http://www.mysite.com/tag/hashtag
This is a http://www.mysite.com/tag/hashtag and this one http://www.mysite.com/tag/too.
This is a [url=http://www.mysite.com/#name]named anchor[/url]
This isn't a pretty way to use quotes
This probably isn't what t.link() (which I don't know) would have returned, but I hope it's a good starting point.

There is an open-source Ruby gem to do this sort of thing (hashtags and #usernames) called twitter-text. You might get some ideas and regexes from that, or try out this JavaScript port.
Using the JavaScript port, you'll want to just do:
var linked = TwitterText.auto_link_hashtags(text, {hashtag_url_base: "http://www.mysite.come/tag/"});

Tim, your solution was almost perfect. Here's what I ended up using:
subject.replace(/(^| )#([A-Za-z0-9_-]+)(?![A-Za-z0-9_\]-])/g, "$1#$2");
The only change is the first conditional, changed it to match the beginning of the string or a space character. (I tried \s, but that didn't work at all.)

How to search csv string and return a match by using a Javascript regex

I'm trying to extract the first user-right from semicolon separated string which matches a pattern.
Users rights are stored in format:
LAA;LA_1;LA_2;LE_3;
String is empty if user does not have any rights.
My best solution so far is to use the following regex in regex.replace statement:
.*?;(LA_[^;]*)?.*
(The question mark at the end of group is for the purpose of matching the whole line in case user has not the right and replace it with empty string to signal that she doesn't have it.)
However, it doesn't work correctly in case the searched right is in the first position:
LA_1;LA_2;LE_3;
It is easy to fix it by just adding a semicolon at the beginning of line before regex replace but my question is, why doesn't the following regex match it?
.*?(?:(?:^|;)(LA_[^;]*))?.*
I have tried numerous other regular expressions to find the solution but so far without success.

I am not sure I get your question right, but in regards to the regular expressions you are using, you are overcomplicating them for no clear reason (at least not to me). You might want something like:
function getFirstRight(rights) {
var m = rights.match(/(^|;)(LA_[^;]*)/)
return m ? m[2] : "";
}

You could just split the string first:
function getFirstRight(rights)
{
return rights.split(";",1)[0] || "";
}

To answer the specific question "why doesn't the following regex match it?", one problem is the mix of this at the beginning:
.*?
eventually followed by:
^|;
Which might be like saying, skip over any extra characters until you reach either the start or a semicolon. But you can't skip over anything and then later arrive at the start (unless it involves newlines in a multiline string).
Something like this works:
.*?(\bLA_[^;]).*
Meaning, skip over characters until a word boundary followed by "LA_".