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Javascript regex split for special characters

Have a look at this fiddle http://jsfiddle.net/yeXWv/
Here I want to split the characters which starts with [* or [*# and ends with *]. The current regular expression split the string which starts with [*# but not [*. I have tried the following patterns,
/(\[\*\#*[a-zA-Z0-9]+\*\])/g
/(\[\*\#{0,1}[a-zA-Z0-9]+\*\])/g
Thanks in advance.

Try making the hash character optional:
/(\[\*\#?[a-zA-Z0-9 ]+\*\])/g
Edit: added missing white space :-)

You forgot to allow for spaces, which was the real problem -- not the missing # character.
/(\[\*\#?[a-zA-Z0-9 ]+\*\])/g
That will preserve the [*...*] strings in the output array. To omit them, remove the parentheses:
/\[\*\#?[a-zA-Z0-9 ]+\*\]/g
http://jsfiddle.net/mblase75/zU576/

Alternation operator inside square brackets does not work

I'm creating a javascript regex to match queries in a search engine string. I am having a problem with alternation. I have the following regex:
.*baidu.com.*[/?].*wd{1}=
I want to be able to match strings that have the string 'word' or 'qw' in addition to 'wd', but everything I try is unsuccessful. I thought I would be able to do something like the following:
.*baidu.com.*[/?].*[wd|word|qw]{1}=
but it does not seem to work.

replace [wd|word|qw] with (wd|word|qw) or (?:wd|word|qw).
[] denotes character sets, () denotes logical groupings.

Your expression:
.*baidu.com.*[/?].*[wd|word|qw]{1}=
does need a few changes, including [wd|word|qw] to (wd|word|qw) and getting rid of the redundant {1}, like so:
.*baidu.com.*[/?].*(wd|word|qw)=
But you also need to understand that the first part of your expression (.*baidu.com.*[/?].*) will match baidu.com hello what spelling/handle????????? or hbaidu-com/ or even something like lkas----jhdf lkja$##!3hdsfbaidugcomlaksjhdf.[($?lakshf, because the dot (.) matches any character except newlines... to match a literal dot, you have to escape it with a backslash (like \.)
There are several approaches you could take to match things in a URL, but we could help you more if you tell us what you are trying to do or accomplish - perhaps regex is not the best solution or (EDIT) only part of the best solution?

Struggling with regex to match only two of a character, not three

I need to match all occurrences of // in a string in a Javascript regex
It can't match /// or /
So far I have (.*[^\/])\/{2}([^\/].*)
which is basically "something that isn't /, followed by // followed by something that isn't /"
The approach seems to work apart from when the string I want to match starts with //
This doesn't work:
//example
This does
stuff // example
How do I solve this problem?
Edit: A bit more context - I am trying to replace // with !, so I am then using:
result = result.replace(myRegex, "$1 ! $2");

Replace two slashes that either begin the string or do not follow a slash,
and are followed by anything not a slash or the end of the string.
s=s.replace(/(^|[^/])\/{2}([^/]|$)/g,'$1!$2');

It looks like it wouldn't work for example// either.
The problem is because you're matching // preceded and followed by at least one non-slash character. This can be solved by anchoring the regex, and then you can make the preceding/following text optional:
^(.*[^\/])?\/{2}([^\/].*)?$

Use negative lookahead/lookbehind assertions:
(.*)(?<!/)//(?!/)(.*)

Use this:
/([^/]*)(\/{2})([^/]*)/g
e.g.
alert("///exam//ple".replace(/([^/]*)(\/{2})([^/]*)/g, "$1$3"));
EDIT: Updated the expression as per the comment.
/[/]{2}/
e.g:
alert("//example".replace(/[/]{2}/, ""));

This does not answer the OP's question about using regex, but since some of the original comments suggested using .replaceAll, since not everyone who reads the question in the future wants to use regex, since people might mistakenly assume that regex is the only alternative, and since these details cannot be accommodated by submitting a comment, here's a poor man's non-regex approach:
Temporarily replace the three contiguous characters with something that would never naturally occur — really important when dealing with user-entered values.
Replace the remaining two contiguous characters using .replaceAll().
Return the original three contiguous characters.
For instance, let's say you wanted to remove all instances of ".." without affecting occurrences of "...".
var cleansedText = $(this).text().toString()
.replaceAll("...", "☰☸☧")
.replaceAll("..", "")
.replaceAll("☰☸☧", "...")
;
$(this).text(cleansedText);
Perhaps not as fast as regex for longer strings, but works great for short ones.

Creating Slugs from Titles?

I have everything in place to create slugs from titles, but there is one issue. My RegEx replaces spaces with hyphens. But when a user types "Hi there" (multiple spaces) the slug ends up as "Hi-----there". When really it should be "Hi-there".
Should I create the regular expression so that it only replaces a space when there is a character either side?
Or is there an easier way to do this?

I use this:
yourslug.replace(/\W+/g, '-')
This replaces all occurrences of one or more non-alphanumeric characters with a single dash.

Just match multiple whitespace characters.
s/\s+/-/g

Daniel's answer is correct.
However if somebody is looking for complete solution I like this function,
http://dense13.com/blog/2009/05/03/converting-string-to-slug-javascript/
Thanks to "dense13"!

It might be the easiest to fold repeated -s into one - as the last step:
replace /-{2,}/ by "-"
Or if you only want this to affect spaces, fold spaces instead (before the other steps, obviously)

I would replace [\s]+ with '-' and then replace [^\w-] with ''

You may want to trim the string first, to avoid leading and trailing hyphens.
function hyphenSpace(s){
s= (s.trim)? s.trim(): s.replace(/^\s+|\s+$/g,'');
return s.split(/\s+/).join('-');
}

JavaScript regex replace - but only part of matched string?

I have the following replace function
myString.replace(/\s\w(?=\s)/,"$1\xA0");
The aim is to take single-letter words (e.g. prepositions) and add a non-breaking space after them, instead of standard space.
However the above $1 variable doesn't work for me. It inserts text "$1 " instead of a part of original matched string + nbsp.
What is the reason for the observed behaviour? Is there any other way to achieve it?

$1 doesn't work because you don't have any capturing subgroups.
The regular expression should be something like /\b(\w+)\s+/.

Seems you want to do something like this:
myString.replace(/\s(\w)\s/,"$1\xA0");
but that way you will loose the whitespace before your single-letter word. So you probably want to also include the first \s in the capturing group.