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How to replace all instance of string '&user_story=x' using regex in javascript?

I have string like this -
var string = 'callback&user_story=1&user_story=2&user_story=100&user_story=a&user_story=john';
&user_story=x (Here x can be anything) can repeat n number of times.
How to replace this '&user_story=x' with a blank value.
What would be regex for it in JS?

The regex looks like this:
var string = 'callback&user_story=1&user_story=2&user_story=100&user_story=a&user_story=john';
var re = /&user_story=.*?(?=&|$)/g
console.log(string.replace(re, ""));
Broken down:
/&user_story=.*?(?=&|$)/g
&user_story= - checks whether the match starts with "&user_story="
.*? - matches any number of any characters, but the ? makes it non-greedy, so it will find as few of these characters before finding the next part of the regex
(?=&|$) - the brackets make this a group, and ?= means it's a lookahead, i.e. it won't actually add matches to the regex match, but just checks to see they're there. It will match either another &, or the end of the string (symbolised by $).
g - is a flag which tells the regex to check the entire string, and not stop after just finding one match.

How to achieve this result using Regex?

Given the input below, what's the regex expression that gives the desired output in javascript? I must achieve this without using a multiline flag.
input
\n
\n
abc def.\n
\n
*\n
\n
desired output (maintain same number of rows but insert = into blank rows)
=\n
=\n
abc def.\n
=\n
*\n
=\n
actual output (using regex /[^a-zA-Z0-9.*]+\n/ replaced with =\n; it somehow removes one of two consecutive `\n`s)
=\n
abc def.=\n
*=\n

You could try a combination of replace functions like so:
str = "\n\nabc def.\n\n*\n\n";
str = str.replace(/\n/g, "=\n");
str = str.replace(/(.)=\n/g, "$1\n");
console.log(str);
Explanation -
After the first replacement/s, the output looks like:
=
=
abc def.=
=
*=
=
Then, you replace any characters followed by a =\n and replace it with that same character (given by $1), followed by a newline.

Your desired outcome is "maintain same number of rows but insert = into blank rows".
An empty ("blank") row is a row that matches the regex: ^$.
^ means the beginning of the input string, $ means the end of the input string but if the m modifier is specified (it means "multi-line"), ^ matches the beginning of a line and $ matches the end of a line.
Your code should be as simple as:
input = "\n\nabc def.\n\n*\n\n";
output = str.replace(/^$/mg, '=');
The m modifier changes the meaning of ^ and $ as explained above. The newline characters are not matched by the regex above and consequently they do not need to be present in the replacement string.
The g modifier tells String.replace() to find and replace all the matching substrings, not only the first one (the default behaviour of String.replace()).
Read more about regular expressions in JavaScript.

This should work with two replace :
value.replace(/^\n/, '=\n').replace(/\n\n/g, '\n=\n')
The first replace takes care of the first line if it starts with a blank row.
The second replace takes care of other lines : adding = in blank rows is the same than inserting = between two consecutives \n

Escape single backslash inbetween non-backslash characters only

I have some input coming in a web page which I will re display and submit elsewhere. The current issue is that I want to double up all single backslashes that are sandwiched inbetween non-backslash characters before submitting the input elsewhere.
Test string "domain\name\\nonSingle\\\WontBe\\\\Returned", I want to only get the first single backslash, between domain and name.
This string should get nothing "\\get\\\nothing\\\\"
My current pattern that I can get closest with is [\w][\\](?!\\) however this will get the "\n" from the 1st test string i have listed. I would like to use lookbehind for the regex however javascript does not have such a thing for the version I am using. Here is the site I have been testing my regexs on http://www.regexpal.com/
Currently I am inefficiently using this regex [\w][\\](?!\\) to extract out all single backslashes sandwiched between non-backslash characters and the character before them (which I don't want) and then replacing it with the same string plus a backslash at the end of it.
For example given domain\name\\bl\\\ah my current regex [\w][\\]\(?!\\) will return "n\". This results in my code having to do some additional processing rather than just using replace.
I don't care about any double, triple or quadruple backslashes present, they can be left alone.

For example given domain\name\\bl\\\ah my current regex [\w][\\]\(?!\\) will return "n\". This results in my code having to do some additional processing rather than just using replace.
It will do just using replace, since you can insert the matched substring with $&, see:
console.log(String.raw`domain\name\\bl\\\ah`.replace(/\w\\(?!\\)/g, "$&\\"))

Easiest method of matching escapes, is to match all escaped characters.
\\(.)
And then in the replacement, decide what to do with it based on what was captured.
var s = "domain\\name\\\\backslashesInDoubleBackslashesWontBeReturned";
console.log('input:', s);
var r = s.replace(/\\(.)/g, function (match, capture1) {
return capture1 === '\\' ? match : '$' + capture1;
});
console.log('result:', r);
The closest you can get to actually matching the unescaped backslashes is
((?:^|[^\\])(?:\\\\)*)\\(?!\\)
It will match an odd number of backslashes, and capture all but the last one into capture group 1.
var re = /((?:^|[^\\])(?:\\\\)*)\\(?!\\)/g;
var s = "domain\\name\\\\escapedBackslashes\\\\\\test";
var parts = s.split(re);
console.dir(parts);
var cleaned = [];
for (var i = 1; i < parts.length; i += 2)
{
cleaned.push(parts[i-1] + parts[i]);
}
cleaned.push(parts[parts.length - 1]);
console.dir(cleaned);
The even-numbered (counting from zero) items will be unmatched text. The odd-numbered items will be the captured text.
Each captured text should be considered part of the preceding text.

JS regexp to match special characters

I'm trying to find a JavaScript regexp for this string: ![](). It needs to be an exact match, though, so:
`!()[]` // No match
hello!()[] // No match
!()[]hello // No Match
!()[] // Match
!()[] // Match (with a whitespace before and/or after)
I tried this: \b![]()\b. It works for words, like \bhello\b, but not for those characters.

The characters specified are control characters and need to be escaped also user \s if you want to match whitespace. Try the following
\s?!(?:\[\]\(\)|\(\)\[\])\s?
EDIT: Added a capture group to extract ![]() if needed
EDIT2: I missed that you wanted order independant for [] and () I've added it in this fiddle http://jsfiddle.net/MfFAd/3/

This matches your example:
\s*!\[\]\(\)\s*
Though the match also includes the spaces before and after !()[].
I think \b does not work here because ![]() is not a word. Check out this quote from MDN:
\b - Matches a word boundary. A word boundary matches the position where a word character is not followed or preceeded by another word-character. Note that a matched word boundary is not included in the match. In other words, the length of a matched word boundary is zero.

Let's create a function for convenience :
function find(r, s) {
return (s.match(r) || []).slice(-1);
}
The following regular expression accepts only the searched string and whitespaces :
var r = /^\s*(!\[\]\(\))\s*$/;
find(r, '![]() '); // ["![]()"]
find(r, '!()[] '); // []
find(r, 'hello ![]()'); // []
This one searches a sub-string surrounded by whitespaces or string boundaries :
var r = /(?:^|\s)(!\[\]\(\))(?:\s|$)/;
find(r, '![]() '); // ["![]()"]
find(r, 'hello ![]()'); // ["![]()"]
find(r, 'hello![]()'); // []

To match all characters except letters and numbers you can use this regex
/[^A-Z0-9]/gi
g - search global [ mean whole text, not just first match ]
i -case insensitive
to remove any other sign for example . and ,
/[^A-Z0-9\.\,]/gi
In order to match exact string you need to group it and global parameter
/(\!\[\]\(\))/g
so it will search for all matches

Remove entire word from string if it contains numeric value

What I'm trying to accomplish is to auto-generate tags/keywords for a file upload, basing these keywords from the filename.
I have accomplished auto-generating titles for each upload, as shown here:
But I have now moved on to trying to auto-generate keywords. Similar to titles, but with more formatting. First, I run the string through this to remove commonly used words from the filename (such as this,that,there... etc)
I am happy with it, but I need to not include words that have numbers in it. I have not found a solution on how to remove a word entirely if it contains a number. The solutions I have found like here only works for a certain match, while this one removes numbers alone. I would like to remove the entire word if it contains ANY numeric digit.

To remove all words which contain a number, use:
string = string.replace(/[a-z]*\d+[a-z]*/gi, '');

Try this expression:
var regex = /\b[^\s]*\d[^\s]*\b/g;
Example:
var str = "normal 5digit dig555it digit5 555";
console.log( str.replace(regex,'') );​ //Result-> normal

Apply a simple regular expression to you current filename strings, replacing all occurrences with the empty string. The regular expression matches "words" containing any digits.
Javascript example:
'asdf 8bit jawesome234 mayhem 234'.replace(/\s*\b\w*\d\w*\b/g, '')
Evaluates to:
"asdf mayhem"
Here the regular expression is /\s*\b\w*\d\w*\b/g, which matches maximal sequences consisting of zero or more whitespace characters (\s*) followed by a word-boundary transition (\b), followed by zero or more alphanum characters (\w*), followed by a digit (\d), followed by zero or more alphanum characters, followed by a word-boundary transition (\b). \b matches the empty string at the transition to an alphanumeric character from either the beginning or end of the word or a non-alphanumeric character. The g after the final / of the regular expression means replace all occurrences, not just the first.
Once the digit-words are removed, you can split the string into keywords however you want (by whitespace, for example).
"asdf mayhem".split(/\s+/);
Evaluates to:
["asdf", "mayhem"]

('Apple Cover Photo 23s423 of your 543634 moms').match(/\b([^\d]+)\b/g, '')
returns
Apple Cover Photo , of your , moms
http://jsfiddle.net/awBPX/2/

use this to Remove words containing numeric :
string.replace("[0-9]","");
hope this helps.
Edited :
check this :
var str = 'one 2two three3 fo4ur 5 six';
var result = str.match(/(^[\D]+\s|\s[\D]+\s|\s[\D]+$|^[\D]+$)+/g).join('');