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regex custom lenght but no whitespace allowed [duplicate]

I have a username field in my form. I want to not allow spaces anywhere in the string. I have used this regex:
var regexp = /^\S/;
This works for me if there are spaces between the characters. That is if username is ABC DEF. It doesn't work if a space is in the beginning, e.g. <space><space>ABC. What should the regex be?

While you have specified the start anchor and the first letter, you have not done anything for the rest of the string. You seem to want repetition of that character class until the end of the string:
var regexp = /^\S*$/; // a string consisting only of non-whitespaces

Use + plus sign (Match one or more of the previous items),
var regexp = /^\S+$/

If you're using some plugin which takes string and use construct Regex to create Regex Object i:e new RegExp()
Than Below string will work
'^\\S*$'
It's same regex #Bergi mentioned just the string version for new RegExp constructor

This will help to find the spaces in the beginning, middle and ending:
var regexp = /\s/g

This one will only match the input field or string if there are no spaces. If there are any spaces, it will not match at all.
/^([A-z0-9!##$%^&*().,<>{}[\]<>?_=+\-|;:\'\"\/])*[^\s]\1*$/
Matches from the beginning of the line to the end. Accepts alphanumeric characters, numbers, and most special characters.
If you want just alphanumeric characters then change what is in the [] like so:
/^([A-z])*[^\s]\1*$/

Replace the special character and space from the string in javascript

I have a below string. I need to remove all the special character and space.
var Uid = "s/Information Needed1-s84102-p306";
I tried the below code.It didn't replace the space from the string.
console.log(Uid.replace(/[^\w\s]/gi, '')}")
The output is:- sInformation Needed1s84102p306
I want the output as sInformationNeeded1s84102p306

Simply try using
/[\W_]/g
\W match any non-word character [^a-zA-Z0-9_]
Included _ if you also want to remove it then
Regex

You can just use:
console.log(Uid.replace(/\W+/g, '')}")
\W will match any non-word character including a space.
RegEx Demo

You can use this expression for your case
var x = "s/Information Needed1-s84102-p306";
console(x.replace(/[^A-Z0-9]/ig, ""));
Here is the working Link

Regex for both newline and backslash for Replace function

I am using a replace function to escape some characters (both newline and backslash) from a string.
Here is my code:
var str = strElement.replace(/\\/\n/g, "");
I am trying to use regex, so that I can add more special characters if needed. Is this a valid regex or can someone tell me what am I doing wrong here?

You're ending the regex early with an unescaped forward slash. You also want to use a set to match individual characters. Additionally you might want to add "\r" (carriage return) in as well as "\n" (new line).
This should work:
var str = strElement.replace(/[\\\n\r]/g, "");

This is not a valid regex as the slash is a delimiter and ends the regex. What you probably wanted is the pipe (|), which is an alternation:
var str = strElement.replace(/\\|\n/g, "");
In case you need to extend it in the future it may be helpful to use a character class to improve readability:
var str = strElement.replace(/[\\\nabcx]/g, "");
A character class matches a single character from it's body.

This should work. The regular expression replaces both the newline characters and the backslashes in escaped html text:
var str = strElement.replace(/\\n|\\r|\\/g, '');

regexp problem, the dot selects all text

I use some jquery to highlight search results. For some reason if i enter a basis dot, all of the text get selected. I use regex and replace to wrap the results in a tag to give the found matches a color.
the code that i use
var pattern = new.RegExp('('+$.unique(text.split(" ")).join("|")+")","gi");
how can i prevent that the dot selects all text, so i want to leave the point out of the code(the dot has no power)

You may be able to get there by doing this:
var pattern = new.RegExp('('+$.unique(text.replace('.', '\\.').split(" ")).join("|")+")","gi");
The idea here is that you're attempting to escape the period, which acts as a wild card in regex.

This will replace all special RegExp characters (except for | since you're using that to join the terms) with their escaped version so you won't get unwanted matches or syntax errors:
var str = $.unique(text.split(" ")).join("|"),
pattern;
str = str.replace(/[\\\.\+\*\?\^\$\[\]\(\)\{\}\/\'\#\:\!\=]/ig, "\\$&");
pattern = new RegExp('('+str+')', 'gi');

The dot is supposed to match all text (almost everything, really). If you want to match a period, you can just escape it as \..

If you have a period in your RegExp it's supposed to match any character besides newline characters. If you don't want that functionality you need to escape the period.
Example RegExp with period escaped /word\./

You need to escape the text you're putting into the regex, so that special characters don't have unwanted meanings. My code is based on some from phpjs.org:
var words = $.unique(text.split(" ")).join("|");
words = words.replace(/[.\\+*?\[\^\]$(){}=!<>|:\\-]/h, '\\$&'); // escape regex special chars
var pattern = new RegExp('(' + words + ")","gi");
This escapes the following characters: .\+*?[^]$(){}=!<>|:- with a backslash \ so you can safely insert them into your new RegExp construction.

Finding Plus Sign in Regular Expression

var string = 'abcd+1';
var pattern = 'd+1'
var reg = new RegExp(pattern,'');
alert(string.search(reg));
I found out last night that if you try and find a plus sign in a string of text with a Javascript regular expression, it fails. It will not find that pattern, even though it exists in that string. This has to be because of a special character. What's the best way to find a plus sign in a piece of text? Also, what other characters will this fail on?

Plus is a special character in regular expressions, so to express the character as data you must escape it by prefixing it with \.
var reg = /d\+1/;

\-\.\/\[\]\\ **always** need escaping
\*\+\?\)\{\}\| need escaping when **not** in a character class- [a-z*+{}()?]
But if you are unsure, it does no harm to include the escape before a non-word character you are trying to match.
A digit or letter is a word character, escaping a digit refers to a previous match, escaping a letter can match an unprintable character, like a newline (\n), tab (\t) or word boundary (\b), or a a set of characters, like any word-character (\w), any non-word character (\W).
Don't escape a letter or digit unless you mean it.

Just a note,
\ should be \\ in RegExp pattern string, RegExp("d\+1") will not work and Regexp(/d\+1/) will get error.
var string = 'abcd+1';
var pattern = 'd\\+1'
var reg = new RegExp(pattern,'');
alert(string.search(reg));
//3

You should use the escape character \ in front of the + in your pattern. eg. \+

You probably need to escape the plus sign:
var pattern = /d\+1/
The plus sign is used in regular expressions to indicate 1 or more characters in a row.

It should be var pattern = '/d\\+1/'.
The string will escape '\\' as '\' ('\\+' --> '\+') so the regex object init with /d\+1/

if you want to use + (plus sign) or $ (sigil /dollar sign), then use \ (backslash) as a prefix. Like that:
\$ or \+