Develop Reference

Javascript substring from last backslash char

This is probably a basic syntax question, but I am not able to find a syntax for a backslash character. Following and other syntaxes that I tried are not accepted for this char.
var fileNameSubstring = data.FileName.substring(data.FileName.lastIndexOf('\') + 1, data.FileName.length);

When defining a string in Javascript you can use the backslash (called escape character) to indicate special characters like new line \n.
To actually have a backslash in your string you should use double blackslash \\.
var fileNameSubstring = data.FileName.substring(data.FileName.lastIndexOf('\\') + 1, data.FileName.length);

Can Someone explain escaping a string in JavaScript

I know this line escapes a javascript string by adding \ before special characters. But can anyone explain how it does that? And also. Can it cause any problems in further string manipulations.
str = str.replace(/[-\\/\\^$*+?.()|[\]{}]/g, '\\$&');

The regular expression replaces anything it matches with backslash and the matched substring. It matches any single characters of the set between [ and ]. That is:
- simple dash
\\ single backslash (preceded by another backslash, because it server as escape character)
/ slash
\\ single backslash again (it is obsolete here)
^, $, *, +, ?, ., (, ), |, [ single characters
\] closing bracket is preceded by escape character, otherwise it would close set of characters to match
{, } single characters
The replacement string \\$& simply says:
\\ add single backslash
$& add the whole substring matched by the expression

Regexp adds invisible dot character replacing \b

I want this to be my regex: /^word\b/ (word is dynamic)
When I set it up to be dynamic I have to use this:
var word='spoon';
'spoon .table .chair'.match(new RegExp('^'+word+'\b'));
However, this finds null, while this:
var word='spoon';
'spoon .table .chair'.match(/^spoon\b/);
finds ["spoon"].
The interesting part is when I examine the difference between the regex I worte and the regex RegExp wrote:
console.log(/^spoon\b/,new RegExp('^'+word+'\b'))
It shows this:
/^spoon\b/ /^spoon/
If I then copy the second part of the log output (/^spoon/) into my code editor I see this character:
What is that? How do I do RegExp word-ending-with as I am not always guaranteed to have a space at the end when the string might be a one-word string (spoon or another word)
I'd rather just do this without the invisible thing

You've got to escape the \ in the b in the regex string by adding an extra slash:
var regex = new RegExp('^' + word + '\\b')
This is because the RegExp is expecting to see the two characters \ and b, but the string '\b' is one character, ascii 8, the backspace character (in the same way that '\n' is a single newline character).

In Javascript, \b doesn't mean a \ followed by a b. It means the backspace character (ASCII code 8). To get a \ followed by a b, you need to escape the slash so that Javascript doesn't parse it as a backspace:
'^' + word + '\\b'
The same thing applies if you want to use \d or \s or anything else: You need to escape the \ with another one so that Javascript doesn't think it's a Javascript escape code and the RegExp can parse it as what you expect.

How can I replace multiple slashes in javascript?

var str = "Hello\\\World\\\";
var newStr = str.replace("\\\", "");
alert(newStr); // I want this to alert: HelloWorld
The number of slashes is always 3, not more not less. How can I replace them? The code above doesn't work at all. I've played around a bit with the global flag, escaping the slashes etc but can't figure it out.

Firstly, you need to escape each slash with another backslash, as mentioned by #Bathsheba.
Additionally, you want your replacement regex to be global:
var str = "Hello\\\\\\World\\\\\\";
var newStr = str.replace(/\\\\\\/g, "");
alert(newStr); // I want this to alert: HelloWorld

If you want three slashes in a row in a string literal then you need to escape each one in turn:
var str = "Hello\\\\\\World\\\\\\";
var newStr = str.replace("\\\\\\", "");
In your current string, \\\W would be one slash and an error as \W is not a valid sequence. (Some more examples: \\ is a single slash, \t a tab, \" a quotation character).

Try this regex \\\\\\ for replace
\\ indicates \
There are 12 characters with special meanings: the backslash \, the caret ^, the dollar sign $, the period or dot ., the vertical bar or pipe symbol |, the question mark ?, the asterisk or star *, the plus sign +, the opening parenthesis (, the closing parenthesis ), and the opening square bracket [, the opening curly brace {, These special characters are often called "metacharacters".
If you want to use any of these characters as a literal in a regex, you need to escape them with a backslash.

Replace '\' with '-' in a string

I have seen all the questions asked previously but it didn't help me out . I have a string that contains backslash and i want to replace the backslashes with '-'
var s="adbc\sjhf\fkjfh\af";
s = s.replace(/\\/g,'-');
alert(s);
I thought this is the proper way to do it and of course i am wrong because in alert it shows adbcsjhffkjfhaf but i need it to be like adbc-sjhf-fkjfh-af.
What mistake i do here and what is the reason for it and how to achieve this...??
Working JS Fiddle

Your s is initially adbcsjhffkjfhaf. You meant
var s="adbc\\sjhf\\fkjfh\\af";

You need to double-up the backslashes in your input string:
var s="adbc\\sjhf\\fkjfh\\af";
Prefixing a character with '\' in a string literal gives special meaning to that character (eg '\t' means a tab character). If you want to actually include a '\' in your string you must escape it with a second backslash: '\\'

Javascript is ignoring the \ in \s \f \a in your string. Do a console.log(s) after assigning, you will understand.
You need to escape \ with \\. Like: "adbc\\sjhf\\fkjfh\\af"

The string doesn't contain a backslash, it contains the \a, \s and \f (escape sequence for Form Feed).
if you change your string to adbc\\sjhf\\fkjfh\\af
var s="adbc\\sjhf\\fkjfh\\af";
s = s.replace(/\\/g,'-');
alert(s);
you will be able to replace it with -