We Keep Coding

In JavaScript, string '\m' is fully equals to string 'm', why? [duplicate]

This question already has answers here:
How can I use backslashes (\) in a string?
(4 answers)
Closed 2 years ago.
console.log('\d' === 'd'); // true
Character 'd' is not a special character, why javascript want to slice the escape notation.
It's better to keep the escape notation in my view.
When I want to fully match string-'\d' using regular expression, it just impossible!
Taking the following code as an example.
console.log(RE.test('\d')); // it should log true
console.log(RE.test('d')); // it should log false
Unfortunately, you just cannot figure out a regular expression pattern.

You have no reason to escape d in a string and JavaScript ignores it. If you need \d you need to escape the escape character: \\d.
See also Why do linters pick on useless escape character?

\d has a special meaning in regular expressions (a digit character), but also in strings (escaped 'd' character, which is exactly like 'd').
Any / creates an escape sequence in a string. Some are "useful" (\n === new line) and some arguably useless (`'\d' === 'd').
If you want the regex \d, you could
1 - use a regex literal instead : /\d/
2 - escape the \ in the string : '\\d', so that the string containing the two characters \ and d is correctly understood by Javascript.

Single Backslash to double backslash Conversion [duplicate]

This question already has answers here:
JavaScript backslash (\) in variables is causing an error
(5 answers)
Escaping backslash in string - javascript
(5 answers)
Closed 3 years ago.
I want all single backslashes to be converted into double backslash
"C:\Users\MyName\ringtone.mp3" --> "C:\\Users\\MyName\\ringtone.mp3"
But for some reason it returns "C:UsersMyNameingtone.mp3"
So far I have tried the escape() function and the encodeURI() function but they don't work either. Partial of the string comes from nodejs OS Module which only returns with a single backslash on windows (homedir() function).
Here is what I have so far in the function
function normalize(path: string): string {
return path.normalize().replace(/\\/g, '\\');
}
Thanks in Advance

This should work:
var original = 'C:\\Users\\MyName\\ringtone.mp3';
var replaced = original.normalize().replace(/\\/g, '\\\\');
console.log('Original: ' + original);
console.log('Replaced: ' + replaced);
From what I see you had 2 problems:
First, it seems you were initializing your string like this:
var original = 'C:\Users\MyName\ringtone.mp3'
This would make your actual string value C:UsersMyNameingtone.mp3 because a \ character in javascript symbolizes an escape character.
Second, is because the \ character is an escape character, so the '\\' in your replace function is only looking to replace the matching pattern with a single backslash.

Switching \n with \\n in javascript [duplicate]

This question already has answers here:
How do I handle newlines in JSON?
(10 answers)
Closed 4 years ago.
So I have a string:
var s = "foo\nbar\nbob";
I want the string to become:
"foo\\nbar\\nbob"
How can I replace every \n with a \\n?
I've tried using some for loops, but I can't figure it out.

A simple .replace would work - search for \n, and replace with \\n:
var s = "foo\nbar\nbob";
console.log(
s.replace(/\n/g, '\\\n')
// ^^ double backslash needed to indicate single literal backslash
);
Note that this results in "a single backslash character, followed by a literal newline character" - there will not be two backslashes in a row in the actual string. It might be a bit less confusing to use String.raw, which will interpret every character in the template literal literally:
var s = "foo\nbar\nbob";
console.log(
s.replace(/\n/g, String.raw`\
`) // template literal contains one backslash, followed by one newline
);

Javascript: Regex .test() for whitespace [duplicate]

This question already has answers here:
Why do regex constructors need to be double escaped?
(5 answers)
Closed 5 years ago.
Strange regex result when looking for any white space characters
new RegExp('^[^\s]+$').test('aaa');
=> true
That's expected, but then...
new RegExp('^[^\s]+$').test('aaa ');
=> true
How does that return true?

You need to escape \ in the string by \\. Otherwise, the generated regex would be /^[^s]+$/, which matches anything other than a string includes s.
new RegExp('^[^\\s]+$').test('aaa ');
Or you can use \S for matching anything other than whitespace.
new RegExp('^\\S+$').test('aaa ');
It would be better to use regex directly instead of parsing a regex string pattern.
/^[^\s]+$/.test('aaa ');
// or
/^\S+$/.test('aaa ');

Why have two '\' in Regex? [duplicate]

This question already has answers here:
Why do regex constructors need to be double escaped?
(5 answers)
Extra backslash needed in PHP regexp pattern
(4 answers)
Regex to replace single backslashes, excluding those followed by certain chars
(3 answers)
Closed 7 years ago.
function trim(str) {
var trimer = new RegExp("(^[\\s\\t\\xa0\\u3000]+)|([\\u3000\\xa0\\s\\t]+\x24)", "g");
return String(str).replace(trimer, "");
}
why have two '\' before 's' and 't'?
and what's this "[\s\t\xa0\u3000]" mean?

You're using a literal string.
In a literal string, the \ character is used to escape some other chars, for example \n (a new line) or \" (a double quote), and it must be escaped itself as \\. So when you want your string to have \s, you must write \\s in your string literal.
Thankfully JavaScript provides a better solution, Regular expression literals:
var trimer = /(^[\s\t\xa0\u3000]+)|([\u3000\xa0\s\t]+\x24)/g

why have two '\' before 's' and 't'?
In regex the \ is an escape which tells regex that a special character follows. Because you are using it in a string literal you need to escape the \ with \.
and what's this "[\s\t\xa0\u3000]" mean?
It means to match one of the following characters:
\s white space.
\t tab character.
\xa0 non breaking space.
\u3000 wide space.
This function is inefficient because each time it is called it is converting a string to a regex and then it is compiling that regex. It would be more efficient to use a Regex literal not a string and compile the regex outside the function like the following:
var trimRegex = /(^[\s\t\xa0\u3000]+)|([\u3000\xa0\s\t]+$)/g;
function trim(str) {
return String(str).replace(trimRegex, "");
}
Further to this \s will match any whitespace which includes tabs, the wide space and the non breaking space so you could simplify the regex to the following:
var trimRegex = /(^\s+)|(\s+$)/g;
Browsers now implement a trim function so you can use this and use a polyfill for older browsers. See this Answer