Replace characters in string with *, but only characters not spaces JavaScript regex - javascript

I am wanting to mask some paragraph text out until it is hovered, so if I have a string like,
Hello World
I would want that to be, Hell* ***** how can I replace all characters with with a * after 4 characters/letters?
I know I can do all the string like this,
str.replace(/./g, '*')
but how can I limit it to after the first 4 letters?

var str = "Hello World"
var after4= str.substr(0,4)+str.substr(4).replace(/[^\s]/g, '*')
console.log(after4)

I can do this by Splitting String in Two Parts:
First Part which would remain as it is.
Second Part : I would replace all the Characters by required Character.
Concat both the Strings
let text = 'Hello World';
let str1 = text.substring(0, 4);
let str2 = text.substring(4).replace(/\S/g, '*');
let updatedStr = str1.concat(str2);
console.log(updatedStr);
/*
Single Line
let updatedStr = text.substring(0, 4) + text.substring(4).replace(/\S/g, '*');
*/
Regex Info :
\S : Any other Character other then Space (Capital 'S')
You can combine them in single line Code :

Another option could be capturing the first 4 non whitespace chars using ^(\S{4}) in group 1 and use that in the replacement.
Other non whitespace chars will be matched using an alternation followed by a single non whitspace char |\S, and those matches will be returned as *
^(\S{4})|\S
Regex demo
let str = "Hello World"
.replace(/^(\S{4})|\S/g, (_, g1) => g1 ? g1 : '*');
console.log(str);

Related

JavaScript Regex - add ~ symbol between lettes except letter after space

how can I put ~ symbol between letters except the first letter after space? Here is what I got from this site by trying here and there.
txt = text.split(/ +?/g).join(" ");
txt.replace(/(.)(?=.)/g, "$1~")
this regex output as
input = "hello friend"
output = "h~e~l~l~o~ ~f~r~i~e~n~d"
How to output "h~e~l~l~o~ f~r~i~e~n~d"?
Use \S instead of . when matching the character you want to insert a ~ next to - \S matches any non-space character, whereas . matches any non-newline character.
const text = 'hello friend';
const singleSpaced = text.split(/ +/g).join(" ");
const output = singleSpaced.replace(/(\S)(?=.)/g, "$1~");
console.log(output);
or
const text = 'hello friend';
const output = text
.replace(/ +/g, ' ')
.replace(/(\S)(?=.)/g, "$1~");
console.log(output);
You can do this in one operation with a replace using this regex:
(?<=\S)(?!$)
It matches the position after a non-whitespace character (?<=\S) except for the position at end of string (?!$). You can then insert a ~ at those positions:
text = "hello friend"
txt = text.replace(/(?<=\S)(?!$)/g, '~')
console.log(txt)
You could match a single non whitespace char, and assert that to the right is not optional whitespace chars followed by the end of the string.
In the replacement use the full match followed by a tilde $&~
\S(?!\s*$)
See a regex demo.
input = "hello friend"
output = input.replace(/\S(?!\s*$)/g, '$&~')
console.log(output)

Regex match apostrophe inside, but not around words, inside a character set

I'm counting how many times different words appear in a text using Regular Expressions in JavaScript. My problem is when I have quoted words: 'word' should be counted simply as word (without the quotes, otherwise they'll behave as two different words), while it's should be counted as a whole word.
(?<=\w)(')(?=\w)
This regex can identify apostrophes inside, but not around words. Problem is, I can't use it inside a character set such as [\w]+.
(?<=\w)(')(?=\w)|[\w]+
Will count it's a 'miracle' of nature as 7 words, instead of 5 (it, ', s becoming 3 different words). Also, the third word should be selected simply as miracle, and not as 'miracle'.
To make things even more complicated, I need to capture diacritics too, so I'm using [A-Za-zÀ-ÖØ-öø-ÿ] instead of \w.
How can I accomplish that?
1) You can simply use /[^\s]+/g regex
const str = `it's a 'miracle' of nature`;
const result = str.match(/[^\s]+/g);
console.log(result.length);
console.log(result);
2) If you are calculating total number of words in a string then you can also use split as:
const str = `it's a 'miracle' of nature`;
const result = str.split(/\s+/);
console.log(result.length);
console.log(result);
3) If you want a word without quote at the starting and at the end then you can do as:
const str = `it's a 'miracle' of nature`;
const result = str.match(/[^\s]+/g).map((s) => {
s = s[0] === "'" ? s.slice(1) : s;
s = s[s.length - 1] === "'" ? s.slice(0, -1) : s;
return s;
});
console.log(result.length);
console.log(result);
You might use an alternation with 2 capture groups, and then check for the values of those groups.
(?<!\S)'(\S+)'(?!\S)|(\S+)
(?<!\S)' Negative lookbehind, assert a whitespace boundary to the left and match '
(\S+) Capture group 1, match 1+ non whitespace chars
'(?!\S) Match ' and assert a whitespace boundary to the right
| Or
(\S+) Capture group 2, match 1+ non whitespace chars
See a regex demo.
const regex = /(?<!\S)'(\S+)'(?!\S)|(\S+)/g;
const s = "it's a 'miracle' of nature";
Array.from(s.matchAll(regex), m => {
if (m[1]) console.log(m[1])
if (m[2]) console.log(m[2])
});

Regex replace all character except last 5 character and whitespace with plus sign

I wanted to replace all characters except its last 5 character and the whitespace with +
var str = "HFGR56 GGKDJ JGGHG JGJGIR"
var returnstr = str.replace(/\d+(?=\d{4})/, '+');
the result should be "++++++ ++++ +++++ JGJGIR" but in the above code I don't know how to exclude whitespace
You need to match each character individually, and you need to allow a match only if more than six characters of that type follow.
I'm assuming that you want to replace alphanumeric characters. Those can be matched by \w. All other characters will be matched by \W.
This gives us:
returnstr = str.replace(/\w(?=(?:\W*\w){6})/g, "+");
Test it live on regex101.com.
The pattern \d+(?=\d{4}) does not match in the example string as is matches 1+ digits asserting what is on the right are 4 digits.
Another option is to match the space and 5+ word characters till the end of the string or match a single word character in group 1 using an alternation.
In the callback of replace, return a + if you have matched group 1, else return the match.
\w{5,}$|(\w)
Regex demo
let pattern = / \w{5,}$|(\w)/g;
let str = "HFGR56 GGKDJ JGGHG JGJGIR"
.replace(pattern, (m, g1) => g1 ? '+' : m);
console.log(str);
Another way is to replace a group at a time where the number of +
replaced is based on the length of the characters matched:
var target = "HFGR56 GGKDJ JGGHG JGJGIR";
var target = target.replace(
/(\S+)(?!$|\S)/g,
function( m, g1 )
{
var len = parseInt( g1.length ) + 1;
//return "+".repeat( len ); // Non-IE (quick)
return Array( len ).join("+"); // IE (slow)
} );
console.log ( target );
You can use negative lookahead with string end anchor.
\w(?!\w{0,5}$)
Match any word character which is not followed by 0 to 5 characters and end of string.
var str = "HFGR56 GGKDJ JGGHG JGJGIR"
var returnstr = str.replace(/\w(?!\w{0,5}$)/g, '+');
console.log(returnstr)

How to append a string to another string after every N char?

I am trying to create a program that adds "gav" after every second letter, when the string is written.
var string1 = "word"
Expected output:
wogavrdgav
You can use the modulus operator for this -
var string1 = "word";
function addGav(str){
var newStr = '';
var strArr = str.split('');
strArr.forEach(function(letter, index){
index % 2 == 1
? newStr += letter + 'gav'
: newStr += letter
})
return newStr;
}
console.log(addGav(string1)); // gives wogavrdgav
console.log(addGav('gavgrif')) //gives gagavvggavrigavf....
RegEx
Here, we can add a quantifier to . (which matches all chars except for new lines) and design an expression with one capturing group ($1):
(.{2})
Demo
JavaScript Demo
const regex = /(.{2})/gm;
const str = `AAAAAA`;
const subst = `$1bbb`;
// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);
console.log('Substitution result: ', result);
RegEx Circuit
You can also visualize your expressions in jex.im:
If you wish to consider new lines as a char, then this expression would do that:
([\s\S]{2})
RegEx Demo
JavaScript Demo
const regex = /([\s\S]{2})/gm;
const str = `ABCDEF
GHIJK
LMNOP
QRSTU
VWXYZ
`;
const subst = `$1bbb`;
// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);
console.log('Substitution result: ', result);
Try this:
const string1 = 'word'
console.log('Input:', string1)
const newStr = string1.replace(/(?<=(^(.{2})+))/g, 'gav')
console.log('Output:', newStr)
.{2}: 2 any character
(.{2})+: match 2 4 6 8 any character
^(.{2})+: match 2 4 6 8 any character from start, if don't have ^, this regex will match from any position
?<=(regex_group): match something after regex_group
g: match all
This way is finding 2,4,6, etc character from the start of the string and don't match this group so it will match '' before 2,4,6, etc character and replace with 'gav'
Example with word:
match wo, word and ignore it, match something before that('') and replace with 'gav' with method replace

Regex - ignoring text between quotes / HTML(5) attribute filtering

So I have this Regular expression, which basically has to filter the given string to a HTML(5) format list of attributes. It currently isn't doing my fulfilling, but that's about to change! (I hope so)
I'm trying to achieve that whenever an occurrence is found, it selects the text until the next occurrence OR the end of the string, as the second match. So if you'd take a look at the current regular expression:
/([a-zA-Z]+|[a-zA-Z]+-[a-zA-Z0-9]+)=["']/g
A string like this: hey="hey world" hey-heyhhhhh3123="Hello world" data-goed="hey"
Would be filtered / matched out like this:
MATCH 1. [0-3] `hey`
MATCH 2. [16-32] `hey-heyhhhhh3123`
MATCH 3. [47-56] `data-goed`
This has to be seen as the attribute-name(s), and now.. we just have to fetch the attribute's value(s). So the mentioned string has to have an outcome like this:
MATCH 1.
1 [0-3] `hey`
2 [6-14] `hey world`
MATCH 2.
1 [16-32] `hey-heyhhhhh3123`
2 [35-45] `Hello world`
MATCH 3.
1 [47-56] `data-goed`
2 [59-61] `hey`
Could anyone try and help me to get my fulfilling? It would be appericiated a lot!
You can use
/([^\s=]+)=(?:"([^"\\]*(?:\\.[^"\\]*)*)"|(\S+))/g
See regex demo
Pattern details:
([^\s=]+) - Group 1 capturing 1 or more characters other than whitespace and = symbol
= - an equal sign
(?:"([^"\\]*(?:\\.[^"\\]*)*)"|(\S+)) - a non-capturing group of 2 alternatives (one more '([^'\\]*(?:\\.[^'\\]*)*)' alternative can be added to account for single quoted string literals)
"([^"\\]*(?:\\.[^"\\]*)*)" - a double quoted string literal pattern:
" - a double quote
([^"\\]*(?:\\.[^"\\]*)*) - Group 2 capturing 0+ characters other than \ and ", followed with 0+ sequences of any escaped symbol followed with 0+ characters other than \ and "
" - a closing dlouble quote
| - or
(\S+) - Group 3 capturing one or more non-whitespace characters
JS demo (no single quoted support):
var re = /([^\s=]+)=(?:"([^"\\]*(?:\\.[^"\\]*)*)"|(\S+))/g;
var str = 'hey="hey world" hey-heyhhhhh3123="Hello \\"world\\"" data-goed="hey" more=here';
var res = [];
while ((m = re.exec(str)) !== null) {
if (m[3]) {
res.push([m[1], m[3]]);
} else {
res.push([m[1], m[2]]);
}
}
console.log(res);
JS demo (with single quoted literal support)
var re = /([^\s=]+)=(?:"([^"\\]*(?:\\.[^"\\]*)*)"|'([^'\\]*(?:\\.[^'\\]*)*)'|(\S+))/g;
var str = 'pseudoprefix-before=\'hey1"\' data-hey="hey\'hey" more=data and="more \\"here\\""';
var res = [];
while ((m = re.exec(str)) !== null) {
if (m[2]) {
res.push([m[1], m[2]])
} else if (m[3]) {
res.push([m[1], m[3]])
} else if (m[4]) {
res.push([m[1], m[4]])
}
}
console.log(res);

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