Develop Reference

Replace a specific character from a string with HTML tags

Having a text input, if there is a specific character it must convert it to a tag. For example, the special character is *, the text between 2 special characters must appear in italic.
For example:
This is *my* wonderful *text*
must be converted to:
This is <i>my</i> wonderful <i>text</i>
So I've tried like:
const arr = "This is *my* wonderful *text*";
if (arr.includes('*')) {
arr[index] = arr.replace('*', '<i>');
}
it is replacing the star character with <i> but doesn't work if there are more special characters.
Any ideas?

You can simply create wrapper and thereafter use regular expression to detect if there is any word that is surrounded by * and simply replace it with any tag, in your example is <i> tag so just see the following
Example
let str = "This is *my* wonderful *text*";
let regex = /(?<=\*)(.*?)(?=\*)/;
while (str.includes('*')) {
let matched = regex.exec(str);
let wrap = "<i>" + matched[1] + "</i>";
str = str.replace(`*${matched[1]}*`, wrap);
}
console.log(str);

here you go my friend:
var arr = "This is *my* wonderful *text*";
const matched = arr.match(/\*(?:.*?)\*/g);
for (let i = 0; i < matched.length; i++) {
arr = arr.replace(matched[i], `<i>${matched[i].replaceAll("*", "")}</i>`);
}
console.log(arr);
an explanation first of all we're matching the regex globaly by setting /g NOTE: that match with global flag returns an array.
secondly we're looking for any character that lies between two astrisks and we're escaping them because both are meta characters.
.*? match everything in greedy way so we don't get something like this my*.
?: for non capturing groups, then we're replacing every element we've matched with itself but without astrisk.

Regex that allows a pattern to start with a an optional, specific character, but no other character

How can I write a regex that allows a pattern to start with a specific character, but that character is optional?
For example, I would like to match all instances of the word "hello" where "hello" is either at the very start of the line or preceded by an "!", in which case it does not have to be at the start of the line. So the first three options here should match, but not the last:
hello
!hello
some other text !hello more text
ahello
I'm specfically interested in JavaScript.

Match it with: /^hello|!hello/g
The ^ will only grab the word "hello" if it's at the beginning of a line.
The | works as an OR.
var str = "hello\n!hello\n\nsome other text !hello more text\nahello";
var regex = /^hello|!hello/g;
console.log( str.match(regex) );
Edit:
If you're trying to match the whole line beginning with "hello" or containing "!hello" as suggested in the comment below, then use the following regex:
/^.*(^hello|!hello).*$/gm
var str = "hello\n!hello\n\nsome other text !hello more text\nahello";
var regex = /^.*(^hello|!hello).*$/gm;
console.log(str.match(regex));

Final solution (hopefully)
Looks like, catching the groups is only available in ECMAScript 2020. Link 1, Link 2.
As a workaround I've found the following solution:
const str = `hello
!hello
some other text !hello more text
ahello
this is a test hello !hello
JvdV is saying hello
helloing or helloed =).`;
function collectGroups(regExp, str) {
const groups = [];
str.replace(regExp, (fullMatch, group1, group2) => {
groups.push(group1 || group2);
});
return groups;
}
const regex = /^(hello)|(?:!)(hello\b)/g;
const groups = collectGroups(regex, str)
console.log(groups)
/(?=!)?(\bhello\b)/g should do it. Playground.
Example:
const regexp = /(?=!)?(\bhello\b)/g;
const str = `
hello
!hello
some other text !hello more text
ahello
`;
const found = str.match(regexp)
console.log(found)
Explanation:
(?=!)?
(?=!) positive lookahead for !
? ! is optional
(\bhello\b): capturing group
\b word boundary ensures that hello is not preceded or succeeded by a character
Note: If you also make sure, that hello should not be succeeded by !, then you could simply add a negative lookahead like so /(?=!)?(\bhello\b)(?!!)/g.
Update
Thanks to the hint of #JvdV in the comment, I've adapted the regex now, which should meet your requirements:
/(^hello\b)|(?:!)(hello\b)/gm
Playground: https://regex101.com/r/CXXPHK/4 (The explanation can be found on the page as well).
Update 2:
Looks like the non-capturing group (?:!) doesn't work well in JavaScript, i.e. I get a matching result like ["hello", "!hello", "!hello", "!hello"], where ! is also included. But who cares, here is a workaround:
const regex = /(^hello\b)|(?:!)(hello\b)/gm;
const found = (str.match(regex) || []).map(m => m.replace(/^!/, ''));

Regex to extract text followed with parentheses (Multiple one in one String)

I want to extract text followed with parentheses of multiple one in one String.
Example:
text(text1) password(password1) in table(table_name) with some random text
So, i want extract each of them in table like this:
COL1 COL2
-------------------
text text1
password password1
table table_name
So in table i mean just the possiblity to use them and call them when needed.
What i have tried:
This regex allow me only to extract the first parenthese but without "text" included and is not what i want:
"text(text1) password(password1) in table(table_name) with some random text".match(/\(([^)]+)\)/)[1]
return:
text1
I want "text" will be included in regex like explained in the example in the top of this post.
Thank's in advance.

You can do something like this:
var str = "text(text1) password(password1) in table(table_name) with some random text";
var exp = new RegExp("([a-z]+)\\(([^)]+)\\)", "ig");
var groups = []
var matches
while(true){
matches = exp.exec(str)
if (!matches) {
break;
}
groups.push([matches[1], matches[2]])
}
console.log(groups)
You should probably change the regular expression, because now, the part before the parentheses can only contain letters.

Your regex for the second part is ok but you should use exec instead of match and use the /g (Global) flag.
For the first capturing group your could match not a whitespace character \S
(\S+)\(([^)]+)\)
const regex = /(\S+)\(([^)]+)\)/g;
const str = `text(text1) password(password1) in table(table_name) with some random text`;
let m;
while ((m = regex.exec(str)) !== null) {
if (m.index === regex.lastIndex) {
regex.lastIndex++;
}
console.log(m[1], m[2]);
}

JS : Remove all strings which are starting with specific character

I have an array contains names. Some of them starting with a dot (.), and some of them have dot in the middle or elsewhere. I need to remove all names only starting with dot. I seek help for a better way in JavaScript.
var myarr = 'ad, ghost, hg, .hi, jk, find.jpg, dam.ark, haji, jive.pdf, .find, home, .war, .milk, raj, .ker';
var refinedArr = ??

You can use the filter function and you can access the first letter of every word using item[0]. You do need to split the string first.
var myarr = 'ad, ghost, hg, .hi, jk, find.jpg, dam.ark, haji, jive.pdf, .find, home, .war, .milk, raj, .ker'.split(", ");
var refinedArr = myarr.filter(function(item) {
return item[0] != "."
});
console.log(refinedArr)

Use filter and startsWith:
let myarr = ['ad', 'ghost', 'hg', '.hi', 'jk'];
let res = myarr.filter(e => ! e.startsWith('.'));
console.log(res);

You can use the RegEx \B\.\w+,? ? and replace with an empty String.
\B matches a non word char
\. matches a dot
\w+ matches one or more word char
,? matches 0 or 1 ,
[space]? matches 0 or 1 [space]
Demo:
const regex = /\B\.\w+,? ?/g;
const str = `ad, ghost, hg, .hi, jk, find.jpg, dam.ark, haji, jive.pdf, .find, home, .war, .milk, raj, .ker`;
const subst = ``;
// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);
console.log('Substitution result: ', result);

Javascript regular expression for matching whole words including special characters

I am trying to match whole exact words using a javascript regular expression.
Given the strings: 1) "I know C++." and 2) "I know Java."
I have tried using new Regex('\\b' + text + '\\b', 'gi') and that works great for words without special characters like example #2.
I've also taken a look at this url:
Regular expression for matching exact word affect the special character matching
and implemented the:
escaped = escaped.replace(/^(\w)/, "\\b$1");
escaped = escaped.replace(/(\w)$/, "$1\\b");
and that will match text = 'C++' (it will match both examples)
However, if someone types a typo, and the string is "I know C++too.", the latter regex will still match the C++ when I don't want it to because the word "C++too" is not an exact match for text = 'C++'.
What changes can I make so that it will not match unless C++ is both the front of the word and the end of the word.

You can add a range of accepted characters([+#]) after word characters:
str = 'I know C++too. I know Java and C#.';
console.log(str.match(/(\w[+#]+|\w+)/g));
NB: \w[+#]+ must be placed first in the alternation expression to take precedence over the more generic \w+.

If whole words including special characters means everything but [\r\n\t\f\v ], you can simply do:
const REGEX = /([^\s]+)+/g;
function selectWords(string) {
const REGEX = /([^\s]+)+/g;
return string
// remove punctuation
.replace(/[^a-z0-9\s+#]/ig, "")
// perform the match
.match(REGEX)
// prevent null returns
|| []
;
}
var text = "Hello World"
var [first, second, ...rest] = selectWords(text);
console.log(1, {first, second, rest});
// example with punctuation
var text = "I can come today, she said, but not tomorrow."
var [first, second, third, ...rest] = selectWords(text);
console.log(2, {first, second, third, rest});
// example with possible throw
var text = ",.'\"` \r"
var [first, second, third, ...rest] = selectWords(text);
console.log(3, {first, second, third, rest});
// example with a specific word to be matched
function selectSpecificWord(string, ...words) {
return selectWords(string)
.filter(word => ~words.indexOf(word))
;
}
var expected = "C++";
var test = "I know C++";
var test1 = "I know C++AndJava";
console.log("Test Case 1", selectSpecificWord(test, expected));
console.log("Test Case 2", selectSpecificWord(test1, expected));

Use this ((?:(?:\w)+?)(?=\b|\w[-+]{2,2})(?:[-+]{2,2})?)
I've included a - symbol for an example also. See it in life.