Develop Reference

How do I create a function that takes a string and returns dashes on the left and right side of every vowel (a e i o u)

Examples
dashed("Carpe Diem") ➞ "C-a-rp-e- D-i--e-m"
dashed("Fight for your right to party!") ➞ "F-i-ght f-o-r y-o--u-r r-i-ght t-o- p-a-rty!"
Notes

There are already a couple of solutions using regex; therefore, I am adding an answer which can be useful to someone who is not familiar with regex.
The following function, referenced by formatVowels, iterates the parameter string and uses String#indexOf to check if any of the characters is a vowel; if yes, it prepends and appends - to it and then adds the resulting text to the variable, result, otherwise, adds the character as it is to the variable, result.
const formatVowels = (str) => {
if (str == undefined) return "";
let result = "";
for (const ch of str) {
result += "AEIOU".indexOf(ch.toUpperCase()) != -1 ? "-" + ch + "-" : ch;
}
return result;
};
// Test
console.log(formatVowels('Hello'));

const my_replacer = (str)=> str.replace(/[aeiouAEIOU]/g,'-$&-')
let a = 'Carpe Diem'
console.log(my_replacer(a))
let b = 'Fight for your right to party!'
console.log(my_replacer(b))

coolString = `Cookke`;
function dashed(coolString) {
const regex = /a|e|i|o|u/gi;
console.log(coolString.replace(regex, `-$&-`));
}
dashed(coolString);

Recombine capture groups in single regexp?

I am trying to handle input groups similar to:
'...A.B.' and want to output '.....AB'.
Another example:
'.C..Z..B.' ==> '......CZB'
I have been working with the following:
'...A.B.'.replace(/(\.*)([A-Z]*)/g, "$1")
returns:
"....."
and
'...A.B.'.replace(/(\.*)([A-Z]*)/g, "$2")
returns:
"AB"
but
'...A.B.'.replace(/(\.*)([A-Z]*)/g, "$1$2")
returns
"...A.B."
Is there a way to return
"....AB"
with a single regexp?
I have only been able to accomplish this with:
'...A.B.'.replace(/(\.*)([A-Z]*)/g, "$1") + '...A.B.'.replace(/(\.*)([A-Z]*)/g, "$2")
==> ".....AB"

If the goal is to move all of the . to the beginning and all of the A-Z to the end, then I believe the answer to
with a single regexp?
is "no."
Separately, I don't think there's a simpler, more efficient way than two replace calls — but not the two you've shown. Instead:
var str = "...A..B...C.";
var result = str.replace(/[A-Z]/g, "") + str.replace(/\./g, "");
console.log(result);
(I don't know what you want to do with non-., non-A-Z characters, so I've ignored them.)
If you really want to do it with a single call to replace (e.g., a single pass through the string matters), you can, but I'm fairly sure you'd have to use the function callback and state variables:
var str = "...A..B...C.";
var dots = "";
var nondots = "";
var result = str.replace(/\.|[A-Z]|$/g, function(m) {
if (!m) {
// Matched the end of input; return the
// strings we've been building up
return dots + nondots;
}
// Matched a dot or letter, add to relevant
// string and return nothing
if (m === ".") {
dots += m;
} else {
nondots += m;
}
return "";
});
console.log(result);
That is, of course, incredibly ugly. :-)

JavaScript RegExp - find all prefixes up to a certain character

I have a string which is composed of terms separated by slashes ('/'), for example:
ab/c/def
I want to find all the prefixes of this string up to an occurrence of a slash or end of string, i.e. for the above example I expect to get:
ab
ab/c
ab/c/def
I've tried a regex like this: /^(.*)[\/$]/, but it returns a single match - ab/c/ with the parenthesized result ab/c, accordingly.
EDIT :
I know this can be done quite easily using split, I am looking specifically for a solution using RegExp.

NO, you can't do that with a pure regex.
Why? Because you need substrings starting at one and the same location in the string, while regex matches non-overlapping chunks of text and then advances its index to search for another match.
OK, what about capturing groups? They are only helpful if you know how many /-separated chunks you have in the input string. You could then use
var s = 'ab/c/def'; // There are exact 3 parts
console.log(/^(([^\/]+)\/[^\/]+)\/[^\/]+$/.exec(s));
// => [ "ab/c/def", "ab/c", "ab" ]
However, it is unlikely you know that many details about your input string.
You may use the following code rather than a regex:
var s = 'ab/c/def';
var chunks = s.split('/');
var res = [];
for(var i=0;i<chunks.length;i++) {
res.length > 0 ? res.push(chunks.slice(0,i).join('/')+'/'+chunks[i]) : res.push(chunks[i]);
}
console.log(res);
First, you can split the string with /. Then, iterate through the elements and build the res array.

I do not think a regular expression is what you are after. A simple split and loop over the array can give you the result.
var str = "ab/c/def";
var result = str.split("/").reduce(function(a,s,i){
var last = a[i-1] ? a[i-1] + "/" : "";
a.push(last + s);
return a;
}, []);
console.log(result);
or another way
var str = "ab/c/def",
result = [],
parts=str.split("/");
while(parts.length){
console.log(parts);
result.unshift(parts.join("/"));
parts.pop();
}
console.log(result);
Plenty of other ways to do it.

You can't do it with a RegEx in javascript but you can split parts and join them respectively together:
var array = "ab/c/def".split('/'), newArray = [], key = 0;
while (value = array[key++]) {
newArray.push(key == 1 ? value : newArray[newArray.length - 1] + "/" + value)
}
console.log(newArray);

May be like this
var str = "ab/c/def",
result = str.match(/.+?(?=\/|$)/g)
.map((e,i,a) => a[i-1] ? a[i] = a[i-1] + e : e);
console.log(result);

Couldn't you just split the string on the separator character?
var result = 'ab/c/def'.split(/\//g);

How to remove the end of a string, starting from a given pattern?

Let's say I have a string like this:
var str = "/abcd/efgh/ijkl/xxx-1/xxx-2";
How do I, using Javascript and/or jQuery, remove the part of str starting with xxx, till the end of str?

str.substring( 0, str.indexOf( "xxx" ) );

Just:
s.substring(0, s.indexOf("xxx"))
A safer version handling invalid input and lack of matching patterns would be:
function trump(str, pattern) {
var trumped = ""; // default return for invalid string and pattern
if (str && str.length) {
trumped = str;
if (pattern && pattern.length) {
var idx = str.indexOf(pattern);
if (idx != -1) {
trumped = str.substring(0, idx);
}
}
}
return (trumped);
}
which you'd call with:
var s = trump("/abcd/efgh/ijkl/xxx-1/xxx-2", "xxx");

Try using string.slice(start, end):
If you know the exact number of characters you want to remove, from your example:
var str = "/abcd/efgh/ijkl/xxx-1/xxx-2";
new_str = str.slice(0, -11);
This would result in str_new == '/abcd/efgh/ijkl/'
Why this is useful:
If the 'xxx' refers to any string (as the OP said), i.e: 'abc', '1k3', etc, and you do not know beforehand what they could be (i.e: Not constant), the accepted answers, as well as most of the others will not work.

Try this:
str.substring(0, str.indexOf("xxx"));
indexOf will find the position of xxx, and substring will cut out the piece you want.

This will take everything from the start of the string to the beginning of xxx.
str.substring(0,str.indexOf("xxx"));

Get everything after the dash in a string in JavaScript

What would be the cleanest way of doing this that would work in both IE and Firefox?
My string looks like this sometext-20202
Now the sometext and the integer after the dash can be of varying length.
Should I just use substring and index of or are there other ways?

How I would do this:
// function you can use:
function getSecondPart(str) {
return str.split('-')[1];
}
// use the function:
alert(getSecondPart("sometext-20202"));

A solution I prefer would be:
const str = 'sometext-20202';
const slug = str.split('-').pop();
Where slug would be your result

var testStr = "sometext-20202"
var splitStr = testStr.substring(testStr.indexOf('-') + 1);

var the_string = "sometext-20202";
var parts = the_string.split('-', 2);
// After calling split(), 'parts' is an array with two elements:
// parts[0] is 'sometext'
// parts[1] is '20202'
var the_text = parts[0];
var the_num = parts[1];

With built-in javascript replace() function and using of regex (/(.*)-/), you can replace the substring before the dash character with empty string (""):
"sometext-20202".replace(/(.*)-/,""); // result --> "20202"

AFAIK, both substring() and indexOf() are supported by both Mozilla and IE. However, note that substr() might not be supported on earlier versions of some browsers (esp. Netscape/Opera).
Your post indicates that you already know how to do it using substring() and indexOf(), so I'm not posting a code sample.

myString.split('-').splice(1).join('-')

I came to this question because I needed what OP was asking but more than what other answers offered (they're technically correct, but too minimal for my purposes). I've made my own solution; maybe it'll help someone else.
Let's say your string is 'Version 12.34.56'. If you use '.' to split, the other answers will tend to give you '56', when maybe what you actually want is '.34.56' (i.e. everything from the first occurrence instead of the last, but OP's specific case just so happened to only have one occurrence). Perhaps you might even want 'Version 12'.
I've also written this to handle certain failures (like if null gets passed or an empty string, etc.). In those cases, the following function will return false.
Use
splitAtSearch('Version 12.34.56', '.') // Returns ['Version 12', '.34.56']
Function
/**
* Splits string based on first result in search
* #param {string} string - String to split
* #param {string} search - Characters to split at
* #return {array|false} - Strings, split at search
* False on blank string or invalid type
*/
function splitAtSearch( string, search ) {
let isValid = string !== '' // Disallow Empty
&& typeof string === 'string' // Allow strings
|| typeof string === 'number' // Allow numbers
if (!isValid) { return false } // Failed
else { string += '' } // Ensure string type
// Search
let searchIndex = string.indexOf(search)
let isBlank = (''+search) === ''
let isFound = searchIndex !== -1
let noSplit = searchIndex === 0
let parts = []
// Remains whole
if (!isFound || noSplit || isBlank) {
parts[0] = string
}
// Requires splitting
else {
parts[0] = string.substring(0, searchIndex)
parts[1] = string.substring(searchIndex)
}
return parts
}
Examples
splitAtSearch('') // false
splitAtSearch(true) // false
splitAtSearch(false) // false
splitAtSearch(null) // false
splitAtSearch(undefined) // false
splitAtSearch(NaN) // ['NaN']
splitAtSearch('foobar', 'ba') // ['foo', 'bar']
splitAtSearch('foobar', '') // ['foobar']
splitAtSearch('foobar', 'z') // ['foobar']
splitAtSearch('foobar', 'foo') // ['foobar'] not ['', 'foobar']
splitAtSearch('blah bleh bluh', 'bl') // ['blah bleh bluh']
splitAtSearch('blah bleh bluh', 'ble') // ['blah ', 'bleh bluh']
splitAtSearch('$10.99', '.') // ['$10', '.99']
splitAtSearch(3.14159, '.') // ['3', '.14159']

For those trying to get everything after the first occurrence:
Something like "Nic K Cage" to "K Cage".
You can use slice to get everything from a certain character. In this case from the first space:
const delim = " "
const name = "Nic K Cage"
const result = name.split(delim).slice(1).join(delim) // prints: "K Cage"
Or if OP's string had two hyphens:
const text = "sometext-20202-03"
// Option 1
const opt1 = text.slice(text.indexOf('-')).slice(1) // prints: 20202-03
// Option 2
const opt2 = text.split('-').slice(1).join("-") // prints: 20202-03

Efficient, compact and works in the general case:
s='sometext-20202'
s.slice(s.lastIndexOf('-')+1)

Use a regular expression of the form: \w-\d+ where a \w represents a word and \d represents a digit. They won't work out of the box, so play around. Try this.

You can use split method for it. And if you should take string from specific pattern you can use split with req. exp.:
var string = "sometext-20202";
console.log(string.split(/-(.*)/)[1])

Everyone else has posted some perfectly reasonable answers. I took a different direction. Without using split, substring, or indexOf. Works great on i.e. and firefox. Probably works on Netscape too.
Just a loop and two ifs.
function getAfterDash(str) {
var dashed = false;
var result = "";
for (var i = 0, len = str.length; i < len; i++) {
if (dashed) {
result = result + str[i];
}
if (str[i] === '-') {
dashed = true;
}
}
return result;
};
console.log(getAfterDash("adfjkl-o812347"));
My solution is performant and handles edge cases.
The point of the above code was to procrastinate work, please don't actually use it.

To use any delimiter and get first or second part
//To divide string using deimeter - here #
//str: full string that is to be splitted
//delimeter: like '-'
//part number: 0 - for string befor delimiter , 1 - string after delimiter
getPartString(str, delimter, partNumber) {
return str.split(delimter)[partNumber];
}