Develop Reference

How to get odd and even position characters from a string?

I'm trying to figure out how to remove every second character (starting from the first one) from a string in Javascript.
For example, the string "This is a test!" should become "hsi etTi sats!"
I also want to save every deleted character into another array.
I have tried using replace method and splice method, but wasn't able to get them to work properly. Mostly because replace only replaces the first character.
function encrypt(text, n) {
if (text === "NULL") return n;
if (n <= 0) return text;
var encArr = [];
var newString = text.split("");
var j = 0;
for (var i = 0; i < text.length; i += 2) {
encArr[j++] = text[i];
newString.splice(i, 1); // this line doesn't work properly
}
}

You could reduce the characters of the string and group them to separate arrays using the % operator. Use destructuring to get the 2D array returned to separate variables
let str = "This is a test!";
const [even, odd] = [...str].reduce((r,char,i) => (r[i%2].push(char), r), [[],[]])
console.log(odd.join(''))
console.log(even.join(''))
Using a for loop:
let str = "This is a test!",
odd = [],
even = [];
for (var i = 0; i < str.length; i++) {
i % 2 === 0
? even.push(str[i])
: odd.push(str[i])
}
console.log(odd.join(''))
console.log(even.join(''))

It would probably be easier to use a regular expression and .replace: capture two characters in separate capturing groups, add the first character to a string, and replace with the second character. Then, you'll have first half of the output you need in one string, and the second in another: just concatenate them together and return:
function encrypt(text) {
let removedText = '';
const replacedText1 = text.replace(/(.)(.)?/g, (_, firstChar, secondChar) => {
// in case the match was at the end of the string,
// and the string has an odd number of characters:
if (!secondChar) secondChar = '';
// remove the firstChar from the string, while adding it to removedText:
removedText += firstChar;
return secondChar;
});
return replacedText1 + removedText;
}
console.log(encrypt('This is a test!'));

Pretty simple with .reduce() to create the two arrays you seem to want.
function encrypt(text) {
return text.split("")
.reduce(({odd, even}, c, i) =>
i % 2 ? {odd: [...odd, c], even} : {odd, even: [...even, c]}
, {odd: [], even: []})
}
console.log(encrypt("This is a test!"));
They can be converted to strings by using .join("") if you desire.

I think you were on the right track. What you missed is replace is using either a string or RegExp.
The replace() method returns a new string with some or all matches of a pattern replaced by a replacement. The pattern can be a string or a RegExp, and the replacement can be a string or a function to be called for each match. If pattern is a string, only the first occurrence will be replaced.
Source: String.prototype.replace()
If you are replacing a value (and not a regular expression), only the first instance of the value will be replaced. To replace all occurrences of a specified value, use the global (g) modifier
Source: JavaScript String replace() Method
So my suggestion would be to continue still with replace and pass the right RegExp to the function, I guess you can figure out from this example - this removes every second occurrence for char 't':
let count = 0;
let testString = 'test test test test';
console.log('original', testString);
// global modifier in RegExp
let result = testString.replace(/t/g, function (match) {
count++;
return (count % 2 === 0) ? '' : match;
});
console.log('removed', result);

like this?
var text = "This is a test!"
var result = ""
var rest = ""
for(var i = 0; i < text.length; i++){
if( (i%2) != 0 ){
result += text[i]
} else{
rest += text[i]
}
}
console.log(result+rest)

Maybe with split, filter and join:
const remaining = myString.split('').filter((char, i) => i % 2 !== 0).join('');
const deleted = myString.split('').filter((char, i) => i % 2 === 0).join('');

You could take an array and splice and push each second item to the end of the array.
function encrypt(string) {
var array = [...string],
i = 0,
l = array.length >> 1;
while (i <= l) array.push(array.splice(i++, 1)[0]);
return array.join('');
}
console.log(encrypt("This is a test!"));

function encrypt(text) {
text = text.split("");
var removed = []
var encrypted = text.filter((letter, index) => {
if(index % 2 == 0){
removed.push(letter)
return false;
}
return true
}).join("")
return {
full: encrypted + removed.join(""),
encrypted: encrypted,
removed: removed
}
}
console.log(encrypt("This is a test!"))
Splice does not work, because if you remove an element from an array in for loop indexes most probably will be wrong when removing another element.

I don't know how much you care about performance, but using regex is not very efficient.
Simple test for quite a long string shows that using filter function is on average about 3 times faster, which can make quite a difference when performed on very long strings or on many, many shorts ones.
function test(func, n){
var text = "";
for(var i = 0; i < n; ++i){
text += "a";
}
var start = new Date().getTime();
func(text);
var end = new Date().getTime();
var time = (end-start) / 1000.0;
console.log(func.name, " took ", time, " seconds")
return time;
}
function encryptREGEX(text) {
let removedText = '';
const replacedText1 = text.replace(/(.)(.)?/g, (_, firstChar, secondChar) => {
// in case the match was at the end of the string,
// and the string has an odd number of characters:
if (!secondChar) secondChar = '';
// remove the firstChar from the string, while adding it to removedText:
removedText += firstChar;
return secondChar;
});
return replacedText1 + removedText;
}
function encrypt(text) {
text = text.split("");
var removed = "";
var encrypted = text.filter((letter, index) => {
if(index % 2 == 0){
removed += letter;
return false;
}
return true
}).join("")
return encrypted + removed
}
var timeREGEX = test(encryptREGEX, 10000000);
var timeFilter = test(encrypt, 10000000);
console.log("Using filter is faster ", timeREGEX/timeFilter, " times")
Using actually an array for storing removed letters and then joining them is much more efficient, than using a string and concatenating letters to it.
I changed an array to string in filter solution to make it the same like in regex solution, so they are more comparable.

How to find sum of integers in a string using JavaScript

I created a function with a regular expression and then iterated over the array by adding the previous total to the next index in the array.
My code isn't working. Is my logic off? Ignore the syntax
function sumofArr(arr) { // here i create a function that has one argument called arr
var total = 0; // I initialize a variable and set it equal to 0
var str = "12sf0as9d" // this is the string where I want to add only integers
var patrn = \\D; // this is the regular expression that removes the letters
var tot = str.split(patrn) // here i add split the string and store it into an array with my pattern
arr.forEach(function(tot) { // I use a forEach loop to iterate over the array
total += tot; // add the previous total to the new total
}
return total; // return the total once finished
}

var patrn = \\D; // this is the regular expression that removes the letters
This is not a valid regular expression in JavaScript.
You are also missing a closing bracket in the end of your code.
A simpler solution would be to find all integers in the string, to convert them into numbers (e.g. using the + operator) and summing them up (e.g. using a reduce operation).
var str = "12sf0as9d";
var pattern = /\d+/g;
var total = str.match(pattern).reduce(function(prev, num) {
return prev + +num;
}, 0);
console.log(str.match(pattern)); // ["12", "0", "9"]
console.log(total); // 21

you have some errors :
change var patrn = \\D with var patrn = "\\D"
use parseInt : total += parseInt(tot);
function sumofArr(arr){ // here i create a function that has one argument called arr
var total = 0; // I initialize a variable and set it equal to 0
var str = "12sf0as9d" // this is the string where I want to add only integers
var patrn = "\\D"; // this is the regular expression that removes the letters
var tot = str.split(patrn) // here i add split the string and store it into an array with my pattern
arr.forEach(function(tot){ // I use a forEach loop to iterate over the array
total += parseInt(tot); // add the previous total to the new total
})
return total; // return the total once finished
}
alert(sumofArr(["1", "2", "3"]));
https://jsfiddle.net/efrow9zs/

function sumofArr(str) {
var tot = str.replace(/\D/g,'').split('');
return tot.reduce(function(prev, next) {
return parseInt(prev, 10) + parseInt(next, 10);
});}
sumofArr("12sf0as9d");

How to make a single string into a multitude of strings?

I have a string called e3 which holds the string 1,2,4,5,3,6. I want to add up all of those numbers up to make the number 21 I was considering doing a for loop for this however I do not know how to turn part of a string into its own value.
I anyone has any better idea of what to do please comment, or answer.

You could use String#split for the string and use Array#reduce for summing.
var e3 = '1,2,4,5,3,6',
sum = e3.split(',').reduce(function (a, b) {
return a + +b; // +b forces b to number
}, 0);
console.log(sum);

If you are sure that it is always a comma separated list of numbers, you could split it on the comma into an array and then use array.reduce() to sum them
var asString = '1,2,4,5,3,6';
var asArray = asString.split(',');
var total = asArray.reduce(function(prev, current){
return prev + parseInt(current, 10);
}, 0);
console.log(total) // outputs 21;

You can do it like this:
var e3 = "1,2,4,5,3,6";
// Split by separator ','
var stringsArr = e3.split(',');
var sum = 0;
// Loop through array of string numbers
stringsArr.forEach(function(str) {
// get Int from a string
var strVal = parseInt(str, 10);
sum += strVal;
});
here's the fiddle

Here is working code to do what you need: https://plnkr.co/edit/8LSkZi0oC8msbHI0qOrz?p=preview
At first you use the split method - this separates a string into an array of strings, based on some separator value. In our case, the separator is a comma, but it could be a blank space or something else:
var testString = '1,2,4,5,3,6';
var separator = ',';
function splitStringOnCommasAndGetArray(string, separator){
var arrayOfStrings = string.split(separator);
return arrayOfStrings;
}
After that, we loop through the array and turn each value into a number. We add the numbers, like so:
function addUpArray(arrayOfStrings){
var totalNumber = 0;
for(var i = 0; i < arrayOfStrings.length; i++){
var currentNum = parseInt(arrayOfStrings[i]);
console.log(currentNum);
totalNumber += currentNum;
}
return totalNumber;
}

Why can't I swap characters in a javascript string?

I am trying to swap first and last characters of array.But javascript is not letting me swap.
I don't want to use any built in function.
function swap(arr, first, last){
var temp = arr[first];
arr[first] = arr[last];
arr[last] = temp;
}

Because strings are immutable.
The array notation is just that: a notation, a shortcut of charAt method. You can use it to get characters by position, but not to set them.
So if you want to change some characters, you must split the string into parts, and build the desired new string from them:
function swapStr(str, first, last){
return str.substr(0, first)
+ str[last]
+ str.substring(first+1, last)
+ str[first]
+ str.substr(last+1);
}
Alternatively, you can convert the string to an array:
function swapStr(str, first, last){
var arr = str.split('');
swap(arr, first, last); // Your swap function
return arr.join('');
}

Let me offer my side of what I understood: swapping items of an array could be something like:
var myFish = ["angel", "clown", "mandarin", "surgeon"];
var popped = myFish.pop();
myFish.unshift(popped) // results in ["surgeon", "angel", "clown", "mandarin"]
Regarding swaping first and last letters of an strings could be done using Regular Expression using something like:
"mandarin".replace(/^(\w)(.*)(\w)$/,"$3$2$1")// outputs nandarim ==> m is last character and n is first letter

I just ran your code right out of Chrome, and it seemed to work find for me. Make sure the indices you pass in for "first" and "last" are correct (remember JavaScript is 0-index based). You might want to also try using console.log in order to print out certain variables and debug if it still doesn't work for you.
EDIT: I didn't realize you were trying to manipulate a String; I thought you just meant an array of characters or values.

function swapStr(str, first, last) {
if (first == last) {
return str;
}
if (last < first) {
var temp = last;
last = first;
first = temp;
}
if (first >= str.length) {
return str;
}
return str.substring(0, first) +
str[last] +
str.substring(first + 1, last) +
str[first] +
str.substring(last + 1);
}

Swap characters inside a string requires the string to convert into an array, then the array can be converted into string again:
function swap(arr, first, last){
arr = arr.split(''); //to array
var temp = arr[first];
arr[first] = arr[last];
arr[last] = temp;
arr = arr.join("").toString() //to string
return arr;
}
Following usage:
str = "ABCDE"
str = swap(str,1,2)
console.log(str) //print "ACBDE"

I hope this piece of code will help somebody.
var word = "DED MOROZ";
var arr = word.split('');
for (var i = 0; i < arr.length/2; i++) {
var temp = arr[i];
arr[i] = arr[arr.length - i - 1];
arr[word.length - i - 1] = temp;
}
console.log(arr.join(""));
Run this code and your characters will be swapped.

how to parse string to int in javascript

i want int from string in javascript how i can get them from
test1 , stsfdf233, fdfk323,
are anyone show me the method to get the integer from this string.
it is a rule that int is always in the back of the string.
how i can get the int who was at last in my string

var s = 'abc123';
var number = s.match(/\d+$/);
number = parseInt(number, 10);
The first step is a simple regular expression - \d+$ will match the digits near the end.
On the next step, we use parseInt on the string we've matched before, to get a proper number.

You can use a regex to extract the numbers in the string via String#match, and convert each of them to a number via parseInt:
var str, matches, index, num;
str = "test123and456";
matches = str.match(/\d+/g);
for (index = 0; index < matches.length; ++index) {
num = parseInt(matches[index], 10);
display("Digit series #" + index + " converts to " + num);
}
Live Example
If the numbers really occur only at the ends of the strings or you just want to convert the first set of digits you find, you can simplify a bit:
var str, matches, num;
str = "test123";
matches = str.match(/\d+/);
if (matches) {
num = parseInt(matches[0], 10);
display("Found match, converts to: " + num);
}
else {
display("No digits found");
}
Live example
If you want to ignore digits that aren't at the end, add $ to the end of the regex:
matches = str.match(/\d+$/);
Live example

var str = "stsfdf233";
var num = parseInt(str.replace(/\D/g, ''), 10);

var match = "stsfdf233".match(/\d+$/);
var result = 0; // default value
if(match != null) {
result = parseInt(match[0], 10);
}

Yet another alternative, this time without any replace or Regular Expression, just one simple loop:
function ExtractInteger(sValue)
{
var sDigits = "";
for (var i = sValue.length - 1; i >= 0; i--)
{
var c = sValue.charAt(i);
if (c < "0" || c > "9")
break;
sDigits = c + sDigits;
}
return (sDigits.length > 0) ? parseInt(sDigits, 10) : NaN;
}
Usage example:
var s = "stsfdf233";
var n = ExtractInteger(s);
alert(n);

This might help you
var str = 'abc123';
var number = str.match(/\d/g).join("");

Use my extension to String class :
String.prototype.toInt=function(){
return parseInt(this.replace(/\D/g, ''),10);
}
Then :
"ddfdsf121iu".toInt();
Will return an integer : 121

First positive or negative number:
"foo-22bar11".match(/-?\d+/); // -22

javascript:alert('stsfdf233'.match(/\d+$/)[0])
Global.parseInt with radix is overkill here, regexp extracted decimal digits already and rigth trimmed string