I would like to convert the string with a mix of upper and lower case. For example, if I have string Mark, I need an output as mARk, or Lewis to convert as lEwIs. How can I achieve this in vanilla JavaScript?
Note: The rule of conversion is random.
I've tried Camelize function, but that not giving me an expected output:
function camelize(str) {
return str.replace(/(?:^\w|[A-Z]|\b\w)/g, function(letter, index) {
return index === 0 ? letter.toLowerCase() : letter.toUpperCase();
}).replace(/\s+/g, '');
}
This will do the trick. Edited to make random
function flipCase(str) {
var flip = '';
for (var i = 0; i < str.length; i++) {
if(Math.random() > .5){
flip += str.charAt(i).toUpperCase();
} else {
flip += str.charAt(i).toLowerCase();
}
}
return flip;
}
I know you mentioned in comments that you wanted it to be random, but for other readers who are looking for a function to invert the casing, here is one way of doing it:
function camelize(str) {
var inverted = "";
for (var i = 0; i < str.length; i++) {
if (str.charCodeAt(i) >= 65 && str.charCodeAt(i) <= 90)
inverted += str.charAt(i).toLowerCase();
else
inverted += str.charAt(i).toUpperCase();
}
return inverted;
}
console.log(camelize("Mark"));
console.log(camelize("Lwies"));
And here is another way using Array.map():
function camelize(str) {
return Array.from(str).map(l => (l.charCodeAt(0) >= 65 && l.charCodeAt(0) <= 90) ? l.toLowerCase() : l.toUpperCase()).join('');
}
console.log(camelize("Mark"));
console.log(camelize("Lwies"));
Related
I am beginner and want to make my own function.
I want to hash the password by shifting every character by given x
positions and reverse to lowercase/uppercase.
I think the code below should return "EFGH7654" but it return 55 with no error message.
How can I fix it? Is it because of I put a function in a function?
Or I type wrong any thing?
function hashPassword(password, x) {
// password is a string, x is a number
// return a string
// (ex. password = 'ab1By', x = 3 so it should return "DE4eB")
function shift(text, s) {
result = "";
for (let i = 0; i < text.length; i++) {
let char = text[i];
if (char.toUpperCase(text[i])) {
let ch = String.fromCharCode((char.charCodeAt(0) + s - 65) % 26 + 65);
result += ch;
} else {
let ch = String.fromCharCode((char.charCodeAt(0) + s - 97) % 26 + 97);
result += ch;
}
}
return result;
}
function reversecase(x) {
var output = '';
for (var i = 0, len = x.length; i < len; i++) {
var character = x[i];
if (character == character.toLowerCase()) {
// The character is lowercase
output = output + character.toUpperCase();
} else {
// The character is uppercase
output = output + character.toLowerCase();
}
}
return output
}
var str = "";
var result = "";
var charcode = "";
for (var i = 0; i < password.length; i++) {
if (typeof password[i] === typeof str) {
char = shift(password[i], x)
charcode = reversecase(char)
result += charcode;
} else {
num = password[i] + x
number = num % 10
result += number.toString()
}
}
return result
};
console.log(hashPassword("abcd4321", 4))
There a quite some problems in your code.
The first problem here is not only the nesting, but the fact that you're defining the result variable in the outer function scope using the var keyword. Then you use (read/write) that variable in different places.
In function shift() (also in return statement)
In the outer function (also in return statement)
The thing you have to understand is, that you're referring to the same variable result every time. To ensure that your variables are scoped, i.e. are only valid within a block (if statement, function body, etc.), you should use the let or const keywords. This makes your code a lot safer.
The second problem are some assumptions you make regarding data types. If you have a string let s = "my string 123", the expression typeof s[x] === 'string' will be true for every x in s.
Another problem is the algorithm itself. The outer function hashPassword() iterates over all characters of the input string. Within that loop you call function shift(password[i], x), passing a single character. The first parameter of shift() is called text - and there is another for loop (which is confusing and does not make sense).
To make things short, please have a look at this simplified version:
function shift(char, x) {
let result;
const code = char.charCodeAt(0);
if (code >= 65 && code < 91) {
result = String.fromCharCode((code + x - 65) % 26 + 65);
}
else if (code >= 48 && code <= 57) {
result = String.fromCharCode((code + x - 48) % 10 + 48);
}
else {
result = String.fromCharCode((code + x - 97) % 26 + 97);
}
return result;
}
function reverseCase(character) {
if (character === character.toLowerCase()) {
return character.toUpperCase();
}
else {
return character.toLowerCase();
}
}
function hashPassword(password, x) {
let result = "";
for (let i = 0; i < password.length; i++) {
const char = shift(password[i], x);
result += reverseCase(char);
}
return result;
}
console.log(hashPassword("abcd4321", 4)); // Output: EFGH8765
I'm learning Javascript and trying to make a function that compares a current letter of a string with the one succeeding it, repeating through the entire string to then determine if it is alphabetical. My if condition is what's wrong, but I'm not sure how I'd change it to make it work.
Also, I'd like to keep the structure similar to what I have if possible.
function is_alphabetical(string) {
let result;
for ( i = 0; i < string.length; i++ ) {
if ( string[i] <= string[i + 1] ) {
result = true;
} else {
result = false;
break;
}
}
return result;
}
Have a look into localCompare function e.g.
"a".localeCompare("b")
>> -1
"a".localeCompare("a")
>> 0
"b".localeCompare("a")
>> 1
in your case:
if((string[i + 1]).localCompare(string[i]))
Hope that helps :)
By keeping the structure similar as you asked, I think this solution works:
function is_alphabetical(string) {
for ( i = 0; i < string.length; i++ ) {
if ( string[i] >= string[i + 1] ) {
return false;
}
}
return true;
}
Here's how I would do it.
function isAlphabetical(str) {
// we iterate only while i<length-1 to avoid going out of bounds when doing str[i+1]
for (let i = 0, n = str.length - 1; i < n; i++) {
if (str[i] > str[i + 1])
return false; // at least one letter is making it not ordered
// else keep checking other letters
}
return true; // if we reached here, its ordered
}
/Write a function called weave that accepts an input string and number. The function should return the string with every xth character replaced with an 'x'./
function weave(word,numSkip) {
let myString = word.split("");
numSkip -= 1;
for(let i = 0; i < myString.length; i++)
{
numSkip += numSkip;
myString[numSkip] = "x";
}
let newString = myString.join();
console.log(newString);
}
weave("weave",2);
I keep getting an infinite loop. I believe the answer I am looking for is "wxaxe".
Here's another solution, incrementing the for loop by the numToSkip parameter.
function weave(word, numToSkip) {
let letters = word.split("");
for (let i=numToSkip - 1; i < letters.length; i = i + numToSkip) {
letters[i] = "x"
}
return letters.join("");
}
Well you need to test each loop to check if it's a skip or not. Something as simple as the following will do:
function weave(word,numSkip) {
var arr = word.split("");
for(var i = 0; i < arr.length; i++)
{
if((i+1) % numSkip == 0) {
arr[i] = "x";
}
}
return arr.join("");
}
Here is a working example
Alternatively, you could use the map function:
function weave(word, numSkip) {
var arr = word.split("");
arr = arr.map(function(letter, index) {
return (index + 1) % numSkip ? letter : 'x';
});
return arr.join("");
}
Here is a working example
Here is a more re-usable function that allows specifying the character used for substitution:
function weave(input, skip, substitute) {
return input.split("").map(function(letter, index) {
return (index + 1) % skip ? letter : substitute;
}).join("");
}
Called like:
var result = weave('weave', 2, 'x');
Here is a working example
You dont need an array, string concatenation will do it, as well as the modulo operator:
function weave(str,x){
var result = "";
for(var i = 0; i < str.length; i++){
result += (i && (i+1)%x === 0)?"x":str[i];
}
return result;
}
With arrays:
const weave = (str,x) => str.split("").map((c,i)=>(i&&!((i+1)%x))?"x":c).join("");
You're getting your word greater in your loop every time, so your loop is infinite.
Try something like this :
for(let k = 1; k <= myString.length; k++)
{
if(k % numSkip == 0){
myString[k-1]='x';
}
}
Looking at what you have, I believe the reason you are getting an error is because the way you update numSkip, it eventually becomes larger than
myString.length. In my code snippet, I make i increment by numSkip which prevents the loop from ever executing when i is greater than myString.length. Please feel free to ask questions, and I will do my best to clarify!
JSFiddle of my solution (view the developer console to see the output.
function weave(word,numSkip) {
let myString = word.split("");
for(let i = numSkip - 1; i < myString.length; i += numSkip)
{
myString[i] = "x";
}
let newString = myString.join();
console.log(newString);
}
weave("weave",2);
Strings are immutable, you need a new string for the result and concat the actual character or the replacement.
function weave(word, numSkip) {
var i, result = '';
for (i = 0; i < word.length; i++) {
result += (i + 1) % numSkip ? word[i] : 'x';
}
return result;
}
console.log(weave("weave", 2));
console.log(weave("abcd efgh ijkl m", 5));
You can do this with fewer lines of code:
function weave(word, numSkip) {
word = word.split("");
for (i = 0; i < word.length; i++) {
word[i] = ((i + 1) % numSkip == 0) ? "x" : word[i];
}
return word.join("");
}
var result = weave("weave", 2);
console.log(result);
I wrote a simple script for a website called Codewars (here: https://www.codewars.com/kata/57814d79a56c88e3e0000786). The purpose of the function was to encrypt a string such that every second character would appear first, and then the rest of them. I tested many random strings of text; it worked for a while. But then, I tested a specific case with 17 characters: "maybe do i really", and it resulted in a character being dropped (notably a space). Initially, I thought the issue was that the .join method didn't allow a double space in a row, so I attempted to make my own function to mimic its functionality: it did not solve the problem. Could anyone answer why this specific string loses a character and returns a wrong encryption? My jsfiddle is here: https://jsfiddle.net/MCBlastoise/fwz62j2g/
Edit: I neglected to mention that it runs a certain number of times based on parameter n, encrypting the string multiple times per that value.
And my code is here:
function encrypt(text, n) {
if (n <= 0 || isNaN(n) === true || text === "" || text === null) {
return text;
}
else {
for (i = 1; i <= n; i++) {
if (i > 1) {
text = encryptedString;
}
var evenChars = [];
var oddChars = [];
for (j = 0; j < text.length; j++) {
if (j % 2 === 0) {
evenChars.push(text.charAt(j));
}
else {
oddChars.push(text.charAt(j));
}
}
var encryptedString = oddChars.join("") + evenChars.join("");
}
return encryptedString;
}
}
function decrypt(encryptedText, n) {
if (n <= 0 || encryptedText === "" || encryptedText === null) {
return encryptedText;
}
else {
for (i = 1; i <= n; i++) {
if (i > 1) {
encryptedText = decryptedString;
}
var oddChars = [];
var evenChars = [];
for (j = 0; j < encryptedText.length; j++) {
if (j < Math.floor(encryptedText.length / 2)) {
oddChars.push(encryptedText.charAt(j));
}
else {
evenChars.push(encryptedText.charAt(j));
}
}
var convertedChars = []
for (k = 0; k < evenChars.length; k++) {
convertedChars.push(evenChars[k]);
convertedChars.push(oddChars[k]);
}
var decryptedString = convertedChars.join("");
}
return decryptedString;
}
}
document.getElementById("text").innerHTML = encrypt("maybe do i really", 1);
document.getElementById("text2").innerHTML = decrypt("ab oiralmyed ely", 1)
<p id="text"></p>
<p id="text2"></p>
Nothing wrong with the code itself. Basically HTML doesn't allow 2 or more spaces. You can use <pre> tag for the case like this.
document.getElementById("text").innerHTML = "<pre>" + encrypt("maybe do i really", 1) + "</pre>";
console.log("HİNDİ".toLocaleLowerCase() == "hindi");
console.log("HİNDİ" == "hindi");
console.log("HİNDİ".toLowerCase());
console.log("HİNDİ".toLocaleLowerCase())
console.log("HİNDİ".toLowerCase())
I am building a search functionality but i come across a thing:
"HİNDİ".toLocaleLowerCase() // "hindi"
"hindi" == "HİNDİ".toLocaleLowerCase() //false
What the heck is going on here?
Solution:
#pmrotule's answer seems to work:
function to_lower(s)
{
var n = "";
for (var i = 0; i < s.length; i++) // do it for one character at a time
{
var c = s[i].toLowerCase();
// call replace() only if the character has a length > 1
// after toLowerCase()
n += c.length > 1 ? c[0].replace(/[^ -~]/g,'') : c;
}
return n;
}
Thanks,
It is a problem of string format. toLocaleLowerCase is meant for human-readable display only. However, there is still a trick you can do:
if ("hindi" == "HİNDİ".toLowerCase().replace(/[^ -~]/g,''))
{
alert("It works!");
}
EDIT
If you want to make it works with all special characters:
function to_lower(s)
{
var n = "";
for (var i = 0; i < s.length; i++) // do it for one character at a time
{
var c = s[i].toLowerCase();
// call replace() only if the character has a length > 1
// after toLowerCase()
n += c.length > 1 ? c.replace(/[^ -~]/g,'') : c;
}
return n;
}
console.log("gök" == to_lower("GÖK"));
console.log("hindi" == to_lower("HİNDİ"));
function to_low(s) // shorter version
{
var n = "";
for (var i = 0; i < s.length; i++)
{ n += s[i].toLowerCase()[0]; }
return n;
}
console.log("hindi" == to_low("HİNDİ"));
The problem is that your character İ is composed by 2 characters.
You have the I and then the 'dot' at the top (UTF-8 decimal code: 775).
Try this:
"HİNDİ".toLocaleLowerCase().split('').map((_,v)=>console.log(_.charCodeAt(0)))
Compare it with this:
"hindi".toLocaleLowerCase().split('').map((_,v)=>console.log(_.charCodeAt(0)))