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
Related
I'm trying to format a string to show two sig figs after the decimal. I want to do this without using built-in methods like 'Math' and such. I have something where I can do that, but the issue comes when the number is just '4.5' for example, I need to add a 0 onto it, but I'm having trouble. Here's my code so far.
function toFixed(num){
let elem = "";
// let x = "";
for (let i = 0; i < num.length; i++) {
if(num[i - 2] === '.'){ // num[i] happens to be the third 8
break;
}
// elem += num[i];
elem += num[i];
elem += "0";
}
return elem
}
console.log(toFixed("7.88888888")) // "7.88"
console.log(toFixed("77645345.987654")) //77645345.88
console.log(toFixed("1")) // "1.00"
console.log(toFixed("4.5")) // "4.50"
Right now I'm getting 0's after each number it seems.
Without Math:
function toFixed(num) {
let [a, b = ""] = num.split(".");
b += "00";
return `${a}.${b.slice(0, 2)}`;
}
console.log(toFixed("7.88888888")) // "7.88"
console.log(toFixed("77645345.987654")) // "77645345.98"
console.log(toFixed("1")) // "1.00"
console.log(toFixed("4.5")) // "4.50"
Without any builtin methods:
function toFixed(num) {
const len = num.length;
let result = "";
for (let i = 0; i < len; i++) {
const c = num[i];
if (c === ".") {
return `${result}.${num[i + 1] || 0}${num[i + 2] || 0}`;
}
result += c;
}
return `${result}.00`;
}
console.log(toFixed("7.88888888")) // "7.88"
console.log(toFixed("77645345.987654")) // "77645345.98"
console.log(toFixed("1")) // "1.00"
console.log(toFixed("4.5")) // "4.50"
function toFixedNumber(num) {
let elem = +num;
return elem.toFixed(2)
}
console.log(toFixedNumber("7.88888888")) // "7.88"
console.log(toFixedNumber("77645345.987654")) //77645345.88
console.log(toFixedNumber("1")) // "1.00"
console.log(toFixedNumber("4.5")) // "4.50"
/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>";
I am learning js now..
I am trying to write a simple js programme..
what I am trying to do is to print all valid combinations of n-pair
of parenthesis(properly opened and closed)
eg (), (()()),(())
i have written the logic can you tell me whether its correct or not
https://jsfiddle.net/e7mcp6xb/
module.exports = Parentheses = (function() {
var _isParenthesesMatch = function(str) {
var parentheses = str.length;
var rightParentheses = '(';
var leftParentheses = ')';
var rightCount = 0;
var leftCount = 0;
for(i=0;i<=str.length;i++){
if(rightParentheses == str.charAt(i))
{
rightCount++;
}
else if(leftParentheses == str.charAt(i))
{
leftCount++;
}
}
if(rightCount == leftCount){
return true;
}
else(rightCount != leftCount){
return false;
}
}
}());
The check is wrong, but You can fix it easily: In each step of the for loop the number of opening parenthesis cannot be smaller than the number of closing ones:
if (rightCount < leftCount)
return false;
The whole function should look like this:
function(str) {
var rightParentheses = '(';
var leftParentheses = ')';
var rightCount = 0;
var leftCount = 0;
for (var i = 0; i <= str.length; i++) {
if (rightParentheses == str.charAt(i))
rightCount++;
else if (leftParentheses == str.charAt(i))
leftCount++;
if (rightCount < leftCount)
return false;
}
return rightCount == leftCount;
}
If You'd like to generate all valid strings, you can use this function:
function nPair(n) {
if (n == 0)
return [""];
var result = [];
for (var i = 0; i < n; ++i) {
var lefts = nPair(i);
var rights = nPair(n - i - 1);
for (var l = 0; l < lefts.length; ++l)
for (var r = 0; r < rights.length; ++r)
result.push("(" + lefts[l] + ")" + rights[r]);
}
return result;
}
// result of nPair(3):
// ["()()()", "()(())", "(())()", "(()())", "((()))"]
Try this, i have modified your code a little bit. Modification and its explanation is marked in comments.
module.exports = Parentheses = (function() {
var _isParenthesesMatch = function(str) {
var parentheses = str.length;
var rightParentheses = '(';
var leftParentheses = ')';
var count=0;
for(i=0;i<str.length;i++){
//this is to check valid combination start always from ( and end with )
if(str.charAt(0)==rightParentheses && str.length-1==leftParentheses)
{
if(rightParentheses == str.charAt(i))
{
count++; //this will calculate how many times rightParentheses is present & increment count by 1
}
else if(leftParentheses == str.charAt(i))
{
count--; //this will simply decrement count to match valid sequence
}
}
if(count==0){
return true;
}
}
}());
Your function is wrong, try checking if left and right parenthesis and balanced:
function isValid(str){
var stripedStr = str.replace(/[^\(\)]+/g, '');
return stripedStr.split('').reduce(function(a, b){
return a > -1 ? b === '(' ? a + 1 : a - 1 : -1;
}, 0) === 0;
}
stripedStr - use replace() to remove any characters that are not ( or ).
split('') - returns an array so we can use reduce.
reduce() - applies a function against an accumulator and each value of the array (from left-to-right) has to reduce it to a single value.
The reduce starts with 0 as initial value and in the reduce function we count parenthesis
(+1 for (, -1 for ) )
Our string is valid if our counter never goes below 0 and we end up with 0.
You can write the reduce function like this too:
function(previousValue, currentValue){
if (previousValue > -1){
if (currentValue === '('){
return previousValue + 1;
} else {
return previousValue - 1;
}
}
return -1;
}
This is equivalent to:
function(a, b){
return a > -1 ? b === '(' ? a + 1 : a - 1 : -1;
}
It is wrong, because your function will return true for this example ))(( or this ())(()
I need to count how long in bytes a textarea is when UTF8 encoded using javascript. Any idea how I would do this?
thanks!
encodeURIComponent(text).replace(/%[A-F\d]{2}/g, 'U').length
Combining various answers, the following method should be fast and accurate, and avoids issues with invalid surrogate pairs that can cause errors in encodeURIComponent():
function getUTF8Length(s) {
var len = 0;
for (var i = 0; i < s.length; i++) {
var code = s.charCodeAt(i);
if (code <= 0x7f) {
len += 1;
} else if (code <= 0x7ff) {
len += 2;
} else if (code >= 0xd800 && code <= 0xdfff) {
// Surrogate pair: These take 4 bytes in UTF-8 and 2 chars in UCS-2
// (Assume next char is the other [valid] half and just skip it)
len += 4; i++;
} else if (code < 0xffff) {
len += 3;
} else {
len += 4;
}
}
return len;
}
[June 2020: The previous answer has been replaced due to it returning incorrect results].
Most modern JS environments (browsers and Node) now support the TextEncoder API, which may be used as follows to count UTF8 bytes:
const textEncoder = new TextEncoder();
textEncoder.encode('⤀⦀⨀').length; // => 9
This is not quite as fast as the getUTF8Length() function mentioned in other answers, below, but should suffice for all but the most demanding use cases. Moreover, it has the benefit of leveraging a standard API that is well-tested, well-maintained, and portable.
If you have non-bmp characters in your string, it's a little more complicated...
Because javascript does UTF-16 encode, and a "character" is a 2-byte-stack (16 bit) all multibyte characters (3 and more bytes) will not work:
<script type="text/javascript">
var nonBmpString = "foo€";
console.log( nonBmpString.length );
// will output 5
</script>
The character "€" has a length of 3 bytes (24bit). Javascript does interpret it as 2 characters, because in JS, a character is a 16 bit block.
So to correctly get the bytesize of a mixed string, we have to code our own function fixedCharCodeAt();
function fixedCharCodeAt(str, idx) {
idx = idx || 0;
var code = str.charCodeAt(idx);
var hi, low;
if (0xD800 <= code && code <= 0xDBFF) { // High surrogate (could change last hex to 0xDB7F to treat high private surrogates as single characters)
hi = code;
low = str.charCodeAt(idx + 1);
if (isNaN(low)) {
throw 'Kein gültiges Schriftzeichen oder Speicherfehler!';
}
return ((hi - 0xD800) * 0x400) + (low - 0xDC00) + 0x10000;
}
if (0xDC00 <= code && code <= 0xDFFF) { // Low surrogate
// We return false to allow loops to skip this iteration since should have already handled high surrogate above in the previous iteration
return false;
/*hi = str.charCodeAt(idx-1);
low = code;
return ((hi - 0xD800) * 0x400) + (low - 0xDC00) + 0x10000;*/
}
return code;
}
Now we can count the bytes...
function countUtf8(str) {
var result = 0;
for (var n = 0; n < str.length; n++) {
var charCode = fixedCharCodeAt(str, n);
if (typeof charCode === "number") {
if (charCode < 128) {
result = result + 1;
} else if (charCode < 2048) {
result = result + 2;
} else if (charCode < 65536) {
result = result + 3;
} else if (charCode < 2097152) {
result = result + 4;
} else if (charCode < 67108864) {
result = result + 5;
} else {
result = result + 6;
}
}
}
return result;
}
By the way...
You should not use the encodeURI-method, because, it's a native browser function ;)
More stuff:
Code on GitHub
More on Mozilla Developer Networks
Cheers
frankneff.ch / #frank_neff
Add Byte length counting function to the string
String.prototype.Blength = function() {
var arr = this.match(/[^\x00-\xff]/ig);
return arr == null ? this.length : this.length + arr.length;
}
then you can use .Blength() to get the size
I have been asking myself the same thing. This is the best answer I have stumble upon:
http://www.inter-locale.com/demos/countBytes.html
Here is the code snippet:
<script type="text/javascript">
function checkLength() {
var countMe = document.getElementById("someText").value
var escapedStr = encodeURI(countMe)
if (escapedStr.indexOf("%") != -1) {
var count = escapedStr.split("%").length - 1
if (count == 0) count++ //perverse case; can't happen with real UTF-8
var tmp = escapedStr.length - (count * 3)
count = count + tmp
} else {
count = escapedStr.length
}
alert(escapedStr + ": size is " + count)
}
but the link contains a live example of it to play with. "encodeURI(STRING)" is the building block here, but also look at encodeURIComponent(STRING) (as already point out on the previous answer) to see which one fits your needs.
Regards
encodeURI(text).split(/%..|./).length - 1
How about simple:
unescape(encodeURIComponent(utf8text)).length
The trick is that encodeURIComponent seems to work on characters while unescape works on bytes.
Try the following:
function b(c) {
var n=0;
for (i=0;i<c.length;i++) {
p = c.charCodeAt(i);
if (p<128) {
n++;
} else if (p<2048) {
n+=2;
} else {
n+=3;
}
}return n;
}
set meta UTF-8 just & it's OK!
<meta charset="UTF-8">
<meta http-equiv="content-type" content="text/html;charset=utf-8">
and js:
if($mytext.length > 10){
// its okkk :)
}