Develop Reference

No response from recursive function

I want to create a function that is able to determine if a number is same or palindrome. if a given number is palindrome or same then return 2 otherwise if it is not palindrome or same then i need check it twice by increment the given number by 1. after that if it palindrome or same then return 1. if no palindrome or same number found then return 0. i write the function which is giving me the exact result when i give the number as 11211 but the function don't show any response if i enter 1122 or other random value. please help me to find where the error of my function.
function sameOrPalindrome(num) {
var c = 0;
var al = 0;
var normalArray = num.toString().split("");
var revArray = num.toString().split("").reverse();
for (var i = 0; i < normalArray.length; i++) {
if (normalArray[i] != revArray[i]) {
c++;
}
}
if (c == 0) {
return 2;
} else {
num++;
al = sameOrPalindrome(num);
if (al == 2) {
return 1;
} else {
num++;
al = sameOrPalindrome(num);
if (al == 2) {
return 1;
}
}
}
return 0;
}
console.log("1233",sameOrPalindrome(1233))

here is my solution to this problem:
function reversedNum(num) {
return (
parseFloat(
num
.toString()
.split('')
.reverse()
.join('')
) * Math.sign(num)
)
}
function sameOrPalindrome(num) {
if (num === reversedNum(num)) {
return 2;
} else {
num++;
if (num === reversedNum(num)) {
return 1;
} else {
num++;
if (num === reversedNum(num)) {
return 1;
}
}
}
return 0;
}
console.log("1233",sameOrPalindrome(1233))

Perhaps not using recurse - I think your function loops
const allEqual = arr => arr.every( v => v === arr[0] )
const sameOrPalin = num => {
const str = String(num);
let arr = str.split("")
if (allEqual(arr)) return 2
arr.reverse();
if (arr.join("") === str) return 1;
return 0
};
console.log("1111",sameOrPalin(1111));
console.log("2111",sameOrPalin(2111));
console.log("2112",sameOrPalin(2112));
console.log("1234",sameOrPalin(1234));
for (let i = 2111; i<=2113; i++) console.log(i,sameOrPalin(i));

Question: I assumed if palindrome test is true at first time then return 2. if not try incrementing by one and test the palindrome again . if true return 1 else try incrementing for last time and check the palindrome if true return 1 else 0.
Store string into array first and do arr.reverse().join("") to compare
let arr=num.toString().split("");
if(num.toString() == arr.reverse().join(""))
function sameOrPalindrome(num, times) {
let arr = num.toString().split("");
if (num.toString() == arr.reverse().join("")) {
if (times == 3) return 2
else return 1;
} else if (times > 0) {
num++; times--;
return sameOrPalindrome(num, times);
} else return 0
}
console.log(sameOrPalindrome(123321, 3));
console.log(sameOrPalindrome(223321, 3));
console.log(sameOrPalindrome(323321, 3));

Your function needs to know if it should not call itself any more, e.g. when it's doing the second and third checks:
function sameOrPalindrome(num,stop) { // <-- added "stop"
var c = 0;
var al = 0;
var normalArray = num.toString().split("");
var revArray = num.toString().split("").reverse();
for (var i = 0; i < normalArray.length; i++) {
if (normalArray[i] != revArray[i]) {
c++;
}
}
if (c == 0) {
return 2;
} else if(!stop) { // <-- check of "stop"
num++;
al = sameOrPalindrome(num,true); // <-- passing true here
if (al == 2) {
return 1;
} else {
num++;
al = sameOrPalindrome(num,true); // <-- and also here
if (al == 2) {
return 1;
}
}
}
return 0;
}
for(let i=8225;i<8230;i++)
console.log(i,sameOrPalindrome(i));

function check_palindrom(num){
var c1 = 0;
var normalArray = num.toString().split("");
var revArray = num.toString().split("").reverse();
for (var i = 0; i < normalArray.length; i++) {
if (normalArray[i] == revArray[i]) {
c1++;
}
}
if(c1==0){
return 2;
}else{
return 1;
}
}//check_palindrom
function my_fun_check_palindrome(mynum){
//console.log(mynum);
var num = mynum;
var c2 = 0;
var al = 0;
var normalArray = mynum.toString().split("");
var revArray = mynum.toString().split("").reverse();
for (var j = 0; j < normalArray.length; j++) {
if (normalArray[j] == revArray[j]) {
c2++;
}
}
if(c2==0){
console.log('Number is palindrome. Return Value :'+ 2);
}
if(1){
console.log('checking again with incremeting value my one');
num = parseInt(num)+1;
al = check_palindrom(num);
if(al==2){
console.log('Number is palindrome. Return Value :'+ 1);
}else{
console.log('Number is not palindrome. Return Value :'+ 0);
}
}
}//my_fun_check_palindrome
console.log(my_fun_check_palindrome(1122));
console.log(my_fun_check_palindrome(11221));

We should always strive to make function more effiecient... you dont need to run full loop. plus actual checking of palindrome can me modularized
function isSameOrPalindrome(num) {
var normalArray = num.toString().split("");
var revArray = num.toString().split("").reverse(),
i;
for (i = 0; i < normalArray.length / 2; i++) {
if (normalArray[i] !== revArray[i]) {
break;
}
}
if (i >= normalArray.length/2) {
return "Palindrome";
} else {
return "Not Palindrome";
}
}
function doCheck(num) {
var isPalindrome = isSameOrPalindrome(num);
console.log(isPalindrome);
if(isPalindrome === "Palindrome") {
return 2;
} else {
num++;
isPalindrome = isSameOrPalindrome(num);
if(isPalindrome === "Palindrome") {
return 1;
} else {
return 0
}
}
}
console.log("100",doCheck(100));

making custom validation for password field in react

I am making a custom registration page with only 2 values Email and Password, later I will add confirm password as well, for my password field I have some restrictions and I am using some regex and also some custom made code to make the validation.
this is my validateField:
validateField(fieldName, value) {
let fieldValidationErrors = this.state.formErrors;
let emailValid = this.state.emailValid;
let passwordValid = this.state.passwordValid;
//let passwordValidConfirm = this.state.passwordConfirmValid;
switch(fieldName) {
case 'email':
emailValid = value.match(/^([\w.%+-]+)#([\w-]+\.)+([\w]{2,})$/i);
fieldValidationErrors.email = emailValid ? '' : ' is invalid';
break;
case 'password':
passwordValid = (value.length >= 5 && value.length <= 32) && (value.match(/[i,o,l]/) === null) && /^[a-z]+$/.test(value) && this.check4pairs(value) && this.check3InRow(value);
fieldValidationErrors.password = passwordValid ? '': ' is not valid';
break;
default:
break;
}
this.setState({formErrors: fieldValidationErrors,
emailValid: emailValid,
passwordValid: passwordValid,
//passwordValidConfirm: passwordValidConfirm
}, this.validateForm);
}
as you can see for
passwordValid
I have made some methods, this one
check3InRow
doesnt work the way I want it to work, this one makes sure, you have at least 3 letters in your string that are in a row so like "abc" or "bce" or "xyz".
check3InRow(value){
var counter3 = 0;
var lastC = 0;
for (var i = 0; i < value.length; i++) {
if((lastC + 1) === value.charCodeAt(i)){
counter3++;
if(counter3 >= 3){
alert(value);
return true;
}
}
else{
counter3 = 0;
}
lastC = value.charCodeAt(i);
}
return false;
}
this doesnt work correctly so it should accept this:
aabcc
as a password but not:
aabbc

You are starting your counter from 0 and looking for greater than equal to 3 which will never be 3 for 3 consecutive characters. Rest everything is fine with your code.
check3InRow(value) {
var counter3 = 1;
var lastC = 0;
for (var i = 0; i < value.length; i++) {
if ((lastC + 1) === value.charCodeAt(i)) {
counter3++;
if (counter3 >= 3) {
alert(value);
return true;
}
} else {
counter3 = 1;
}
lastC = value.charCodeAt(i);
}
return false;
}

Can we not do a simple version of that function? Like
function check3InRow2(value){
for (var i = 0; i < value.length-2; i++) {
const first = value.charCodeAt(i);
const second = value.charCodeAt(i+1);
const third = value.charCodeAt(i+2);
if(Math.abs(second - first) === 1 && Math.abs(third-second) === 1){
return true;
}
}
return false;
}
I mean complexity wise it is O(N) so maybe we can give this a try
Also adding the your function. When you are AT a char then you should consider counter with 1. Because if another one matches it will be 2 consecutive values.
function check3InRow(value) {
var counter3 = 1;
var lastC = value.charCodeAt(0);
for (var i = 1; i < value.length; i++) {
if ((lastC + 1) === value.charCodeAt(i)) {
counter3++;
if (counter3 >= 3) {
return true;
}
} else {
counter3 = 1;
}
lastC = value.charCodeAt(i);
}
return false;
}

How do I get a toggle button to toggle an array back and forth from descending to ascending?

I am using bubbleSort, and I can get the array to toggle from its original order to descending, but I am having trouble getting it to go from descending back to ascending. Should I just copy the bubbleSort code and flip the greater than/less than signs? Any help is appreciated!
var myStuff = [];
function myfunctionA() {
var enteredvalue = document.getElementById("numbers").value;
// alert(typeof Number(document.getElementById('numbers').value));
if (enteredvalue == "") {
alert("Input is not a number");
} else if (isNaN(enteredvalue)) {
alert('You need to enter a valid number!');
}
var elementExists = false;
var x = document.getElementById('numbers').value;
for (var i = 0; i < myStuff.length; i++) {
if (myStuff[i] == Number(x)) {
elementExists = true;
}
}
if(elementExists != true) {
myStuff.push(Number(enteredvalue));
alert('Thank You for entering a valid number.');
} else {
alert('Element is here');
}
}
function myfunctionB() {
window.alert(myStuff.length);
}
function myfunctionC() {
var sum = 0;
for(var i = 0; i < myStuff.length; i++) {
sum+=myStuff[i];
}
alert(sum);
}
function myfunctionD() {
if (myStuff.length == 0) {
alert("already empty");
} else {
myStuff = [];
}
alert("Array Empty");
}
function myfunctionE() {
alert(myStuff.join('\n'));
{
if (myStuff == []) {
alert("Enter something into Array")
}
}
}
function bubbleSort() {
var sorted = true;
var temp;
while(sorted) {
sorted = false;
for(var i = 0; i < myStuff.length-1; i++) {
if(myStuff[i] < myStuff[i+1]) {
temp = myStuff[i];
myStuff[i] = myStuff[i+1];
myStuff[i+1] = temp;
sorted = true;
}
}
}
}

First you'll need a toggle to tell which way you are going.
var isAscending = false;
Then in your bubbleSort function inside the for-statement, above the if-statement.
var sortComparison;
if (isAscending) sortComparison = myStuff[i] > myStuff[i];
if (!isAscending) sortComparison = myStuff[i] < myStuff[i];
Then replace your if-statement with:
if (sortComparison)
Finally, once you have finished sorting, you can toggle your variable:
isAscending = !isAscending;
Though, I'd recommend using a toggled variable and simply using sort() and reverse() instead.
https://jsfiddle.net/ytcax0qc/

stop for execution javascript?

Well, I'm trying to get him to make some fields, if any of them is equal, it's the message and, though my message is giving non-stop, direct, how can I make it stop giving a lot of alert?
function sendAll(){
for (i = 1;i <=10;i ++) {
for (o = 1;o <=10;o ++) {
if(document.getElementById("table" + i).value==document.getElementById("table" + o).value){
alert("Some field is equal, check again.");
break;
return false;
}
}
}
}

You don't need break statement, just remove it. return statement will do the rest
use
function sendAll() {
for (i = 1; i <= 10; i++) {
for (o = 1; o <= 10; o++) {
if (document.getElementById("table" + i).value == document.getElementById("table" + o).value) {
alert("Some field is equal, check again.");
//break;
return false;
}
}
}
//If you are using for validation purpose return true
// as default
return true;
}

Your break simply stops the inner-most loop, but allows the outer loop to continue. To break out of the function entirely, just use a return:
function sendAll(){
for (i = 1;i <=10;i ++) {
for (o = 1;o <=10;o ++) {
if(document.getElementById("table" + i).value==document.getElementById("table" + o).value){
alert("Some field is equal, check again.");
return false;
}
}
}
return true; // suggested by Mike W
}

Try something like this
function sendAll(){
var bEqual = false;
for (i = 1;i <=10;i ++) {
for (o = 1;o <=10;o ++) {
if(document.getElementById("table" + i).value==document.getElementById("table" + o).value){
bEqual = true;
}
}
}
if (bEqual) {
alert("Some field is equal, check again.");
}
return bEqual;
}

Perhaps you could do:
function sendAll(){
if (!fieldsOk()) {
alert('some message');
}
}
function fieldsOk() {
for (i = 1;i <=10;i ++) {
for (o = 1;o <=10;o ++) {
if(document.getElementById("table" + i).value==document.getElementById("table" + o).value){
return false;
}
}
}
return true;
}

Javascript Phone number validation Paratheses sign

I did some searching and there where others asking this question and answers to it but none that seemed to fit what I was trying to do. Basically I'm working on a validation of the phone entry that accepts (123)4567890 as an entry. I've already implemented one that accepts a simple number string such as 1234567890 and one with dashes 123-456-7890. I know I'm making a simple mistake somewehre but I can't figure out what I'm doing wrong.
Here's the phone number with dashes form that is working:
//Validates phone number with dashes.
function isTwelveAndDashes(phone) {
if (phone.length != 12) return false;
var pass = true;
for (var i = 0; i < phone.length; i++) {
var c = phone.charAt(i);
if (i == 3 || i == 7) {
if (c != '-') {
pass = false;
}
}
else {
if (!isDigit(c)) {
pass = false;
}
}
}
return pass;
}​
and this is the one I can't manage to work out.
function isTwelveAndPara(phone) {
if (phone.length != 12) return false;
var pass = true;
for (var i = 0; i < phone.length; i++) {
var c = phone.charAt(i);
if (i == 0) {
if (c != '(') {
pass = false;
}
}
if (i == 4) {
if (c != ')') {
pass = false;
}
}
else {
if (!isDigit(c)) {
pass = false;
}
}
}
return pass;
}​

You can do it very easily with regex:
return !!phone.match(/\(\d{3}\)\d{7}/g)
Live DEMO
Update:
The code you had didn't work because you forgot the else if:
else if (i == 4) { // Added the "else" on the left.

Checking phone number with RegEx is certainly the way to go. Here is the validation
function that ignores spaces, parentheses and dashes:
check_phone(num) {
return num.replace(/[\s\-\(\)]/g,'').match(/^\+?\d{6,10}$/) != null}
You can vary the number of digits to accept with the range in the second regular expression {6,10}. Leading + is allowed.

Something like that (a RegExp rule) can make sure it matches either rule.
var numbers = ['(1234567890','(123)4567890','123-456-7890','1234567890','12345678901'];
var rule = /^(\(\d{3}\)\d{7}|\d{3}-\d{3}-\d{4}|\d{10})$/;
for (var i = 0; i < numbers.length; i++) {
var passed = rule.test(numbers[i].replace(/\s/g,''));
console.log(numbers[i] + '\t-->\t' + (passed ? 'passed' : 'failed'));
}
EDIT:
function isDigit(num) {
return !isNaN(parseInt(num))
}
function isTwelveAndPara(phone) {
if (phone.length != 12) return false;
for (var i = 0; i < phone.length; i++) {
var c = phone.charAt(i);
if (i == 0) {
if (c != '(') return false;
} else if (i == 4) {
if (c != ')') return false;
} else if (!isDigit(c)) return false;
}
return true;
}
// or...
function isTwelveAndPara(phone) {
if (phone.length != 12 || phone.charAt(0) != '(' || phone.charAt(4) != ')') return false;
for (var i = 1; i < phone.length, i != 4; i++) {
if (!isDigit(phone.charAt(i))) return false;
}
return true;
}