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;
}
Related
I am trying to check my user inputted password with a series of if statements and boolean variables within a function. It seems like my if statements are not modifying my boolean variables. Could someone tell me why?
I was trying to use (/[a-zA-z]/).test(pValue.charAt(0))) as a boolean to see if the first character entry was a lower or upper case letter, but that didn't work either.
document.querySelector("#enter").addEventListener("click", validate);
function validate(e) {
var count = false;
var firstChar = false;
var hasNum = false;
var special = false;
var pValue = document.querySelector("#passwrd").value;
var pLength = pValue.length;
console.log(pValue);
console.log(pLength);
if(pLength > 4 && pLength <= 8) {
count = true;
}
if(pValue.search(e.charCode === [65 - 90]) === 0) {
firstChar = true;
}
console.log(firstChar);
for(var j = 0; j < pLength; j++) {
if(pValue.charAt(j) == "$" || pValue.charAt(j) == "%" || pValue.charAt(j) == "#") {
special = true;
}
}
for(var i = 0; i < pLength; i++) {
if(!isNaN(pValue.charAt(i))) {
hasNum = true;
}
}
if(count && firstChar && hasNum && special) {
document.querySelector("#show_word").textContent = pValue;
}
}
I am trying to run a program that makes the user guess the word by typing in the letter. The same letter should not be allowed to be inputted twice. I was thinking of storing each letter inputted in a list and then I would loop through the list checking each element against the user input. If it is inputted twice and error message like try again will occur. This is what I have so far.
var x = ["Football", "Pie", "Red", "Amber", "Purple", "Blue"];
var y = x[Math.floor(Math.random() * x.length)].toLowerCase();
var answerArray = [];
var lettersUsed = [];
var numberOfGuesses = 10;
for (var i = 0; i < y.length; i++)
{
answerArray[i] = "_";
}
var remainingLetters = y.length;
while (remainingLetters > 0 && numberOfGuesses > 0)
{
console.log(answerArray.join(" "));
var guess = prompt("Guess a letter\n");
if (guess === null)
{
console.log("Game over");
break;
}
else if (guess.length !== 1)
{
console.log("Enter a single letter\n");
}
else
{
numberOfGuesses--;
lettersUsed.push(guess);
for (var j = 0; j < y.length; j++)
{
if (y[j] === guess.toLowerCase() && answerArray[j] === "_")
{
answerArray[j] = guess.toLowerCase();
remainingLetters--;
}
}
}
}
console.log(answerArray.join(" "));
if (numberOfGuesses > 0)
{
console.log("Well done! You've won! Your stick guy has been saved!\n");
}
else
{
console.log("Game over! The word was " + y);
}
///console.log(lettersUsed);
How do I write the for loop? Any help would be appreciated.
You can use the Array.includes(item) method. It will return true if the item is in the array. For example:
let lettersUsed = ['a', 'b', 'c'];
let newLetter = 'b';
if (lettersUsed.includes(newLetter)) {
console.log('letter already used');
} else {
console.log('letter not used yet');
}
Edit in response to question from OP:
You could add this after you check the input length:
else if (guess.length !== 1)
{
console.log("Enter a single letter\n");
}
else if (lettersUsed.includes(guess))
{
console.log('Letter already used');
}
else
{
numberOfGuesses--;
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));
I have the next function:
function solveSudoku(prev_tab, fila, columna) {
let tab = _.cloneDeep(prev_tab);
let sig_fila = fila;
let sig_col = columna;
if (fila === 8 && columna === 8) {
//console.log(tab);
return tab;
}
if (columna === 8) {
sig_col = 0;
sig_fila = sig_fila + 1
} else {
sig_col = sig_col + 1;
}
if ((tab[fila][columna]) !== '') {
solveSudoku(tab, sig_fila, sig_col)
} else {
for (let num = 1; num <= 9; num++) {
if (numeroValido(tab, num, fila, columna)) {
tab[fila][columna] = num;
//tab.toString();
solveSudoku(tab, sig_fila, sig_col)
}
}
}
}
it returns undefined instead of a 2D array, i already try to add return in every recursive call =>
return solveSudoku( tab, sig_fila, sig_col )
but now that doesn't work either
I'm not really familiar with algorithms for solving sudoku, so I don't know if the algorithm below is correct.
But you need to ensure that the result of the recursion is returned. In my update below, I return the first recursive call. In the loop, I only return it if the recursion successfully found a solution, otherwise the loop continues trying other numbers in the column.
function solveSudoku(prev_tab, fila, columna) {
let tab = _.cloneDeep(prev_tab);
let sig_fila = fila;
let sig_col = columna;
if (fila === 8 && columna === 8) {
//console.log(tab);
return tab;
}
if (columna === 8) {
sig_col = 0;
sig_fila = sig_fila + 1
} else {
sig_col = sig_col + 1;
}
if ((tab[fila][columna]) !== '') {
return solveSudoku(tab, sig_fila, sig_col)
} else {
for (let num = 1; num <= 9; num++) {
if (numeroValido(tab, num, fila, columna)) {
tab[fila][columna] = num;
//tab.toString();
let result = solveSudoku(tab, sig_fila, sig_col);
if (result) { // continue searching if the recursion failed
return result;
}
}
}
}
}
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;
}