I'm searching for how can I check if word has a same end and start. There is my code:
JavaScript:
function Check() {
var value = document.getElementById("input").value;
if (value == startsWith(endsWith()) && value == endsWith(startsWith())) {
return alert("Yes");
} else {
return alert("No");
}
}
<input id="input" style="margin:20px;" type="text"><br>
<button id="button" onclick="Check();">Check</button>
<p id="demo" style="color:white;">Answer</p>
Just check the first and last character of the string:
if(value[0] == value[value.length-1])
{
alert('Yes');
}
else
{
alert('No');
}
Try below function.
I am comparing first character and last character of your input string.
function Check() {
var value = document.getElementById("input").value;
if (value.charAt(0) === value.charAt(value.length - 1)) {
return alert("Yes");
} else {
return alert("No");
}
}
Use str.length - 1 to get the index of the last character of the string.
function check(str) {
if (str[0] === str[str.length - 1]) {
alert('yes')
} else {
alert('no');
}
}
check('test');
check('tests');
To check first/last n characters you can do this.
function checkMore(str, amount) {
if (str[0] !== str[str.length - 1]) {
alert('no');
return;
} else if (amount !== 1) {
return checkMore(str.slice(1, str.length - 1), amount - 1);
}
alert('yes');
}
checkMore('stests', 1);
checkMore('stests', 2);
checkMore('stests', 3);
And to check the whole string (basically checking if the string is palindrome).
function checkPalindrome(str) {
if (str.length === 0 || str.length === 1) {
alert('yes');
return;
}
if (str[0] !== str[str.length - 1]) {
alert('no');
return;
}
checkPalindrome(str.slice(1, str.length - 1));
}
checkPalindrome('car');
checkPalindrome('carac');
checkPalindrome('carrac');
You can use match() or indexOf(). Both work, but both search the whole string. It is more efficient to extract the substring in the relevant place and compare it with the one you expect there:
function Check ( word ) {
return word.charAt(0) == word.charAt( word.length - 1 )
}
<button tye="button" onclick="alert( Check('Hello') )">Check 'Hello'</button>
<button tye="button" onclick="alert( Check('HelloH') )">Check 'HelloH'</button>
Of course, you might as well use a regular expression, with a smart regex engine it should be efficient as well (and your code is more concise):
function Check ( word ) {
return /^(.).*\1$|^.$/.test( word )
}
<button tye="button" onclick="alert( Check('Hello') )">Check 'Hello'</button>
<button tye="button" onclick="alert( Check('WoW') )">Check 'WoW'</button>
var value = "wow"
return (value[0] === value[value.length - 1] ? true : false)
0th index will contain starting character and value.length - 1 will contain ending character. If they are equal it will return true and false otherwise.
You can use .split('' ) to split a string by spaces, so into words. It splits it into an array, so you can compare index [0] to index [length-1]
For example
function Check(string) {
let arr = string.split(' ');
return arr[0] == arr[arr.length-1];
}
Use regex:
if (/^(.).*\1$|^.$/.test(value))
// first and last are same letter
I've started reading Eloquent Javascript, and there's an exercise about making a recursive function to check for evenness.
I've made it in a couple different ways, it's quite simple, but for some reason I can't get it to work with negative numbers anymore. I had it working, then probably accidentally changed something, and now it only works for positives.
Could you please tell me why this code is 'wrong'?
(textfield.append just prints something to a textfield I've made in an html/css-document, so I can save the exercises in some kind of 'program'.)
function evencheck(n){
if (n == 0){
$('#textfield').append('Even');
}
if (n == 1 || n == -1){
$('#textfield').append('Uneven');
}
else{
if(n > 1){
n -= 2;
evencheck(n);
}
if(n < -1){
n += 2;
evencheck(n);
}
}
}
I know it can be written shorter, I've made a shorter form of it, but that didn't work on negatives either.
I know the problem is a stack overflow, but why is this happening?
not an answer but an extended comment
function evencheck(n){
if (n == 0){
return $('#textfield').append('Even');
}
if (n == 1 || n == -1){
return $('#textfield').append('Uneven');
}
return evencheck(n > 1? n-2 : n+2);
}
The upper code will probably be faster, as the compiler can optimize it to:
function evencheck(n){
while(true){
if (n == 0){
return $('#textfield').append('Even');
}
if (n == 1 || n == -1){
return $('#textfield').append('Uneven');
}
n = n>1? n -2 : n+2;
}
}
So youre not filling the function stack ( really huge numbers possible) , and its actually really fast.
More about that
Your code seems to work, except for large numbers. Try it with something like -12 or 10 works fine. When you input 30000 it hangs itself. Probably because you call the method recursively too many times.
const
inputElement = document.getElementById('number-input'),
checkTrigger = document.getElementById('check-number')
resultLog = document.getElementById('result');
function evencheck(n){
if (n == 0){
resultLog.textContent = `${n} is event.`;
}
if (n == 1 || n == -1){
resultLog.textContent = `${n} is unevent.`;
}
else{
if(n > 1){
n -= 2;
evencheck(n);
}
if(n < -1){
n += 2;
evencheck(n);
}
}
}
function checkInputNumber(event) {
const
numberToCheck = parseInt(inputElement.value);
evencheck(numberToCheck);
}
checkTrigger.addEventListener('click', checkInputNumber);
$('#evenrecursive').click(function(){ $('#textfield').append("<p style ='color:blue'>new command: check if number is even.</p>"); var n = prompt('pick a number', ''); evencheck(n); });
#evenrecursive {
border: 1px solid;
min-height: 20px;
width: 100%;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input id="number-input" type="number" />
<button type="button" id="check-number">check</button>
<p id="result"></p>
<div id="evenrecursive">click me for prompt.</div>
I've added your trigger and the problem you're having is type coercion. When the initial value of n is "-10" the code winds up in n += 2, this comes down to n = "-10" + 2 which is "-102" and thus your code never reaches the end.
now it only works for positives. why is this happening?
You might be passing a string, in which case n += 2 does do string concatenation instead of number addition (whereas -= 2 always casts to a number). Try adding
if (typeof n != "number") throw new TypeError("n must be a number");
in the first line of the function, and make sure you do use parseInt or some other suitable parsing method if you take a string input from the user.
Basically, I want to change a person input from 0######### to (0#)########, the event im using is onblur
function numberChange(){
var check = document.getElementById("number").value;
var regCheck = /^[0-9]+$/;
if (check != 10 || !regCheck)
{
alert("Please input your 10 digit mobile number")
return false;
}
else if (check == 10 || regCheck)
{
return true;
}
}
Your regCheck should be ^\(0\d\)\d{8}$
Demo
Update :
This regex validates the number of characters, you can omit length check in your code :
function numberChange(){
var check = document.getElementById("number").value;
var re = new RegExp(/^\(0\d\)\d{8}$/);
if (re.test(check)) {
return true;
}
else {
alert("Please input your 10 digit mobile number")
return false;
}
}
This will check for 10 digits, first number is zero, and that all inputs are digits. Once all of that passes the code will add the '(' & ')' to it. This approach gives specific errors depending on how the user incorrectly entered the number.
<html>
<body>
Phone number: <input type="text" id="number" onblur="checkNumber()">
<script>
function checkNumber() {
var x = document.getElementById("number").value;
// Check for 10 digits.
if(x.length != 10) {
alert('The number must be 10 digits.');
return false;
}
// Make sure the first digit is 0.
if(x.charAt(0) != 0) {
alert('The first digit must be 0.');
return false;
}
// Make sure all digits are numbers and not characters.
if(isNaN(x) == true) {
alert('All digits must be numbers.');
return false;
}
// If here then everything else passed so add the '(' and ')' to the number.
document.getElementById("number").value = '(0' + x.charAt(1) + ')' + x.substring(2, 10);
}
</script>
</body>
</html>
The JavaScript below just checks if there are 4 digits in a field or not and then warns if there isn't.
But for some reason, if I enter four zeros (0000) it takes that as an empty field and throws the warning.
Any idea how to fix this...?? I am no programmer but have put this code together after weeks of trial-and-error.
function validate(){
// if ( (!isNaN($("#couponstart1").val())) || (!isNaN($("#couponstart2").val())) || (!isNaN($("#couponend1").val())) || (!isNaN($("#couponend2").val())) ) {
var all_ok = true;
var err_msg = "";
var fld = "";
if ( $("#couponstart1").val() == '' || $("#couponstart2").val() == '' || $("#couponend1").val() == '' || $("#couponend2").val() == '' ) {
all_ok = false;
err_msg += "\n - Card Numbers cannot be blank";
fld='couponstart1';
}else{
if ( isNaN($("#couponstart1").val()) || isNaN($("#couponstart2").val()) || isNaN($("#couponend1").val()) || isNaN($("#couponend2").val()) ) {
all_ok = false;
err_msg += "\n - Card Number has to be numeric";
fld='couponstart1';
}else{
if ( $("#couponstart1").val() < 1 || $("#couponstart2").val() < 1 || $("#couponend1").val() < 1 || $("#couponend2").val() < 1 ) {
all_ok = false;
err_msg += "\n - Card Numbers are not correct";
fld='couponstart1';
}else if ($("#couponstart1").val().length != 4 || $("#couponstart2").val().length != 4 || $("#couponend1").val().length != 4 || $("#couponend2").val().length < 4){
all_ok = false;
err_msg += "\n - Card Numbers are not correct";
fld='couponstart1';
}
}
}
if (all_ok == false){
alert("The following errors have taken place" + err_msg);
setFocus(fld);
}
return all_ok;
}
It's sad that computers do what we ask, not what we want.
If your input is filled with 0000, and you compare < it with the number 1: ($("#couponstart1").val() < 1, javascript will try to parse the value 0000 to the operand type which accounts to 0. So it is right to say that 0 is lower than 1and you'll get your error message Card number are not correct.
Let's try a different approach:
<!-- html -->
<form>
<p>
<label for="couponstart1">Coupon Start Part 1</label>
<input type="text" maxlength="4" id="couponstart1" class="validateme" />
</p>
<p>
<label for="couponstart2">Coupon Start Part 2</label>
<input type="text" maxlength="4" id="couponstart2" class="validateme" />
</p>
<p>
<label for="couponend1">Coupon End Part 1</label>
<input type="text" maxlength="4" id="couponend1" class="validateme" />
</p>
<p>
<label for="couponstart2">Coupon Start End 2</label>
<input type="text" maxlength="4" id="couponend2" class="validateme" />
</p>
<button type="button" id="testit">Test</button>
</form>
and:
/* javascript/jQuery */
$(document).ready(function(){
$("#testit").click(function(){
var rgx = /\d{4}/;
var allgood = true;
$("input.validateme").each(function(){
if( ! rgx.test( $(this).val() ) ){
alert("Card number for " + ($('label[for="' + $(this).attr("id") + '"]').text()) + " is not correct!\nPlease use 4 digits all numbers!");
$(this).select();
allgood = false;
return false;
}
});
if(allgood) alert("All good!");
});
});
Basically we have a form with 4 inputs with "validateme" class.
With jQuery we loop over the inputs that have this class an run it against the regex /\d{4}/ which basically will test for a number(\d) exactly 4 digits long ({4}) meaning from 0000 to 9999. Note that we use the label of the input to identify which field is not right.
Otherwise, all good.
You can fiddle with this code here: http://jsfiddle.net/L8dcgcrs/1/
Assuming the four digits correspond to $("#couponstart1").val(), $("#couponstart2").val(), $("#couponend1").val(), $("#couponend2").val() then the problem is that in JavaScript there are two types of comparison operators.
In your first if statement you choose to evaluate if $("#coupontstart1").val() == '' || ...). In this case you are using an equality operator which converts the operands if they are not the same type. So using this operator 0 == '' returns true. Instead you could use the strict equality operator === in the following manner:
if ( $("#couponstart1").val() === '' || ...)
In which case 0 === '' would return false. This most likely will help you in the first if statement you write as entering 0000 would make all the conditions false and you would proceed to the else block.
Hope this helps!
Can anyone point me to some code to determine if a number in JavaScript is even or odd?
Use the below code:
function isOdd(num) { return num % 2;}
console.log("1 is " + isOdd(1));
console.log("2 is " + isOdd(2));
console.log("3 is " + isOdd(3));
console.log("4 is " + isOdd(4));
1 represents an odd number, while 0 represents an even number.
Use the bitwise AND operator.
function oddOrEven(x) {
return ( x & 1 ) ? "odd" : "even";
}
function checkNumber(argNumber) {
document.getElementById("result").innerHTML = "Number " + argNumber + " is " + oddOrEven(argNumber);
}
checkNumber(17);
<div id="result" style="font-size:150%;text-shadow: 1px 1px 2px #CE5937;" ></div>
If you don't want a string return value, but rather a boolean one, use this:
var isOdd = function(x) { return x & 1; };
var isEven = function(x) { return !( x & 1 ); };
You could do something like this:
function isEven(value){
if (value%2 == 0)
return true;
else
return false;
}
function isEven(x) { return (x%2)==0; }
function isOdd(x) { return !isEven(x); }
Do I have to make an array really large that has a lot of even numbers
No. Use modulus (%). It gives you the remainder of the two numbers you are dividing.
Ex. 2 % 2 = 0 because 2/2 = 1 with 0 remainder.
Ex2. 3 % 2 = 1 because 3/2 = 1 with 1 remainder.
Ex3. -7 % 2 = -1 because -7/2 = -3 with -1 remainder.
This means if you mod any number x by 2, you get either 0 or 1 or -1. 0 would mean it's even. Anything else would mean it's odd.
This can be solved with a small snippet of code:
function isEven(value) {
return !(value % 2)
}
Hope this helps :)
In ES6:
const isOdd = num => num % 2 == 1;
Like many languages, Javascript has a modulus operator %, that finds the remainder of division. If there is no remainder after division by 2, a number is even:
// this expression is true if "number" is even, false otherwise
(number % 2 == 0)
Similarly, if there is a remainder of 1 after division by 2, a number is odd:
// this expression is true if "number" is odd, false otherwise
(number % 2 == 1)
This is a very common idiom for testing for even integers.
With bitwise, codegolfing:
var isEven=n=>(n&1)?"odd":"even";
Use my extensions :
Number.prototype.isEven=function(){
return this % 2===0;
};
Number.prototype.isOdd=function(){
return !this.isEven();
}
then
var a=5;
a.isEven();
==False
a.isOdd();
==True
if you are not sure if it is a Number , test it by the following branching :
if(a.isOdd){
a.isOdd();
}
UPDATE :
if you would not use variable :
(5).isOdd()
Performance :
It turns out that Procedural paradigm is better than OOP paradigm .
By the way , i performed profiling in this FIDDLE . However , OOP way is still prettiest .
A simple function you can pass around. Uses the modulo operator %:
var is_even = function(x) {
return !(x % 2);
}
is_even(3)
false
is_even(6)
true
if (X % 2 === 0){
} else {
}
Replace X with your number (can come from a variable). The If statement runs when the number is even, the Else when it is odd.
If you just want to know if any given number is odd:
if (X % 2 !== 0){
}
Again, replace X with a number or variable.
<script>
function even_odd(){
var num = document.getElementById('number').value;
if ( num % 2){
document.getElementById('result').innerHTML = "Entered Number is Odd";
}
else{
document.getElementById('result').innerHTML = "Entered Number is Even";
}
}
</script>
</head>
<body>
<center>
<div id="error"></div>
<center>
<h2> Find Given Number is Even or Odd </h2>
<p>Enter a value</p>
<input type="text" id="number" />
<button onclick="even_odd();">Check</button><br />
<div id="result"><b></b></div>
</center>
</center>
</body>
Many people misunderstand the meaning of odd
isOdd("str") should be false.
Only an integer can be odd.
isOdd(1.223) and isOdd(-1.223) should be false.
A float is not an integer.
isOdd(0) should be false.
Zero is an even integer (https://en.wikipedia.org/wiki/Parity_of_zero).
isOdd(-1) should be true.
It's an odd integer.
Solution
function isOdd(n) {
// Must be a number
if (isNaN(n)) {
return false;
}
// Number must not be a float
if ((n % 1) !== 0) {
return false;
}
// Integer must not be equal to zero
if (n === 0) {
return false;
}
// Integer must be odd
if ((n % 2) !== 0) {
return true;
}
return false;
}
JS Fiddle (if needed): https://jsfiddle.net/9dzdv593/8/
1-liner
Javascript 1-liner solution. For those who don't care about readability.
const isOdd = n => !(isNaN(n) && ((n % 1) !== 0) && (n === 0)) && ((n % 2) !== 0) ? true : false;
You can use a for statement and a conditional to determine if a number or series of numbers is odd:
for (var i=1; i<=5; i++)
if (i%2 !== 0) {
console.log(i)
}
This will print every odd number between 1 and 5.
Just executed this one in Adobe Dreamweaver..it works perfectly.
i used if (isNaN(mynmb))
to check if the given Value is a number or not,
and i also used Math.abs(mynmb%2) to convert negative number to positive and calculate
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
</head>
<body bgcolor = "#FFFFCC">
<h3 align ="center"> ODD OR EVEN </h3><table cellspacing = "2" cellpadding = "5" bgcolor="palegreen">
<form name = formtwo>
<td align = "center">
<center><BR />Enter a number:
<input type=text id="enter" name=enter maxlength="10" />
<input type=button name = b3 value = "Click Here" onClick = compute() />
<b>is<b>
<input type=text id="outtxt" name=output size="5" value="" disabled /> </b></b></center><b><b>
<BR /><BR />
</b></b></td></form>
</table>
<script type='text/javascript'>
function compute()
{
var enter = document.getElementById("enter");
var outtxt = document.getElementById("outtxt");
var mynmb = enter.value;
if (isNaN(mynmb))
{
outtxt.value = "error !!!";
alert( 'please enter a valid number');
enter.focus();
return;
}
else
{
if ( mynmb%2 == 0 ) { outtxt.value = "Even"; }
if ( Math.abs(mynmb%2) == 1 ) { outtxt.value = "Odd"; }
}
}
</script>
</body>
</html>
When you need to test if some variable is odd, you should first test if it is integer. Also, notice that when you calculate remainder on negative number, the result will be negative (-3 % 2 === -1).
function isOdd(value) {
return typeof value === "number" && // value should be a number
isFinite(value) && // value should be finite
Math.floor(value) === value && // value should be integer
value % 2 !== 0; // value should not be even
}
If Number.isInteger is available, you may also simplify this code to:
function isOdd(value) {
return Number.isInteger(value) // value should be integer
value % 2 !== 0; // value should not be even
}
Note: here, we test value % 2 !== 0 instead of value % 2 === 1 is because of -3 % 2 === -1. If you don't want -1 pass this test, you may need to change this line.
Here are some test cases:
isOdd(); // false
isOdd("string"); // false
isOdd(Infinity); // false
isOdd(NaN); // false
isOdd(0); // false
isOdd(1.1); // false
isOdd("1"); // false
isOdd(1); // true
isOdd(-1); // true
Using % will help you to do this...
You can create couple of functions to do it for you... I prefer separte functions which are not attached to Number in Javascript like this which also checking if you passing number or not:
odd function:
var isOdd = function(num) {
return 'number'!==typeof num ? 'NaN' : !!(num % 2);
};
even function:
var isEven = function(num) {
return isOdd(num)==='NaN' ? isOdd(num) : !isOdd(num);
};
and call it like this:
isOdd(5); // true
isOdd(6); // false
isOdd(12); // false
isOdd(18); // false
isEven(18); // true
isEven('18'); // 'NaN'
isEven('17'); // 'NaN'
isOdd(null); // 'NaN'
isEven('100'); // true
A more functional approach in modern javascript:
const NUMBERS = "nul one two three four five six seven ocho nueve".split(" ")
const negate = f=> (...args)=> !f(...args)
const isOdd = n=> NUMBERS[n % 10].indexOf("e")!=-1
const isEven = negate(isOdd)
One liner in ES6 just because it's clean.
const isEven = (num) => num % 2 == 0;
Subtract 2 to it recursively until you reach either -1 or 0 (only works for positive integers obviously) :)
Every odd number when divided by two leaves remainder as 1 and every even number when divided by zero leaves a zero as remainder. Hence we can use this code
function checker(number) {
return number%2==0?even:odd;
}
How about this...
var num = 3 //instead get your value here
var aa = ["Even", "Odd"];
alert(aa[num % 2]);
This is what I did
//Array of numbers
var numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10,32,23,643,67,5876,6345,34,3453];
//Array of even numbers
var evenNumbers = [];
//Array of odd numbers
var oddNumbers = [];
function classifyNumbers(arr){
//go through the numbers one by one
for(var i=0; i<=arr.length-1; i++){
if (arr[i] % 2 == 0 ){
//Push the number to the evenNumbers array
evenNumbers.push(arr[i]);
} else {
//Push the number to the oddNumbers array
oddNumbers.push(arr[i]);
}
}
}
classifyNumbers(numbers);
console.log('Even numbers: ' + evenNumbers);
console.log('Odd numbers: ' + oddNumbers);
For some reason I had to make sure the length of the array is less by one. When I don't do that, I get "undefined" in the last element of the oddNumbers array.
I'd implement this to return a boolean:
function isOdd (n) {
return !!(n % 2);
// or ((n % 2) !== 0).
}
It'll work on both unsigned and signed numbers. When the modulus return -1 or 1 it'll get translated to true.
Non-modulus solution:
var is_finite = isFinite;
var is_nan = isNaN;
function isOdd (discriminant) {
if (is_nan(discriminant) && !is_finite(discriminant)) {
return false;
}
// Unsigned numbers
if (discriminant >= 0) {
while (discriminant >= 1) discriminant -= 2;
// Signed numbers
} else {
if (discriminant === -1) return true;
while (discriminant <= -1) discriminant += 2;
}
return !!discriminant;
}
By using ternary operator, you we can find the odd even numbers:
var num = 2;
result = (num % 2 == 0) ? 'even' : 'odd'
console.log(result);
Another example using the filter() method:
let even = arr.filter(val => {
return val % 2 === 0;
});
// even = [2,4,6]
So many answers here but i just have to mention one point.
Normally it's best to use the modulo operator like % 2 but you can also use the bitwise operator like & 1. They both would yield the same outcome. However their precedences are different. Say if you need a piece of code like
i%2 === p ? n : -n
it's just fine but with the bitwise operator you have to do it like
(i&1) === p ? n : -n
So there is that.
this works for arrays:
function evenOrOdd(numbers) {
const evenNumbers = [];
const oddNumbers = [];
numbers.forEach(number => {
if (number % 2 === 0) {
evenNumbers.push(number);
} else {
oddNumbers.push(number);
}
});
console.log("Even: " + evenNumbers + "\nOdd: " + oddNumbers);
}
evenOrOdd([1, 4, 9, 21, 41, 92]);
this should log out:
4,92
1,9,21,41
for just a number:
function evenOrOdd(number) {
if (number % 2 === 0) {
return "even";
}
return "odd";
}
console.log(evenOrOdd(4));
this should output even to the console
A Method to know if the number is odd
let numbers = [11, 20, 2, 5, 17, 10];
let n = numbers.filter((ele) => ele % 2 != 0);
console.log(n);