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Javascript numbers - split/slice Prime

I am completely new to Javascript and trying to solve a simple problem now for more than two weeks and still not getting it(please help).
TASK ::::
Read a 4 digit Number e.g. 5678
Write a function
Split/separate the numbers and than build (5678, 567, 56, 5), than check if the numbers(5678, 567, 56, 5) are Prime numbers.
Give in Console/Result if 5678 a prime number or not, 567 a prime number or not and so on.
Check "if all numbers are Prime" than show result "All prime" if not show result "Not all prime".
Trying to solve the problem with (if else) but not really getting it, because i know very less about Javascript (arrays, string, split, slice) yet.
please help me understand. Thanks.
var a = 123456789;
var b = a.toString().length; //<<--->> ANTWORT: 9
document.write('ANTWORT: ',a );
for (i=0; i<b; i++) {
var x = a.toString().slice(0, -i);
document.write(x, ",");
}
function isPrime{
for(var i = 2; i < a; i++);
if(num % i === 0) return false;
return num > 1;
}

//integer is a string at the moment
integer = prompt("Enter a integer: ");
//initialize array for dictionary
dictArray = [];
stuff = document.getElementById("stuff");
//loop through all values of the string
for (var i = integer.length; i > 0; i--)
{
//take a substring from 0 to the ith char and turn it into an int
num = parseInt(integer.substring(0, i));
//add a dictionary to the array the tells what the number is
//and if it was prime or not as a bool
dictArray.push({"num": num, "prime": isprime(num)});
(dictArray[integer.length - i]["prime"]) ? stuff.innerHTML += "<br>" + num + " is prime." : stuff.innerHTML += "<br>" + num + " is not prime.";
}
function isprime(num)
{
if (num <= 3) return num >= 1;
if ((num % 2 === 0) || (num % 3 === 0)) return false;
let count = 5;
while (Math.pow(count, 2) <= num) {
if (num % count === 0 || num % (count + 2) === 0) return false;
count += 6;
}
return true;
}
//print the array
(dictArray.find(x => !x.prime) == undefined) ? stuff.innerHTML += "<br>All prime!" : stuff.innerHTML += "<br>Not all prime!";
//console.log(dictArray);
<div id="stuff">
</div>

How to do a combined pyramid with stars and numbers together in a single pattern

Need a pyramid with numbers and stars and there are no examples to review.
*
* 0 *
* 1 0 1 *
* 0 1 0 1 0 *
* 1 0 1 0 1 0 1 *
Can I get some assistance with this?
This is what I have, but I can't figure out how to get the the numbers in there..
function displayPyramid(n) {
for (let i = 0; i < n; i++) {
let str = ''
let count = 0
// Spacing before each row
for (let j = 1; j < n - i; j++) {
str = str + ' '
}
// Row worth of data
const max = 2 * i + 1
for (let k = 1; k <= max; k++) {
str = str + '*'
}
console.log(str);
}
}
displayPyramid(5)

Here's the thought process to get the expected result:
All rows with even indices are called 'even rows' (0, 2, 4, 6, …)
All rows with uneven indices are called 'uneven rows' (1, 3, 5, 7, …)
All stars are either the first or the last column in any row
For all even rows, the even indices display 0s, the uneven indices display 1s
For all uneven rows, the uneven indices display 0s, the even indices display 1s
I added a function called getSymbol that performs the last three checks and returns the appropriate symbol. I also renamed you iterator variables for better readability.
function getSymbol(column, max, row) {
if (column === 0 || column === max - 1) { // if it is the first or last index within the row
return '*'
}
if (row % 2 === 0) { // if it is an uneven row
if (column % 2 === 0) { // if it is an even column
return '0';
}
return '1' // if it is an uneven column within that row
}
// if it's not an even row, it's an uneven row
if (column % 2 === 0) { // if it is an even column within that row
return '1';
}
return '0' // if it is an uneven column within that row
}
function displayPyramid(n) {
for (let row = 0; row < n; row++) {
let str = ''
let count = 0
// Spacing before each row
for (let spaces = 1; spaces < n - row; spaces++) {
str = str + ' '
}
// Row worth of data
const max = 2 * row + 1
for (let column = 0; column < max; column++) {
str = str + getSymbol(column, max, row)
}
console.log(str);
}
}
displayPyramid(5)

function displayPyramid(n) {
for (let i = 0; i < n; i++) {
let str = "";
let count = 0;
// Spacing before each row
let padCount = 2 * n - 2 * i - 1;
for (let j = 1; j < padCount; j++) {
str = str + " ";
}
// Row worth of data
const max = 2 * i + 1;
for (let k = 1; k <= max; k++) {
if (k === 1 || k === max) {
str = str + "* ";
} else {
str += (k % 2).toString() + " ";
}
}
console.log(str);
}
}
displayPyramid(5);

Let's first try to write a function for just one line of the pyramid, any line. We'll ignore the whitespace for now too.
Every line of the pyramid seems to follow a few simple rules:
All lines start with a *.
The 'inner sequence' is just alternating 1 and 0 a certain number of times.
There are 2*n - 1 digits in the inner sequence of each line, except the first.
Even lines (0, 2, 4, ...) begin the inner sequence with a 1.
Odd lines (1, 3, 5, ...) begin the inner sequence with a 0.
All lines end with an additional *, except the first.
In the rules above, everything can be determined from n. So our function only needs one argument:
function GetLine(n) {
// ...
}
We can also determine whether a line is even or odd, and how many characters the inner sequence has:
EvenLine = (n % 2 == 0);
Count = 2*n - 1;
Building the sequence between the two stars can be done with a simple for loop.
Putting that all together, we can build the following function.
function GetLine(n) {
// Create a string for the line that starts with a star
var Line = "*";
// Determine whether the line is even or odd
var EvenLine = (n % 2 == 0);
// Calculate the total number of ones and zeros in the line
var Count = (2 * n - 1);
// We need a variable to store whether the next character should be a one or zero
var One = EvenLine ? true : false; // Even lines start with 1, Odd starts with 0
// Repeat 'Count' times, alternating between ones and zeros
for (var i=0; i<Count; i++)
{
Line += One ? "1" : "0";
One = !One; // Toggle the bool value to alternate on the next iteration
}
// Only add a tail star if we're not on the first line
if (n > 0) Line += "*";
return Line;
}
When we call GetLine() with some consecutive numbers we can see that the pattern is mostly there:
console.log(GetLine(0));
console.log(GetLine(1));
console.log(GetLine(2));
*
*0*
*101*
Now all that needs to be done is insert the whitespace in two steps:
A space between each character.
The leading space to align the pyramid.
function printPyramid(s) {
for (var n=0; n<s; n++) {
// Get the line for n
var line = GetLine(n);
// This one-liner can be used to insert whitespace between each character
// split('') will explode the characters into an array
// join(' ') will turn the array back into a string with ' ' inbetween each element
line = line.split('').join(' ');
// Then we just add the necessary whitespace for alignment
// We need 2 * (s - n - 1) spaces in front of each line.
line = " ".repeat(s - n - 1) + line;
// Print the line
console.log(line);
}
}
Finally,
printPyramid(5);
*
* 0 *
* 1 0 1 *
* 0 1 0 1 0 *
* 1 0 1 0 1 0 1 *

const rows = 5;
const columns = 2 * rows - 1;
// create an array filled with " "
const array = Array(rows).fill().map( () => Array(columns).fill(" "));
// if r and c are the row and column indices of the array, then
// in each row place two askerisks where c = rows -1 ± r (if r is 0, there's only one asterisk)
// and alternating 1's (if c is odd) and 0's (if c is even) for all elements between columns rows - 1 - r and rows - 1 + r
const pyramid = array.map(function(row, r) {
return row.map(function(v, c) {
return c === rows - 1 - r || c === rows - 1 + r ? "*" : c > rows - 1 - r && c < rows - 1 + r ? c % 2 === 1 ? "1" : "0" : v;
});
});
// const pyramid = array.map( (row, r) => row.map( (v, c) => c === rows - 1 - r || c === rows -1 + r ? "*" : c > rows - 1 - r && c < rows - 1 + r ? c % 2 === 1 ? "1" : "0" : v))

function displayPyramid(n)
{
for (let i = 0; i < n; i++)
{
let str = ''
let count = 0
// Spacing before each row
for (let j = 1; j < n - i; j++)
{
str = str + ' '
}
// Row worth of data
const max = 2 * i + 1
str = str + '*'
const innerChars = ['0', '1'];
for (let k = 1; k < max; k++)
{
if (k === max - 1)
str += '*'
else
str += innerChars[(i % 2 + k) % 2];
}
console.log(str);
}
}
displayPyramid(5)
The main observations to make:
The first row we have no innner char
The second row begins with inner char 0
The third row begins with inner char 1 then alternates
The forth row begins with inner char 0 then alternate
We notice a pattern, even rows (or odd if zero based) begin with 0, and vice versa. Alternating pattern should hint to an array that we loop around.

I wasn't sure if you needed the spaces, or if you wanted to change the characters in the tree, so I added them as parameters.
If you want to have more than 2 elements in the array, and have the first element in the center, you can shift the array on every row, instead of reversing it.
function displayPyramid(n, charArray, withSpace) {
let charArrayOrdered = charArray
for (let i = 0; i < n; i++) {
let str = ''
let count = 0
// Spacing before each row
let withSpaceChar = ''
if(withSpace) {
withSpaceChar=' '
}
for (let j = 1; j < n - i; j++) {
str = str + ' ' + withSpaceChar
}
// Row worth of data
const max = 2 * i
charArrayOrdered = charArrayOrdered.reverse()
let charState = 0
for (let k = 0; k <= max; k++) {
let char = '*'
if(k!=0 && k !=max) {
char=charArrayOrdered[charState]
charState++
}
if(charState==charArrayOrdered.length){
charState=0
}
if(k!=max && withSpace) {
char = char+' '
}
str = str + char
}
console.log(str);
}
}
displayPyramid(5,[0,1], true)

You could do something like this but if in the middle is 0 or 1 will depend on the number you provided.
function build(n) {
let grid = ''
let total = n * 2;
for (let i = 0; i < n - 1; i++) {
let row = [];
let start = total / 2 - i - 1
let end = total / 2 + i
if (!i) grid += ' '.repeat(start + 1) + '*' + '\n'
for (let j = 0; j < total; j++) {
if (j < start) row.push(' ');
else if (j > end) row.push(' ');
else {
if (j == start) row.push('*');
if (j == end) row.push('*');
else row.push(j % 2 == 0 ? 1 : 0)
}
}
grid += row.join('') + '\n'
}
return grid;
}
console.log(build(3))
console.log(build(6))
console.log(build(9))

Inserting into a number string

Have the function DashInsert(num) insert dashes ('-') between each two odd numbers in num. For example: if num is 454793 the output should be 4547-9-3. Don't count zero as an odd number.
Here is my code (not working). When I run it, I get the same response as an infinite loop where I have to kill the page but I can't see why. I know there are ways to do this by keeping it as a string but now I'm wondering why my way isn't working. Thanks...
function DashInsert(num) {
num = num.split("");
for (i = 1; i < num.length; i++) {
if (num[i - 1] % 2 != 0 && num[i] % 2 != 0) {
num.splice(i, 0, "-");
}
}
num = num.join("");
return num;
}

Using num.splice you are inserting new entries into the array, therefor increasing its length – and that makes the value of i “running behind” the increasing length of the array, so the break condition is never met.
And apart from that, on the next iteration after inserting a -, num[i-1] will be that - character, and therefor you are practically trying to check if '-' % 2 != 0 … that makes little sense as well.
So, when you insert a - into the array, you have to increase i by one as well – that will a) account for the length of the array having increased by one, and also it will check the next digit after the - on the next iteration:
function DashInsert(num) {
num = num.split("");
for (i = 1; i < num.length; i++) {
if (num[i - 1] % 2 != 0 && num[i] % 2 != 0) {
num.splice(i, 0, "-");
i++; // <- this is the IMPORTANT part!
}
}
num = num.join("");
return num;
}
alert(DashInsert("454793"));
http://jsfiddle.net/37wA9/

Once you insert a dash -, the if statement is checking this '-'%2 != 0 which is always true and thus inserts another dash, ad infinitum.
Here's one way to do it with replace using a regex and function:
function DashInsert(n) {
var f = function(m,i,s) { return m&s[i+1]&1 ? m+'-' : m; };
return String(n).replace(/\d/g,f);
}
DashInsert(454793) // "4547-9-3"

When you are adding a dash, this dash will be processed as a number on the next iteration. You need to forward one step.
function DashInsert(num) {
var num = num.split("");
for (var i = 1; i < num.length; i++) {
if ((num[i - 1] % 2 != 0) && (num[i] % 2 != 0)) {
num.splice(i, 0, "-");
i++; // This is the only thing that needs changing
}
}
num = num.join("");
return num;
}

It's because there are cases when you use the % operator on dash '-' itself, e.g. right after you splice a dash into the array.
You can correct this behavior by using a clone array.
function DashInsert(num) {
num = num.split("");
var clone = num.slice(0);
var offset = 0;
for (i = 1; i < num.length; i++) {
if (num[i - 1] % 2 != 0 && num[i] % 2 != 0) {
clone.splice(i + offset, 0, "-");
offset++;
}
}
return clone.join("");
}
alert(DashInsert("45739"));
Output: 45-7-3-9
Demo: http://jsfiddle.net/262Bf/

To complement the great answers already given, I would like to share an alternative implementation, that doesn't modify arrays in-place:
function DashInsert(num) {
var characters = num.split("");
var numbers = characters.map(function(chr) {
return parseInt(chr, 10);
});
var withDashes = numbers.reduce(function(result, current) {
var lastNumber = result[result.length - 1];
if(lastNumber == null || current % 2 === 0 || lastNumber % 2 === 0) {
return result.concat(current);
} else {
return result.concat("-", current);
}
}, []);
return withDashes.join("");
}
It's longer, but IMHO reveals the intention better, and avoids the original issue.

JavaScript: Get the second digit from a number?

I have a number assigned to a variable, like that:
var myVar = 1234;
Now I want to get the second digit (2 in this case) from that number without converting it to a string first. Is that possible?

So you want to get the second digit from the decimal writing of a number.
The simplest and most logical solution is to convert it to a string :
var digit = (''+myVar)[1];
or
var digit = myVar.toString()[1];
If you don't want to do it the easy way, or if you want a more efficient solution, you can do that :
var l = Math.pow(10, Math.floor(Math.log(myVar)/Math.log(10))-1);
var b = Math.floor(myVar/l);
var digit = b-Math.floor(b/10)*10;
Demonstration
For people interested in performances, I made a jsperf. For random numbers using the log as I do is by far the fastest solution.

1st digit of number from right → number % 10 = Math.floor((number / 1) % 10)
1234 % 10; // 4
Math.floor((1234 / 1) % 10); // 4
2nd digit of number from right → Math.floor((number / 10) % 10)
Math.floor((1234 / 10) % 10); // 3
3rd digit of number from right → Math.floor((number / 100) % 10)
Math.floor((1234 / 100) % 10); // 2
nth digit of number from right → Math.floor((number / 10^n-1) % 10)
function getDigit(number, n) {
return Math.floor((number / Math.pow(10, n - 1)) % 10);
}
number of digits in a number → Math.max(Math.floor(Math.log10(Math.abs(number))), 0) + 1 Credit to: https://stackoverflow.com/a/28203456/6917157
function getDigitCount(number) {
return Math.max(Math.floor(Math.log10(Math.abs(number))), 0) + 1;
}
nth digit of number from left or right
function getDigit(number, n, fromLeft) {
const location = fromLeft ? getDigitCount(number) + 1 - n : n;
return Math.floor((number / Math.pow(10, location - 1)) % 10);
}

Get rid of the trailing digits by dividing the number with 10 till the number is less than 100, in a loop. Then perform a modulo with 10 to get the second digit.
if (x > 9) {
while (x > 99) {
x = (x / 10) | 0; // Use bitwise '|' operator to force integer result.
}
secondDigit = x % 10;
}
else {
// Handle the cases where x has only one digit.
}

A "number" is one thing.
The representation of that number (e.g. the base-10 string "1234") is another thing.
If you want a particular digit in a decimal string ... then your best bet is to get it from a string :)
Q: You're aware that there are pitfalls with integer arithmetic in Javascript, correct?
Q: Why is it so important to not use a string? Is this a homework assignment? An interview question?

You know, I get that the question asks for how to do it without a number, but the title "JavaScript: Get the second digit from a number?" means a lot of people will find this answer when looking for a way to get a specific digit, period.
I'm not bashing the original question asker, I'm sure he/she had their reasons, but from a search practicality standpoint I think it's worth adding an answer here that does convert the number to a string and back because, if nothing else, it's a much more terse and easy to understand way of going about it.
let digit = Number((n).toString().split('').slice(1,1))
// e.g.
let digit = Number((1234).toString().split('').slice(1,1)) // outputs 2
Getting the digit without the string conversion is great, but when you're trying to write clear and concise code that other people and future you can look at really quick and fully understand, I think a quick string conversion one liner is a better way of doing it.

function getNthDigit(val, n){
//Remove all digits larger than nth
var modVal = val % Math.pow(10,n);
//Remove all digits less than nth
return Math.floor(modVal / Math.pow(10,n-1));
}
// tests
[
0,
1,
123,
123456789,
0.1,
0.001
].map(v =>
console.log([
getNthDigit(v, 1),
getNthDigit(v, 2),
getNthDigit(v, 3)
]
)
);

This is how I would do with recursion
function getDigits(n, arr=[]) {
arr.push(n % 10)
if (n < 10) {
return arr.reverse()
}
return getDigits(Math.floor(n/10),arr)
}
const arr = getDigits(myVar)
console.log(arr[2])

I don’t know why you need this logic, but following logic will get you the second number
<script type="text/javascript">
var myVal = 58445456;
var var1 = new Number(myVal.toPrecision(1));
var var2 = new Number(myVal.toPrecision(2));
var rem;
rem = var1 - var2;
var multi = 0.1;
var oldvalue;
while (rem > 10) {
oldvalue = rem;
rem = rem * multi;
rem = rem.toFixed();
}
alert(10-rem);
</script>

function getDigit(number, indexFromRight) {
var maxNumber = 9
for (var i = 0; i < indexFromRight - 2; i++) {
maxNumber = maxNumber * 10 + 9
}
if (number > maxNumber) {
number = number / Math.pow(10, indexFromRight - 1) | 0
return number % 10
} else
return 0
}

Just a simple idea to get back any charter from a number as a string or int:
const myVar = 1234;
String(myVar).charAt(1)
//"2"
parseInt(String(myVar).charAt(1))
//2

you can use this function
index = 0 will give you the first digit from the right (the ones)
index = 1 will give you the second digit from the right (the tens)
and so on
const getDigit = (num, index) => {
if(index === 0) {
return num % 10;
}
let result = undefined;
for(let i = 1; i <= index; i++) {
num -= num % 10;
num /= 10;
result = num % 10;
}
return result;
}
for Example:
getDigit(125, 0) // returns 5
gitDigit(125, 1) // returns 2
gitDigit(125, 2) // returns 1
gitDigit(125, 3) // returns 0

function left(num) {
let newarr = [];
let numstring = num.split('[a-z]').join();
//return numstring;
const regex = /[0-9]/g;
const found = numstring.match(regex);
// return found;
for(i=0; i<found.length; i++){
return found[i];
}
}
//}
console.log(left("TrAdE2W1n95!"))

function getNthDigit(n, number){
return ((number % Math.pow(10,n)) - (number % Math.pow(10,n-1))) / Math.pow(10,n-1);
}
Explanation (Number: 987654321, n: 5):
a = (number % Math.pow(10,n)) - Remove digits above => 54321
b = (number % Math.pow(10,n-1)) - Extract digits below => 4321
a - b => 50000
(a - b) / 10^(5-1) = (a - b) / 10000 => 5

var newVar = myVar;
while (newVar > 100) {
newVar /= 10;
}
if (newVar > 0 && newVar < 10) {
newVar = newVar;
}
else if (newVar >= 10 && newVar < 20) {
newVar -= 10;
}
else if (newVar >= 20 && newVar < 30) {
newVar -= 20;
}
else if (newVar >= 30 && newVar < 40) {
newVar -= 30;
}
else if (newVar >= 40 && newVar < 50) {
newVar -= 40;
}
else if (newVar >= 50 && newVar < 60) {
newVar -= 50;
}
else if (newVar >= 60 && newVar < 70) {
newVar -= 60;
}
else if (newVar >= 70 && newVar < 80) {
newVar -= 70;
}
else if (newVar >= 80 && newVar < 90) {
newVar -= 80;
}
else if (newVar >= 90 && newVar < 100) {
newVar -= 90;
}
else {
newVar = 0;
}
var secondDigit = Math.floor(newVar);
That's how I'd do it :)
And here's a JSFiddle showing it works :) http://jsfiddle.net/Cuytd/
This is also assuming that your original number is always greater than 9... If it's not always greater than 9 then I guess you wouldn't be asking this question ;)

Convert integer to alpha ordered list equivalent

I need to a function to convert an integer to the equivalent alpha ordered list index. For example:
1 = a
2 = b
.
.
.
26 = z
27 = aa
28 = ab
.
.
etc.
Currently I have the following which almost works but there's a small logic error somewhere that makes it not quite get it right (it goes ax, ay, bz, ba, bb, bc...):
function intToAlpha( int ) {
var asciiStart = 97,
alphaMax = 26,
asciiCode,
char,
alpha = '',
place,
num,
i;
for ( i = 0; Math.pow(alphaMax, i) < int; i++ ) {
place = Math.pow(alphaMax, i);
num = Math.floor( ( int / place ) % alphaMax);
asciiCode = ( num == 0 ? alphaMax : num ) + asciiStart - 1;
char = String.fromCharCode(asciiCode);
alpha = char + alpha;
}
return alpha;
}
for (i = 1; i < 300; i++) {
console.log( i + ': ' + intToAlpha(i) );
}

This function is used in NVu/Kompozer/SeaMonkey Composer, with a small tweak to generate lower case directly:
function ConvertArabicToLetters(num)
{
var letters = "";
while (num > 0) {
num--;
letters = String.fromCharCode(97 + (num % 26)) + letters;
num = Math.floor(num / 26);
}
return letters;
}

You need to make sure that you use the correct value when taking the mod.
function intToAlpha( int ) {
var asciiStart = 97,
alphaMax = 26,
asciiCode,
char,
alpha = "";
while(int > 0) {
char = String.fromCharCode(asciiStart + ((int-1) % alphaMax));
alpha = char + alpha;
int = Math.floor((int-1)/26);
}
return alpha;
}

A while back I needed the same thing in SQL, so I asked (and answered) the question Multi-base conversion - using all combinations for URL shortener.
The thing that is making it complicated is that it's not a straight base conversion, as there is no character representing the zero digit.
I converted the SQL function into Javascript:
function tinyEncode(id) {
var code, value, adder;
var chars = 'abcdefghijklmnopqrstuvwxyz';
if (id <= chars.length) {
code = chars.substr(id - 1, 1);
} else {
id--;
value = chars.length;
adder = 0;
while (id >= value * (chars.length + 1) + adder) {
adder += value;
value *= chars.length;
}
code = chars.substr(Math.floor((id - adder) / value) - 1, 1);
id = (id - adder) % value;
while (value > 1) {
value = Math.floor(value / chars.length);
code += chars.substr(Math.floor(id / value), 1);
id = id % value;
}
}
return code;
}
Demo: http://jsfiddle.net/Guffa/mstBe/