We Keep Coding

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 can I make this to remember the last modification statement

How can I make this to remember the last modification statement.
Bacause this code is Always reinitialize the str variable.
But I have to make a loop what is add plus one "*" to my str. This is the reason why I want to "save" the previous statement.
Above I posted the test results.
function padIt(str, n) {
do {
if (n % 2 === 0) {
str + "*";
}
else {
str = "*" + str;
}
} while (n > 5)
return str;
}
I get this:
Test Passed: Value == '\'*a\''
Expected: '\'*a*\'', instead got: '\'a\''
Expected: '\'**a*\'', instead got: '\'*a\''
Expected: '\'**a**\'', instead got: '\'a\''

You are missing += in your if block. It should be str += "*";
function padIt(str, n) {
do {
if (n % 2 === 0) {
str += "*";
} else {
str = "*" + str;
}
} while (n > 5)
return str;
}

I think you really intended the padIt function to be recursive. If so, we can attempt such a solution by adding some padding on each side, in each recursive call. The base case occurs when the n counter reaches one, in which case we just return the cumulatively built padded string.
padIt = function(str, n) {
if (n === 1) return str;
if (n%2 === 0) {
return padIt(str + "*", n-1);
}
else {
return padIt("*" + str, n-1);
}
}
console.log(padIt("a", 5));

padStart() not working in IE11

I'm using angularjs 1.7.2 and kendo ui scheduler. All routes are working fine in almost all browser except when it comes to padStart() part in IE 11.
When padStart code is taken this error shows up
TypeError: Object doesn't support property or method 'padStart'
let ret = '#' + ((r << 16) + (g << 8) + b).toString(16).padStart(6, '0');
Is there a way we can handle this or an alternative way for implementing padStart

IE 11 is not supporting this function. Please take a look here: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart#Browser_compatibility
What you are looking for are polyfills to fill up missing functions of your browser.
The following code also taken from developer.mozilla.org will help you:
// https://github.com/behnammodi/polyfill/blob/master/string.polyfill.js
// https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart
if (!String.prototype.padStart) {
String.prototype.padStart = function padStart(targetLength,padString) {
targetLength = targetLength>>0; //truncate if number or convert non-number to 0;
padString = String((typeof padString !== 'undefined' ? padString : ' '));
if (this.length > targetLength) {
return String(this);
}
else {
targetLength = targetLength-this.length;
if (targetLength > padString.length) {
padString += padString.repeat(targetLength/padString.length); //append to original to ensure we are longer than needed
}
return padString.slice(0,targetLength) + String(this);
}
};
}
Edit: As mentioned in the comments, by #Plaute, the function repeat needs also to be polyfilled which can be found here: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/repeat
Or include this snippet:
if (!String.prototype.repeat) {
String.prototype.repeat = function(count) {
'use strict';
if (this == null) {
throw new TypeError('can\'t convert ' + this + ' to object');
}
var str = '' + this;
count = +count;
if (count != count) {
count = 0;
}
if (count < 0) {
throw new RangeError('repeat count must be non-negative');
}
if (count == Infinity) {
throw new RangeError('repeat count must be less than infinity');
}
count = Math.floor(count);
if (str.length == 0 || count == 0) {
return '';
}
// Ensuring count is a 31-bit integer allows us to heavily optimize the
// main part. But anyway, most current (August 2014) browsers can't handle
// strings 1 << 28 chars or longer, so:
if (str.length * count >= 1 << 28) {
throw new RangeError('repeat count must not overflow maximum string size');
}
var maxCount = str.length * count;
count = Math.floor(Math.log(count) / Math.log(2));
while (count) {
str += str;
count--;
}
str += str.substring(0, maxCount - str.length);
return str;
}
}
Alternatively, to work around the String.prototype.repeat dependency, use the following line:
padString += Array.apply(null, Array(targetLength)).map(function(){ return padString; }).join("");

how to count up to a specific number in javascript, fizzbuzz

I'm trying to solve a problem similar to the popular fizzbuzz or pingpong question. I already figured out how to write this portion of the code, however I also need to make a counter in which the user inputs their number and the page gives back 1 to the number they input, as well as the fizzbuzz/popcorn numbers. I think the code is something along the lines of
for (i = 1; i <= 20; i++) {...}
But I don't know how to combine this with my other code
var pingpong = function(number) {
if (number % 15 === 0) {
return 'pingpong';
} else if (number % 5 === 0) {
return 'pong';
} else if (number % 3 === 0) {
return 'ping';
} else {
return false;
}
};
Sorry, I know this is a super beginner question, but I'm just starting out and am having a hard time figuring out how everything works together.

You are close. To read more about JavaScripts loops.
var pingpong = function(number) {
if (number % 15 === 0) {
return 'pingpong';
} else if (number % 5 === 0) {
return 'pong';
} else if (number % 3 === 0) {
return 'ping';
} else {
return false;
}
};
for (var i = 1; i <= 20; i++) {
var num = pingpong(i);
// Just show it on the screen
document.body.innerHTML += i + ': ' + num + '<br />';
}

Convert column index into corresponding column letter

I need to convert a Google Spreadsheet column index into its corresponding letter value, for example, given a spreadsheet:
I need to do this (this function obviously does not exist, it's an example):
getColumnLetterByIndex(4); // this should return "D"
getColumnLetterByIndex(1); // this should return "A"
getColumnLetterByIndex(6); // this should return "F"
Now, I don't recall exactly if the index starts from 0 or from 1, anyway the concept should be clear.
I didn't find anything about this on gas documentation.. am I blind? Any idea?
Thank you

I wrote these a while back for various purposes (will return the double-letter column names for column numbers > 26):
function columnToLetter(column)
{
var temp, letter = '';
while (column > 0)
{
temp = (column - 1) % 26;
letter = String.fromCharCode(temp + 65) + letter;
column = (column - temp - 1) / 26;
}
return letter;
}
function letterToColumn(letter)
{
var column = 0, length = letter.length;
for (var i = 0; i < length; i++)
{
column += (letter.charCodeAt(i) - 64) * Math.pow(26, length - i - 1);
}
return column;
}

This works good
=REGEXEXTRACT(ADDRESS(ROW(); COLUMN()); "[A-Z]+")
even for columns beyond Z.
Simply replace COLUMN() with your column number. The value of ROW() doesn't matter.

No need to reinvent the wheel here, use the GAS range instead:
var column_index = 1; // your column to resolve
var ss = SpreadsheetApp.getActiveSpreadsheet();
var sheet = ss.getSheets()[0];
var range = sheet.getRange(1, column_index, 1, 1);
Logger.log(range.getA1Notation().match(/([A-Z]+)/)[0]); // Logs "A"

=SUBSTITUTE(ADDRESS(1,COLUMN(),4), "1", "")
This takes your cell, gets it's address as e.g. C1, and removes the "1".
How it works
COLUMN() gives the number of the column of the cell.
ADDRESS(1, ..., <format>) gives an address of a cell, in format speficied by <format> parameter. 4 means the address you know - e.g. C1.
The row doesn't matter here, so we use 1.
See ADDRESS docs
Finally, SUBSTITUTE(..., "1", "") replaces the 1 in the address C1, so you're left with the column letter.

This works on ranges A-Z
formula =char(64+column())
js String.fromCharCode(64+colno)
an google spreadsheet appscript code, based on #Gardener would be:
function columnName(index) {
var cname = String.fromCharCode(65 + ((index - 1) % 26));
if (index > 26)
cname = String.fromCharCode(64 + (index - 1) / 26) + cname;
return cname;
}

In javascript:
X = (n) => (a=Math.floor(n/26)) >= 0 ? X(a-1) + String.fromCharCode(65+(n%26)) : '';
console.assert (X(0) == 'A')
console.assert (X(25) == 'Z')
console.assert (X(26) == 'AA')
console.assert (X(51) == 'AZ')
console.assert (X(52) == 'BA')

Adding to #SauloAlessandre's answer, this will work for columns up from A-ZZ.
=if(column() >26,char(64+(column()-1)/26),) & char(65 + mod(column()-1,26))
I like the answers by #wronex and #Ondra Žižka. However, I really like the simplicity of #SauloAlessandre's answer.
So, I just added the obvious code to allow #SauloAlessandre's answer to work for wider spreadsheets.
As #Dave mentioned in his comment, it does help to have a programming background, particularly one in C where we added the hex value of 'A' to a number to get the nth letter of the alphabet as a standard pattern.
Answer updated to catch the error pointed out by #Sangbok Lee. Thank you!

I was looking for a solution in PHP. Maybe this will help someone.
<?php
$numberToLetter = function(int $number)
{
if ($number <= 0) return null;
$temp; $letter = '';
while ($number > 0) {
$temp = ($number - 1) % 26;
$letter = chr($temp + 65) . $letter;
$number = ($number - $temp - 1) / 26;
}
return $letter;
};
$letterToNumber = function(string $letters) {
$letters = strtoupper($letters);
$letters = preg_replace("/[^A-Z]/", '', $letters);
$column = 0;
$length = strlen($letters);
for ($i = 0; $i < $length; $i++) {
$column += (ord($letters[$i]) - 64) * pow(26, $length - $i - 1);
}
return $column;
};
var_dump($numberToLetter(-1));
var_dump($numberToLetter(26));
var_dump($numberToLetter(27));
var_dump($numberToLetter(30));
var_dump($letterToNumber('-1A!'));
var_dump($letterToNumber('A'));
var_dump($letterToNumber('B'));
var_dump($letterToNumber('Y'));
var_dump($letterToNumber('Z'));
var_dump($letterToNumber('AA'));
var_dump($letterToNumber('AB'));
Output:
NULL
string(1) "Z"
string(2) "AA"
string(2) "AD"
int(1)
int(1)
int(2)
int(25)
int(26)
int(27)
int(28)

Simple way through Google Sheet functions, A to Z.
=column(B2) : value is 2
=address(1, column(B2)) : value is $B$1
=mid(address(1, column(B2)),2,1) : value is B
It's a complicated way through Google Sheet functions, but it's also more than AA.
=mid(address(1, column(AB3)),2,len(address(1, column(AB3)))-3) : value is AB

I also was looking for a Python version here is mine which was tested on Python 3.6
def columnToLetter(column):
character = chr(ord('A') + column % 26)
remainder = column // 26
if column >= 26:
return columnToLetter(remainder-1) + character
else:
return character

A comment on my answer says you wanted a script function for it. All right, here we go:
function excelize(colNum) {
var order = 1, sub = 0, divTmp = colNum;
do {
divTmp -= order; sub += order; order *= 26;
divTmp = (divTmp - (divTmp % 26)) / 26;
} while(divTmp > 0);
var symbols = "0123456789abcdefghijklmnopqrstuvwxyz";
var tr = c => symbols[symbols.indexOf(c)+10];
return Number(colNum-sub).toString(26).split('').map(c=>tr(c)).join('');
}
This can handle any number JS can handle, I think.
Explanation:
Since this is not base26, we need to substract the base times order for each additional symbol ("digit"). So first we count the order of the resulting number, and at the same time count the number to substract. And then we convert it to base 26 and substract that, and then shift the symbols to A-Z instead of 0-P.
Anyway, this question is turning into a code golf :)

Java Apache POI
String columnLetter = CellReference.convertNumToColString(columnNumber);

This will cover you out as far as column AZ:
=iferror(if(match(A2,$A$1:$AZ$1,0)<27,char(64+(match(A2,$A$1:$AZ$1,0))),concatenate("A",char(38+(match(A2,$A$1:$AZ$1,0))))),"No match")

A function to convert a column index to letter combinations, recursively:
function lettersFromIndex(index, curResult, i) {
if (i == undefined) i = 11; //enough for Number.MAX_SAFE_INTEGER
if (curResult == undefined) curResult = "";
var factor = Math.floor(index / Math.pow(26, i)); //for the order of magnitude 26^i
if (factor > 0 && i > 0) {
curResult += String.fromCharCode(64 + factor);
curResult = lettersFromIndex(index - Math.pow(26, i) * factor, curResult, i - 1);
} else if (factor == 0 && i > 0) {
curResult = lettersFromIndex(index, curResult, i - 1);
} else {
curResult += String.fromCharCode(64 + index % 26);
}
return curResult;
}
function lettersFromIndex(index, curResult, i) {
if (i == undefined) i = 11; //enough for Number.MAX_SAFE_INTEGER
if (curResult == undefined) curResult = "";
var factor = Math.floor(index / Math.pow(26, i));
if (factor > 0 && i > 0) {
curResult += String.fromCharCode(64 + factor);
curResult = lettersFromIndex(index - Math.pow(26, i) * factor, curResult, i - 1);
} else if (factor == 0 && i > 0) {
curResult = lettersFromIndex(index, curResult, i - 1);
} else {
curResult += String.fromCharCode(64 + index % 26);
}
return curResult;
}
document.getElementById("result1").innerHTML = lettersFromIndex(32);
document.getElementById("result2").innerHTML = lettersFromIndex(6800);
document.getElementById("result3").innerHTML = lettersFromIndex(9007199254740991);
32 --> <span id="result1"></span><br> 6800 --> <span id="result2"></span><br> 9007199254740991 --> <span id="result3"></span>

In python, there is the gspread library
import gspread
column_letter = gspread.utils.rowcol_to_a1(1, <put your col number here>)[:-1]
If you cannot use python, I suggest looking the source code of rowcol_to_a1() in https://github.com/burnash/gspread/blob/master/gspread/utils.py

Here's a two liner which works beyond ZZ using recursion:
Python
def col_to_letter(n):
l = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
return col_to_letter((n-1)//26) + col_to_letter(n%26) if n > 26 else l[n-1]
Javascript
function colToLetter(n) {
l = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
return n > 26 ? colToLetter(Math.floor((n-1)/26)) + colToLetter(n%26) : l[n-1]
}

If you need a version directly in the sheet, here a solution:
For the colonne 4, we can use :
=Address(1,4)
I keep the row number to 1 for simplicty.
The above formula returns $D$1 which is not what you want.
By modifying the formula a little bit we can remove the dollar signs in the cell reference.
=Address(1,4,4)
Adding four as the third argument tells the formula that we are not looking for absolute cell reference.
Now the returns is : D1
So you only need to remove the 1 to get the colonne lettre if you need, for example with :
=Substitute(Address(1,4,4),"1","")
That returns D.

This is a way to convert column letters to column numbers.
=mmult(ArrayFormula(ifna(vlookup(substitute(mid(rept(" ",3-len(filter(A:A,A:A<>"")))&filter(A:A,A:A<>""),sequence(1,3),1)," ",""),{char(64+sequence(26)),sequence(26)},2,0),0)*{676,26,1}),sequence(3,1,1,0))
Screenshot of the Google Sheet

Don't use 26 radix. Like below.
const n2c = n => {
if (!n) return '';
// Column number to 26 radix. From 0 to p.
// Column number starts from 1. Subtract 1.
return [...(n-1).toString(26)]
// to ascii number
.map(c=>c.charCodeAt())
.map((c,i,arr)=> {
// last digit
if (i===arr.length-1) return c;
// 10 -> p
else if (arr.length - i > 2 && arr[i+1]===48) return c===49 ? null : c-2;
// 0 -> p
else if (c===48) return 112;
// a-1 -> 9
else if (c===97) return 57;
// Subtract 1 except last digit.
// Look at 10. This should be AA not BA.
else return c-1;
})
.filter(c=>c!==null)
// Convert with the ascii table. [0-9]->[A-J] and [a-p]->[K-Z]
.map(a=>a>96?a-22:a+17)
// to char
.map(a=>String.fromCharCode(a))
.join('');
};
const table = document.createElement('table');
table.border = 1;
table.cellPadding = 3;
for(let i=0, row; i<1380; i++) {
if (i%5===0) row = table.insertRow();
row.insertCell().textContent = i;
row.insertCell().textContent = n2c(i);
}
document.body.append(table);
td:nth-child(odd) { background: gray; color: white; }
td:nth-child(even) { background: silver; }

Simple typescript functional approach
const integerToColumn = (integer: number): string => {
const base26 = (x: number): string =>
x < 26
? String.fromCharCode(65 + x)
: base26((x / 26) - 1) + String.fromCharCode(65 + x % 26)
return base26(integer)
}
console.log(integerToColumn(0)) // "A"
console.log(integerToColumn(1)) // "B"
console.log(integerToColumn(2)) // "C"

Here is a general version written in Scala. It's for a column index start at 0 (it's simple to modify for an index start at 1):
def indexToColumnBase(n: Int, base: Int): String = {
require(n >= 0, s"Index is non-negative, n = $n")
require(2 <= base && base <= 26, s"Base in range 2...26, base = $base")
def digitFromZeroToLetter(n: BigInt): String =
('A' + n.toInt).toChar.toString
def digitFromOneToLetter(n: BigInt): String =
('A' - 1 + n.toInt).toChar.toString
def lhsConvert(n: Int): String = {
val q0: Int = n / base
val r0: Int = n % base
val q1 = if (r0 == 0) (n - base) / base else q0
val r1 = if (r0 == 0) base else r0
if (q1 == 0)
digitFromOneToLetter(r1)
else
lhsConvert(q1) + digitFromOneToLetter(r1)
}
val q: Int = n / base
val r: Int = n % base
if (q == 0)
digitFromZeroToLetter(r)
else
lhsConvert(q) + digitFromZeroToLetter(r)
}
def indexToColumnAtoZ(n: Int): String = {
val AtoZBase = 26
indexToColumnBase(n, AtoZBase)
}

In PowerShell:
function convert-IndexToColumn
{
Param
(
[Parameter(Mandatory)]
[int]$col
)
"$(if($col -gt 26){[char][int][math]::Floor(64+($col-1)/26)})$([char](65 + (($col-1) % 26)))"
}

Here is a 0-indexed JavaScript function without a maximum value, as it uses a while-loop:
function indexesToA1Notation(row, col) {
const letterCount = 'Z'.charCodeAt() - 'A'.charCodeAt() + 1;
row += 1
let colName = ''
while (col >= 0) {
let rem = col % letterCount
colName = String.fromCharCode('A'.charCodeAt() + rem)
col -= rem
col /= letterCount
}
return `${colName}${row}`
}
//Test runs:
console.log(indexesToA1Notation(0,0)) //A1
console.log(indexesToA1Notation(37,9)) //J38
console.log(indexesToA1Notation(5,747)) //ABT6
I wrote it for a web-app, so I'm not 100% sure it works in Google Apps Script, but it is normal JavaScript, so I assume it will.
For some reason I cant get the snippet to show its output, but you can copy the code to some online playground if you like

Here's a zero-indexed version (in Python):
letters = []
while column >= 0:
letters.append(string.ascii_uppercase[column % 26])
column = column // 26 - 1
return ''.join(reversed(letters))