Change

format(number, format mask) - Converts a number into a formatted text.

https://ninoxdb.de/en/manual/calculations/reference-of-functions-and-language

This information is available at the link I posted above. If you go to that link and type command+f, your browser should popup a search box where you can enter the search term; format in this case.

Examples:
|  format(42.5, "0") => "42"
|  format(42.5, "000") => "042"
|  format(42.5, "000.00") => "042.50"
|  format(42.5, "0.00") => "42.50"
|  format(42.5, "#,##0.00") => "42.50"
|  format(1042.5, "#,##0.00") => "1,042.50"

I apologize, I misunderstood your original post. That would require some custom coding. I'm sure that's possible, but it would require a time commitment that I can't make right now. Maybe someone else will jump in and help.

Yep, this got stuck in my head. This works for whole numbers upto 9,999.

let arySingles := ["", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"];
let aryMultiples := ["", "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"];
let txtNumber := format(Number, "0");
let numDigits := length(txtNumber);
let numIndex := 0;
let txtNumberWord := "";
let i := numDigits;
while i > 0 do 
switch i do
case 4:
(
numIndex := number(substr(txtNumber, 0, 1));
txtNumberWord := item(arySingles, numIndex) + " Thousand"
)
case 3:
if substr(txtNumber, numDigits - i, 1) != 0 then
numIndex := number(substr(txtNumber, numDigits - i, 1));
txtNumberWord := txtNumberWord + " " + item(arySingles, numIndex) + " Hundred"
end
case 2:
if substr(txtNumber, numDigits - i, 1) != 0 then
if substr(txtNumber, 2, 1) = 1 then
numIndex := number(substr(txtNumber, numDigits - i, 2));
txtNumberWord := txtNumberWord + " " + item(arySingles, numIndex);
i := 0
else
numIndex := number(substr(txtNumber, numDigits - i, 1));
txtNumberWord := txtNumberWord + " " + item(aryMultiples, numIndex)
end
end
case 1:
if substr(txtNumber, numDigits - i, 1) != 0 then
numIndex := number(substr(txtNumber, numDigits - i, 1));
txtNumberWord := txtNumberWord + " " + item(arySingles, numIndex)
end
end;
i := i - 1
end
;
txtNumberWord

You can paste this to a formula field and plug the field name into this line where you see "Number".

let txtNumber := format(Number, "0");

It is currently only able to display numbers upto 9,999. When I get time, I'll try to update for larger numbers. Add a formula field to your table, copy the code and paste it to the formula field.

You will have to replace "Number" with the name of your field as I described at the end of my previous post.

“Number” in my case is a number field. I’m not sure if you can use a formula field in this situation. Try this, add a variable above “txtNumber” that is equal to the same formula you use in your “Net” formula. Don’t use the field “Net”, but instead, it’s formula.

Wait! The only place you replace “Number” is in the line,

let txtNumber := format(Number, “0”);

You only replace it in that one location. You are replacing the number() function and you can’t do that.

After you have restored the number() functions, try

let txtNumber := format(Net, “0”);

I have confirmed this works with a formula field. All you have to do is copy and paste.

let arySingles := ["", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"];
let aryMultiples := ["", "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"];
let txtNumber := format(Net, "0");
let numDigits := length(txtNumber);
let numIndex := 0;
let txtNumberWord := "";
let i := numDigits;
while i > 0 do 
switch i do
case 4:
(
numIndex := number(substr(txtNumber, 0, 1));
txtNumberWord := item(arySingles, numIndex) + " Thousand"
)
case 3:
if substr(txtNumber, numDigits - i, 1) != 0 then
numIndex := number(substr(txtNumber, numDigits - i, 1));
txtNumberWord := txtNumberWord + " " + item(arySingles, numIndex) + " Hundred"
end
case 2:
if substr(txtNumber, numDigits - i, 1) != 0 then
if substr(txtNumber, 2, 1) = 1 then
numIndex := number(substr(txtNumber, numDigits - i, 2));
txtNumberWord := txtNumberWord + " " + item(arySingles, numIndex);
i := 0
else
numIndex := number(substr(txtNumber, numDigits - i, 1));
txtNumberWord := txtNumberWord + " " + item(aryMultiples, numIndex)
end
end
case 1:
if substr(txtNumber, numDigits - i, 1) != 0 then
numIndex := number(substr(txtNumber, numDigits - i, 1));
txtNumberWord := txtNumberWord + " " + item(arySingles, numIndex)
end
end;
i := i - 1
end
;
txtNumberWord

Change the last line from,

txtNumberWord

to

txtNumberWord + " usd"

The large numbers definitely highlighted the shortcomings of my solution so I did some searching online. I found a very nice solution here,

http://www.techiedelight.com/c-program-convert-number-to-words/

and translated it to be used in Ninox. I had to make some changes so that it would work with western numbering and it will convert numbers upto 9,999,999,999.49. I had some fun with the user defined functions and commenting. The only change you have to make is here,

let numAmount := YourField;

Here is the code for the formula field,

let empty := "";
let x := [empty, "One ", "Two ", "Three ", "Four ", "Five ", "Six ", "Seven ", "Eight ", "Nine ", "Ten ", "Eleven ", "Twelve ", "Thirteen ", "Fourteen ", "Fifteen ", "Sixteen ", "Seventeen ", "Eighteen ", "Nineteen "];
let y := [empty, empty, "Twenty ", "Thirty ", "Forty ", "Fifty ", "Sixty ", "Seventy ", "Eighty ", "Ninety "];
"/*                       ";
" * User defined functions";
" */                      ";
function mod(dividend : number,divisor : number) do
dividend - divisor * floor(dividend / divisor)
end;
function makeCentsToMe(num : number) do
if contains(text(num), ".") then
let dpIndex := index(text(num), ".");
"and " + substr(text(num), dpIndex + 1, 2) + "/100 usd"
end
end;
function numberToWord(num : number,suffix : text) do
if floor(num) != 0 then
if num > 19 then
item(y, floor(num / 10)) + item(x, mod(num, 10)) + suffix
else
item(x, num) + suffix
end
else
empty
end
end;
"/*             ";
" * Main program";
" */            ";
"// Number to be converted to words";
let numAmount := Amount;
"// Test for number of digits";
if length(format(numAmount, "0")) < 11 then
"// String to store word representation of given number";
let NtW := "";
"// This handles Cents";
NtW := makeCentsToMe(numAmount);
"// This handles digits at Ones and Tens place";
NtW := numberToWord(mod(numAmount, 100), "") + NtW;
"// This handles digit at Hundreds place";
NtW := numberToWord(mod(numAmount / 100, 10), "Hundred ") + NtW;
"// This handles digits at One Thousands and Ten Thousands place";
NtW := numberToWord(mod(numAmount / 1000, 100), "Thousand ") + NtW;
"// This handles digit at Hundred Thousands place";
NtW := numberToWord(mod(numAmount / 100000, 10), "Hundred ") + NtW;
"// This handles digits at One Millions and Ten Millions place";
NtW := numberToWord(mod(numAmount / 1000000, 100), "Million ") + NtW;
"// This handles digit at Hundred Millions place";
NtW := numberToWord(mod(numAmount / 100000000, 10), "Hundred ") + NtW;
"// This handles digit at One Billions place";
NtW := numberToWord(mod(numAmount / 1000000000, 10), "Billion ") + NtW;
NtW
else
"Number is too large"
end

change the first line to

let Amount := Net;

let numAmount := Amount; is already in the formula. All you have to do is replace "Amount" with "Net".