Convert Number to Words - First, Second, Third and So On

Convert number to words - first, second, third and so on

This seems like a simpler approach, a nice recursive algorithm

CREATE FUNCTION fnIntegerToWords(@Number as BIGINT) 
    RETURNS VARCHAR(1024)
AS

BEGIN
      DECLARE @Below20 TABLE (ID int identity(0,1), Word varchar(32))
      DECLARE @Below100 TABLE (ID int identity(2,1), Word varchar(32))
      INSERT @Below20 (Word) VALUES 
                        ( 'Zero'), ('One'),( 'Two' ), ( 'Three'),
                        ( 'Four' ), ( 'Five' ), ( 'Six' ), ( 'Seven' ),
                        ( 'Eight'), ( 'Nine'), ( 'Ten'), ( 'Eleven' ),
                        ( 'Twelve' ), ( 'Thirteen' ), ( 'Fourteen'),
                        ( 'Fifteen' ), ('Sixteen' ), ( 'Seventeen'),
                        ('Eighteen' ), ( 'Nineteen' ) 

       INSERT @Below100 VALUES ('Twenty'), ('Thirty'),('Forty'), ('Fifty'),
                               ('Sixty'), ('Seventy'), ('Eighty'), ('Ninety')

    declare @belowHundred as varchar(126) 

    if @Number > 99 begin
        select @belowHundred = dbo.fnIntegerToWords( @Number % 100)
    end

    DECLARE @English varchar(1024) = 

    (

      SELECT Case 
        WHEN @Number = 0 THEN  ''

        WHEN @Number BETWEEN 1 AND 19 
          THEN (SELECT Word FROM @Below20 WHERE ID=@Number)

       WHEN @Number BETWEEN 20 AND 99   
         THEN  (SELECT Word FROM @Below100 WHERE ID=@Number/10)+ '-' +
               dbo.fnIntegerToWords( @Number % 10) 

       WHEN @Number BETWEEN 100 AND 999   
         THEN  (dbo.fnIntegerToWords( @Number / 100)) +' Hundred '+
             Case WHEN @belowHundred <> '' THEN 'and ' + @belowHundred else @belowHundred end 

       WHEN @Number BETWEEN 1000 AND 999999   
         THEN  (dbo.fnIntegerToWords( @Number / 1000))+' Thousand '+
             dbo.fnIntegerToWords( @Number % 1000)  

       WHEN @Number BETWEEN 1000000 AND 999999999   
         THEN  (dbo.fnIntegerToWords( @Number / 1000000))+' Million '+
             dbo.fnIntegerToWords( @Number % 1000000) 

       WHEN @Number BETWEEN 1000000000 AND 999999999999   
         THEN  (dbo.fnIntegerToWords( @Number / 1000000000))+' Billion '+
             dbo.fnIntegerToWords( @Number % 1000000000) 

            ELSE ' INVALID INPUT' END
    )

    SELECT @English = RTRIM(@English)

    SELECT @English = RTRIM(LEFT(@English,len(@English)-1))
        WHERE RIGHT(@English,1)='-'

    RETURN (@English)

END 

.NET or 3rd party library to convert number into first , second , third , etc...

Humanizer is a great library for this: Humanizr

Install-Package Humanizer

With Humanizer, you'd do:

int number = 5;
string ordinal = number.ToOrdinalWords()

PHP: increment counter function using words (i.e. First, Second, Third, etc.. )

There is a class from PEAR package can do that:

<?php

// include class
include("Numbers/Words.php");

// create object
$nw = new Numbers_Words();

// convert to string
echo "600 in words is " . $nw->toWords(600);

?>

Source.

How to convert number into word

Can you try this one.

function createFullWordOrdinal($number)
{
    $ord1     = array(1 => "first", 2 => "second", 3 => "third", 5 => "fifth", 8 => "eight", 9 => "ninth", 11 => "eleventh", 12 => "twelfth", 13 => "thirteenth", 14 => "fourteenth", 15 => "fifteenth", 16 => "sixteenth", 17 => "seventeenth", 18 => "eighteenth", 19 => "nineteenth");
    $num1     = array("zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eightteen", "nineteen");
    $num10    = array("zero", "ten", "twenty", "thirty", "fourty", "fifty", "sixty", "seventy", "eighty", "ninety");
    $places   = array(2 => "hundred", 3 => "thousand", 6 => "million", 9 => "billion", 12 => "trillion", 15 => "quadrillion", 18 => "quintillion", 21 => "sextillion", 24 => "septillion", 27 => "octillion");

    $number = array_reverse(str_split($number));

    if ($number[0] == 0) 
    {
        if ($number[1] >= 2)
            $out = str_replace("y", "ieth", $num10[$number[1]]);
        else
            $out = $num10[$number[1]]."th";
    } 
    else if (isset($number[1]) && $number[1] == 1) 
    {
        $out = $ord1[$number[1] . $number[0]];
    } 
    else 
    {
        if (array_key_exists($number[0], $ord1))
            $out = $ord1[$number[0]];
        else
            $out = $num1[$number[0]]."th";
    }

    if((isset($number[0]) && $number[0] == 0) || (isset($number[1]) && $number[1] == 1))
    {
        $i = 2;
    } 
    else 
    {
        $i = 1;
    }

    while ($i < count($number)) 
    {
        if ($i == 1) 
        {
            $out = $num10[$number[$i]] . " " . $out;
            $i++;
        } 
        else if ($i == 2) 
        {
            $out = $num1[$number[$i]] . " hundred " . $out;
            $i++;
        } 
        else 
        {
            if (isset($number[$i + 2])) 
            {
                $tmp = $num1[$number[$i + 2]] . " hundred ";
                $tmpnum = $number[$i + 1].$number[$i];
                if ($tmpnum < 20)
                    $tmp .= $num1[$tmpnum] . " " . $places[$i] . " ";
                else
                    $tmp .= $num10[$number[$i + 1]] . " " . $num1[$number[$i]] . " " . $places[$i] . " ";

                $out = $tmp . $out;
                $i+=3;
            } 
            else if (isset($number[$i + 1])) 
            {
                $tmpnum = $number[$i + 1].$number[$i];
                if ($tmpnum < 20)
                    $out = $num1[$tmpnum] . " " . $places[$i] . " " . $out;
                else
                    $out = $num10[$number[$i + 1]] . " " . $num1[$number[$i]] . " " . $places[$i] . " " . $out;
                $i+=2;
            } 
            else 
            {
                $out = $num1[$number[$i]] . " " . $places[$i] . " " . $out;
                $i++;
            }
        }
    }
    return $out;
}

This will give you the output you want.

createFullWordOrdinal(1) ----> first
createFullWordOrdinal(2) ----> second
createFullWordOrdinal(3) ----> third
createFullWordOrdinal(4) ----> fourth

How to convert numbers to words without using num2word library?

Your first statement logic is incorrect. Unless Number is 1 or smaller, that statement is always True; 200 is greater than 1 as well.

Use and instead, and include 1 in the acceptable values:

if (Number >= 1) and (Number <= 19):

You could use chaining as well:

if 1 <= Number <= 19:

For numbers of 20 or larger, use divmod() to get both the number of tens and the remainder:

tens, remainder = divmod(Number, 10)

Demo:

>>> divmod(42, 10)
(4, 2)

then use those values to build your number from the parts:

return num2words2[tens - 2] + '-' + num2words1[below_ten]

Don't forget to account for cases when the number is above 20 and doesn't have a remainder from the divmod operation:

return num2words2[tens - 2] + '-' + num2words1[remainder] if remainder else num2words2[tens - 2]

All put together:

def number(Number):
    if 0 <= Number <= 19:
        return num2words1[Number]
    elif 20 <= Number <= 99:
        tens, remainder = divmod(Number, 10)
        return num2words2[tens - 2] + '-' + num2words1[remainder] if remainder else num2words2[tens - 2]
    else:
        print('Number out of implemented range of numbers.')

How to replace 1 with first, 2 with second,3 with third etc

There is no inbuilt function for it.

I did write one for up to 99:

var special = ['zeroth','first', 'second', 'third', 'fourth', 'fifth', 'sixth', 'seventh', 'eighth', 'ninth', 'tenth', 'eleventh', 'twelfth', 'thirteenth', 'fourteenth', 'fifteenth', 'sixteenth', 'seventeenth', 'eighteenth', 'nineteenth'];
var deca = ['twent', 'thirt', 'fort', 'fift', 'sixt', 'sevent', 'eight', 'ninet'];

function stringifyNumber(n) {
  if (n < 20) return special[n];
  if (n%10 === 0) return deca[Math.floor(n/10)-2] + 'ieth';
  return deca[Math.floor(n/10)-2] + 'y-' + special[n%10];
}

// TEST LOOP SHOWING RESULTS
for (var i=0; i<100; i++) console.log(stringifyNumber(i));

DEMO: http://jsbin.com/AqetiNOt/1/edit

JavaScript numbers to Words

JavaScript is parsing the group of 3 numbers as an octal number when there's a leading zero digit. When the group of three digits is all zeros, the result is the same whether the base is octal or decimal.

But when you give JavaScript '009' (or '008'), that's an invalid octal number, so you get zero back.

If you had gone through the whole set of numbers from 190,000,001 to 190,000,010 you'd hav seen JavaScript skip '...,008' and '...,009' but emit 'eight' for '...,010'. That's the 'Eureka!' moment.

Change:

for (j = 0; j < finlOutPut.length; j++) {
    finlOutPut[j] = triConvert(parseInt(finlOutPut[j]));
}

to

for (j = 0; j < finlOutPut.length; j++) {
    finlOutPut[j] = triConvert(parseInt(finlOutPut[j],10));
}

Code also kept on adding commas after every non-zero group, so I played with it and found the right spot to add the comma.

Old:

for (b = finlOutPut.length - 1; b >= 0; b--) {
    if (finlOutPut[b] != "dontAddBigSufix") {
        finlOutPut[b] = finlOutPut[b] + bigNumArry[bigScalCntr] + ' , ';
        bigScalCntr++;
    }
    else {
        //replace the string at finlOP[b] from "dontAddBigSufix" to empty String.
        finlOutPut[b] = ' ';
        bigScalCntr++; //advance the counter  
    }
}

    //convert The output Arry to , more printable string 
    for(n = 0; n<finlOutPut.length; n++){
        output +=finlOutPut[n];
    }

New:

for (b = finlOutPut.length - 1; b >= 0; b--) {
    if (finlOutPut[b] != "dontAddBigSufix") {
        finlOutPut[b] = finlOutPut[b] + bigNumArry[bigScalCntr]; // <<<
        bigScalCntr++;
    }
    else {
        //replace the string at finlOP[b] from "dontAddBigSufix" to empty String.
        finlOutPut[b] = ' ';
        bigScalCntr++; //advance the counter  
    }
}

    //convert The output Arry to , more printable string 
    var nonzero = false; // <<<
    for(n = 0; n<finlOutPut.length; n++){
        if (finlOutPut[n] != ' ') { // <<<
            if (nonzero) output += ' , '; // <<<
            nonzero = true; // <<<
        } // <<<
        output +=finlOutPut[n];
    }

Is there a library to convert integer to first/second/third

The python inflect package has a method for converting numerals into ordinals:

import inflect
p = inflect.engine()

for i in range(1,25,5):
    print(p.ordinal(i))

displays:

1st
6th
11th
16th
21st

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