Php Round - aprox to decimals - php

I have the next code for average.
function array_average2(){
$args = func_get_args();
if(isset($args[0])){
if(is_array($args[0])){
$ret = (array_sum($args[0]) / count($args[0]));
}else{
$ret = (array_sum($args) / func_num_args());
}
}else{
$ret = 0;
}
$ret2=0.01 * (int)($ret*100);
return $ret2;
}
i need php round to rezult next:
$ret=1.23 - i need 1
$ret=6.23 - i need 6
$ret=6.70 - i need 7
$ret=5.50 - i need 5.50
$ret=5.49 - i need 5
Conclusion if decimal is next to 0.50 to be next value, else previous but if it is fix 0.50 to stai. 5+6=5.50.. don't change


Well make use of the round() in PHP and apply this logic
function array_average2(){
$args = func_get_args();
if(isset($args[0])){
if(is_array($args[0])){
$ret = (array_sum($args[0]) / count($args[0]));
}else{
$ret = (array_sum($args) / func_num_args());
}
}else{
$ret = 0;
}
$ret2=0.01 * (int)($ret*100);
$str = strval($ret);
if(strpos($str,'.50')!==false)
{
return $ret;
}
else
{
return round($ret2);
}
}


Probably not the best but it should work:
$string = (string)$ret;
if (substr($string, -3) != '.50') {
$ret = round($ret);
}
OR
$string = (string)$ret;
if (strrpos($string, '.50') !== 0) {
$ret = round($ret);
}

Related

Detecting a cycle in an array PHP

I'm running a simple script which puts an integer through the formula of the Collatz conjecture and adds the output of each step into an array.
I want to use a function to detect if there's a cycle in the array, using Floyd's algorithm. And though I feel like I'm not doing a bad job, I don't seem to get it right. At this moment I'm getting the error Trying to get property 'next' of non-object in C:\xampp\htdocs\educom\week3\functions.php on line 12
See my code below. Any feedback is greatly appreciated!
include("functions.php");
$n = $_POST['number'];
$step = 0;
$reeks1 = array();
$cycle = 0;
echo "Your entry is: ". $n ."<br><br>";
while($n!==1 && $cycle==0){
$cycle = detect_cycle(array($reeks1));
if($n % 2 == 0){
$n = $n / 2;
array_push($reeks1, "$n");
$step++;
echo $step .": ". $n ."<br>";
}else{
$n = ($n * 3) + 1;
array_push($reeks1, "$n");
$step++;
echo $step .": ". $n ."<br>";
}
}
functions.php:
function detect_cycle($node){
if ($node==NULL){
return FALSE;
}
$turtle = $node;
$rabbit = $node->next;
while($rabbit != NULL){
if($rabbit === $turtle){
return TRUE;
}elseif($rabbit->next == NULL){
return FALSE;
}else{
$turtle = $turtle->next;
$rabbit = $rabbit->next->next;
}
}
return FALSE;
}

Check this out. IMPORTANT I don't know is this according to your theory. but it won't give you errors if you use like this.
function detect_cycle($node){
if ($node==NULL){
return FALSE;
}
$turtle = $node;
$rabbit = $node[0];
while($rabbit != NULL){
if($rabbit === $turtle){
return TRUE;
}elseif($rabbit[0] == NULL){
return FALSE;
}else{
$turtle = $turtle[0]; // use the number of the element key starting from 0
$rabbit = $rabbit[0][1];
}
}
return FALSE;
}

PHP loop isn't working, what am I doing wrong?

I am building a function that would receive 2 params, a string and a number.
it would basically print the first letters $n of $s.
I need to run a loop, please don't advise other non looping methods.
And yes, I need to keep the loop true throughout the function, it's suppose to close when the if is met.
For some reason the loop isn't closing, even though there's a return in the if condition that is being met when $stringCounter equals $n (=10 in this example.)
function printCounter($s, $n)
{
$stringCaller = '';
$stringCounter = strlen($stringCaller);
while (1) {
$stringCaller .= $s;
if ($stringCounter == $n) {
return $stringCaller;
}
}
}
printCounter('aba', '10');

You should imove the calc of $stringCounter inside the loop otherwise this never change
$stringCaller = '';
while (1) {
$stringCounter = strlen($stringCaller);
$stringCaller .= $s;
if ($stringCounter >= $n) {
return $stringCaller;
}
}

On my oppinion, TS is searching next approach:
function printCounter($s, $n)
{
$result = '';
$str_lenght = strlen($s);
if(!$str_lenght) {
return $result;
}
while (true) {
$result .= $s;
$result_lenght = strlen($result);
if($result_lenght/$str_lenght >= $n) {
return $result_lenght;
}
}
}
echo printCounter('aba', '10');
exit;

You can fix part of the issue by updating $stringCounter inside the loop.
function printCounter($s, $n)
{
$stringCaller = '';
while (1) {
$stringCaller .= $s;
$stringCounter = strlen($stringCaller); // check strlen after each addition of $s
if ($stringCounter == $n) {
return $stringCaller;
}
}
}
However, even after you fix that issue, there will still be cases where your loop will never exit (such as the example in your question) because $stringCounter can only equal $n if $n is a multiple of strlen($s).
For the example printCounter('aba', '10');
Iteration $stringCounter ($stringCounter == $n)
1 3 false
2 6 false
3 9 false
4 12 false
Obviously $stringCounter will only get farther away from $n from that point.
So to ensure that the function will exit, check for inequality instead with:
if ($stringCounter >= $n) {
If you need the function to return exactly $n characters, the function will need to be a little more complex and take a substring of $s to make up the remaining characters when $stringCounter + strlen($s) will be greater than $n, or return an $n character substring of your result like this:
function printCounter($s, $n)
{
$result = '';
while (1) {
$result .= $s;
if (strlen($result) >= $n) {
return substr($result, 0, $n);
}
}
}

PHP URL Shortener error

I have this PHP code which is supposed to increase a URL shortener mask on each new entry.
My problem is that it dosen't append a new char when it hits the last one (z).
(I know incrementing is a safety issue since you can guess earlier entries, but this is not a problem in this instance)
If i add 00, it can figure out 01 and so on... but is there a simple fix to why it won't do it on its own?
(The param is the last entry)
<?php
class shortener
{
public function ShortURL($str = null)
{
if (!is_null($str))
{
for($i = (strlen($str) - 1);$i >= 0;$i--)
{
if($str[$i] != 'Z')
{
$str[$i] = $this->_increase($str[$i]);
#var_dump($str[$i]);
break;
}
else
{
$str[$i] = '0';
if($i == 0)
{
$str = '0'.$str;
}
}
}
return $str;
}
else {
return '0';
}
}
private function _increase($letter)
{
//Lowercase: 97 - 122
//Uppercase: 65 - 90
// 0 - 9 : 48 - 57
$ord = ord($letter);
if($ord == 122)
{
$ord = 65;
}
elseif ($ord == 57)
{
$ord = 97;
}
else
{
$ord++;
}
return chr($ord);
}
}
?>

Effectively, all you are doing is encoding a number into Base62. So if we take the string, decode it into base 10, increment it, and reencode it into Base62, it will be much easier to know what we are doing, and the length of the string will take care of itself.
class shortener
{
public function ShortURL($str = null)
{
if ($str==null) return 0;
$int_val = $this->toBase10($str);
$int_val++;
return $this->toBase62($int_val);
}
public function toBase62($num, $b=62) {
$base='0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';
$r = $num % $b ;
$res = $base[$r];
$q = floor($num/$b);
while ($q) {
$r = $q % $b;
$q =floor($q/$b);
$res = $base[$r].$res;
}
return $res;
}
function toBase10( $num, $b=62) {
$base='0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';
$limit = strlen($num);
$res=strpos($base,$num[0]);
for($i=1;$i<$limit;$i++) {
$res = $b * $res + strpos($base,$num[$i]);
}
return $res;
}
}

Calculating Nth root with bcmath in PHP

We are looking for the Nth root in PHP. We need to do this with a very large number, and the windows calculator returns 2. With the following code we are getting 1. Does anybody have an idea how this works?
echo bcpow(18446744073709551616, 1/64);

Well it seems that PHP and the BC lib has some limits, and after searching on the internet i found this interesting article/code:
So you should use this function:
<?php
function NRoot($num, $n) {
if ($n<1) return 0; // we want positive exponents
if ($num<=0) return 0; // we want positive numbers
if ($num<2) return 1; // n-th root of 1 or 2 give 1
// g is our guess number
$g=2;
// while (g^n < num) g=g*2
while (bccomp(bcpow($g,$n),$num)==-1) {
$g=bcmul($g,"2");
}
// if (g^n==num) num is a power of 2, we're lucky, end of job
if (bccomp(bcpow($g,$n),$num)==0) {
return $g;
}
// if we're here num wasn't a power of 2 :(
$og=$g; // og means original guess and here is our upper bound
$g=bcdiv($g,"2"); // g is set to be our lower bound
$step=bcdiv(bcsub($og,$g),"2"); // step is the half of upper bound - lower bound
$g=bcadd($g,$step); // we start at lower bound + step , basically in the middle of our interval
// while step!=1
while (bccomp($step,"1")==1) {
$guess=bcpow($g,$n);
$step=bcdiv($step,"2");
$comp=bccomp($guess,$num); // compare our guess with real number
if ($comp==-1) { // if guess is lower we add the new step
$g=bcadd($g,$step);
} else if ($comp==1) { // if guess is higher we sub the new step
$g=bcsub($g,$step);
} else { // if guess is exactly the num we're done, we return the value
return $g;
}
}
// whatever happened, g is the closest guess we can make so return it
return $g;
}
echo NRoot("18446744073709551616","64");
?>
Hope this was helpful ...

I had problems with HamZa's solution getting to work with arbitrary precission, so i adopted it a little.
<?php
function NthRoot($Base, $NthRoot, $Precision = 100) {
if ($NthRoot < 1) return 0;
if ($Base <= 0) return 0;
if ($Base < 2) return 1;
$retVal = 0;
$guess = bcdiv($Base, 2, $Precision);
$continue = true;
$step = bcdiv(bcsub($Base, $guess, $Precision), 2, $Precision);
while ($continue) {
$test = bccomp($Base, bcpow($guess, $NthRoot, $Precision), $Precision);
if ($test == 0) {
$continue = false;
$retVal = $guess;
}
else if ($test > 0) {
$step = bcdiv($step, 2, $Precision);
$guess = bcadd($guess, $step, $Precision);
}
else if ($test < 0) {
$guess = bcsub($guess, $step, $Precision);
}
if (bccomp($step, 0, $Precision) == 0) {
$continue = false;
$retVal = $guess;
}
}
return $retVal;
}

How to display Currency in Indian Numbering Format in PHP?

I have a question about formatting the Rupee currency (Indian Rupee - INR).
For example, numbers here are represented as:
1
10
100
1,000
10,000
1,00,000
10,00,000
1,00,00,000
10,00,00,000
Refer Indian Numbering System
I have to do with it PHP.
I have saw this question Displaying Currency in Indian Numbering Format. But couldn't able to get it for PHP my problem.
Update:
How to use money_format() in indian currency format?

You have so many options but money_format can do the trick for you.
Example:
$amount = '100000';
setlocale(LC_MONETARY, 'en_IN');
$amount = money_format('%!i', $amount);
echo $amount;
Output:
1,00,000.00
Note:
The function money_format() is only defined if the system has strfmon capabilities. For example, Windows does not, so money_format() is undefined in Windows.
Pure PHP Implementation - Works on any system:
$amount = '10000034000';
$amount = moneyFormatIndia( $amount );
echo $amount;
function moneyFormatIndia($num) {
$explrestunits = "" ;
if(strlen($num)>3) {
$lastthree = substr($num, strlen($num)-3, strlen($num));
$restunits = substr($num, 0, strlen($num)-3); // extracts the last three digits
$restunits = (strlen($restunits)%2 == 1)?"0".$restunits:$restunits; // explodes the remaining digits in 2's formats, adds a zero in the beginning to maintain the 2's grouping.
$expunit = str_split($restunits, 2);
for($i=0; $i<sizeof($expunit); $i++) {
// creates each of the 2's group and adds a comma to the end
if($i==0) {
$explrestunits .= (int)$expunit[$i].","; // if is first value , convert into integer
} else {
$explrestunits .= $expunit[$i].",";
}
}
$thecash = $explrestunits.$lastthree;
} else {
$thecash = $num;
}
return $thecash; // writes the final format where $currency is the currency symbol.
}

$num = 1234567890.123;
$num = preg_replace("/(\d+?)(?=(\d\d)+(\d)(?!\d))(\.\d+)?/i", "$1,", $num);
echo $num;
// Input : 1234567890.123
// Output : 1,23,45,67,890.123
// Input : -1234567890.123
// Output : -1,23,45,67,890.123

echo 'Rs. '.IND_money_format(1234567890);
function IND_money_format($money){
$len = strlen($money);
$m = '';
$money = strrev($money);
for($i=0;$i<$len;$i++){
if(( $i==3 || ($i>3 && ($i-1)%2==0) )&& $i!=$len){
$m .=',';
}
$m .=$money[$i];
}
return strrev($m);
}
NOTE:: it is not tested on float values and it suitable for only Integer

The example you've linked is making use of the ICU libraries which are available with PHP in the intl Extension­Docs:
$fmt = new NumberFormatter($locale = 'en_IN', NumberFormatter::CURRENCY);
echo $fmt->format(10000000000.1234)."\n"; # Rs 10,00,00,00,000.12
Or maybe better fitting in your case:
$fmt = new NumberFormatter($locale = 'en_IN', NumberFormatter::DECIMAL);
echo $fmt->format(10000000000)."\n"; # 10,00,00,00,000

Simply use below function to format in INR.
function amount_inr_format($amount) {
$fmt = new \NumberFormatter($locale = 'en_IN', NumberFormatter::DECIMAL);
return $fmt->format($amount);
}

Check this code, it works 100% for Indian Rupees format with decimal format.
You can use numbers like :
123456.789
123.456
123.4
123
and 1,2,3,4,5,6,7,8,9,.222
function moneyFormatIndia($num){
$explrestunits = "" ;
$num = preg_replace('/,+/', '', $num);
$words = explode(".", $num);
$des = "00";
if(count($words)<=2){
$num=$words[0];
if(count($words)>=2){$des=$words[1];}
if(strlen($des)<2){$des="$des";}else{$des=substr($des,0,2);}
}
if(strlen($num)>3){
$lastthree = substr($num, strlen($num)-3, strlen($num));
$restunits = substr($num, 0, strlen($num)-3); // extracts the last three digits
$restunits = (strlen($restunits)%2 == 1)?"0".$restunits:$restunits; // explodes the remaining digits in 2's formats, adds a zero in the beginning to maintain the 2's grouping.
$expunit = str_split($restunits, 2);
for($i=0; $i<sizeof($expunit); $i++){
// creates each of the 2's group and adds a comma to the end
if($i==0)
{
$explrestunits .= (int)$expunit[$i].","; // if is first value , convert into integer
}else{
$explrestunits .= $expunit[$i].",";
}
}
$thecash = $explrestunits.$lastthree;
} else {
$thecash = $num;
}
return "$thecash.$des"; // writes the final format where $currency is the currency symbol.
}

When money_format is not available :
function format($amount): string
{
list ($number, $decimal) = explode('.', sprintf('%.2f', floatval($amount)));
$sign = $number < 0 ? '-' : '';
$number = abs($number);
for ($i = 3; $i < strlen($number); $i += 3)
{
$number = substr_replace($number, ',', -$i, 0);
}
return $sign . $number . '.' . $decimal;
}

<?php
$amount = '-100000.22222'; // output -1,00,000.22
//$amount = '0100000.22222'; // output 1,00,000.22
//$amount = '100000.22222'; // output 1,00,000.22
//$amount = '100000.'; // output 1,00,000.00
//$amount = '100000.2'; // output 1,00,000.20
//$amount = '100000.0'; // output 1,00,000.00
//$amount = '100000'; // output 1,00,000.00
echo $aaa = moneyFormatIndia($amount);
function moneyFormatIndia($amount)
{
$amount = round($amount,2);
$amountArray = explode('.', $amount);
if(count($amountArray)==1)
{
$int = $amountArray[0];
$des=00;
}
else {
$int = $amountArray[0];
$des=$amountArray[1];
}
if(strlen($des)==1)
{
$des=$des."0";
}
if($int>=0)
{
$int = numFormatIndia( $int );
$themoney = $int.".".$des;
}
else
{
$int=abs($int);
$int = numFormatIndia( $int );
$themoney= "-".$int.".".$des;
}
return $themoney;
}
function numFormatIndia($num)
{
$explrestunits = "";
if(strlen($num)>3)
{
$lastthree = substr($num, strlen($num)-3, strlen($num));
$restunits = substr($num, 0, strlen($num)-3); // extracts the last three digits
$restunits = (strlen($restunits)%2 == 1)?"0".$restunits:$restunits; // explodes the remaining digits in 2's formats, adds a zero in the beginning to maintain the 2's grouping.
$expunit = str_split($restunits, 2);
for($i=0; $i<sizeof($expunit); $i++) {
// creates each of the 2's group and adds a comma to the end
if($i==0) {
$explrestunits .= (int)$expunit[$i].","; // if is first value , convert into integer
} else {
$explrestunits .= $expunit[$i].",";
}
}
$thecash = $explrestunits.$lastthree;
} else {
$thecash = $num;
}
return $thecash; // writes the final format where $currency is the currency symbol.
}
?>

So if I'm reading that right, the Indian Numbering System separates the thousands, then every power of a hundred past that? Hmm...
Perhaps something like this?
function indian_number_format($num) {
$num = "".$num;
if( strlen($num) < 4) return $num;
$tail = substr($num,-3);
$head = substr($num,0,-3);
$head = preg_replace("/\B(?=(?:\d{2})+(?!\d))/",",",$head);
return $head.",".$tail;
}

$amount=-3000000000111.11;
$amount<0?(($sign='-').($amount*=-1)):$sign=''; //Extracting sign from given amount
$pos=strpos($amount, '.'); //Identifying the decimal point position
$amt= substr($amount, $pos-3); // Extracting last 3 digits of integer part along with fractional part
$amount= substr($amount,0, $pos-3); //removing the extracted part from amount
for(;strlen($amount);$amount=substr($amount,0,-2)) // Now loop through each 2 digits of remaining integer part
$amt=substr ($amount,-2).','.$amt; //forming Indian Currency format by appending (,) for each 2 digits
echo $sign.$amt; //Appending sign

I think this a quick and simplest solution:-
function formatToInr($number){
$number=round($number,2);
// windows is not supported money_format
if(setlocale(LC_MONETARY, 'en_IN')){
return money_format('%!'.$decimal.'n', $number);
}
else {
if(floor($number) == $number) {
$append='.00';
}else{
$append='';
}
$number = preg_replace("/(\d+?)(?=(\d\d)+(\d)(?!\d))(\.\d+)?/i", "$1,", $number);
return $number.$append;
}
}

You should check the number_format function.Here is the link
Separating thousands with commas will look like
$rupias = number_format($number, 2, ',', ',');

I have used different format parameters to money_format() for my output.
setlocale(LC_MONETARY, 'en_IN');
if (ctype_digit($amount) ) {
// is whole number
// if not required any numbers after decimal use this format
$amount = money_format('%!.0n', $amount);
}
else {
// is not whole number
$amount = money_format('%!i', $amount);
}
//$amount=10043445.7887 outputs 1,00,43,445.79
//$amount=10043445 outputs 1,00,43,445

Above Function Not working with Decimal
$amount = 10000034000.001;
$amount = moneyFormatIndia( $amount );
echo $amount;
function moneyFormatIndia($num){
$nums = explode(".",$num);
if(count($nums)>2){
return "0";
}else{
if(count($nums)==1){
$nums[1]="00";
}
$num = $nums[0];
$explrestunits = "" ;
if(strlen($num)>3){
$lastthree = substr($num, strlen($num)-3, strlen($num));
$restunits = substr($num, 0, strlen($num)-3);
$restunits = (strlen($restunits)%2 == 1)?"0".$restunits:$restunits;
$expunit = str_split($restunits, 2);
for($i=0; $i<sizeof($expunit); $i++){
if($i==0)
{
$explrestunits .= (int)$expunit[$i].",";
}else{
$explrestunits .= $expunit[$i].",";
}
}
$thecash = $explrestunits.$lastthree;
} else {
$thecash = $num;
}
return $thecash.".".$nums[1];
}
}
Answer : 10,00,00,34,000.001

It's my very own function to do the task
function bd_money($num) {
$pre = NULL; $sep = array(); $app = '00';
$s=substr($num,0,1);
if ($s=='-') {$pre= '-';$num = substr($num,1);}
$num=explode('.',$num);
if (count($num)>1) $app=$num[1];
if (strlen($num[0])<4) return $pre . $num[0] . '.' . $app;
$th=substr($num[0],-3);
$hu=substr($num[0],0,-3);
while(strlen($hu)>0){$sep[]=substr($hu,-2); $hu=substr($hu,0,-2);}
return $pre.implode(',',array_reverse($sep)).','.$th.'.'.$app;
}
It took 0.0110 Seconds per THOUSAND query while number_format took 0.001 only.
Always try to use PHP native functions only when performance is target issue.

$r=explode('.',12345601.20);
$n = $r[0];
$len = strlen($n); //lenght of the no
$num = substr($n,$len-3,3); //get the last 3 digits
$n = $n/1000; //omit the last 3 digits already stored in $num
while($n > 0) //loop the process - further get digits 2 by 2
{
$len = strlen($n);
$num = substr($n,$len-2,2).",".$num;
$n = round($n/100);
}
echo "Rs.".$num.'.'.$r[1];

If you dont want to use any inbuilt function in my case i was doing on iis server so was unable to use one the function in php so did this
$num = -21324322.23;
moneyFormatIndiaPHP($num);
function moneyFormatIndiaPHP($num){
//converting it to string
$numToString = (string)$num;
//take care of decimal values
$change = explode('.', $numToString);
//taking care of minus sign
$checkifminus = explode('-', $change[0]);
//if minus then change the value as per
$change[0] = (count($checkifminus) > 1)? $checkifminus[1] : $checkifminus[0];
//store the minus sign for further
$min_sgn = '';
$min_sgn = (count($checkifminus) > 1)?'-':'';
//catch the last three
$lastThree = substr($change[0], strlen($change[0])-3);
//catch the other three
$ExlastThree = substr($change[0], 0 ,strlen($change[0])-3);
//check whethr empty
if($ExlastThree != '')
$lastThree = ',' . $lastThree;
//replace through regex
$res = preg_replace("/\B(?=(\d{2})+(?!\d))/",",",$ExlastThree);
//main container num
$lst = '';
if(isset($change[1]) == ''){
$lst = $min_sgn.$res.$lastThree;
}else{
$lst = $min_sgn.$res.$lastThree.".".$change[1];
}
//special case if equals to 2 then
if(strlen($change[0]) === 2){
$lst = str_replace(",","",$lst);
}
return $lst;
}

This for both integer and float values
function indian_money_format($number)
{
if(strstr($number,"-"))
{
$number = str_replace("-","",$number);
$negative = "-";
}
$split_number = #explode(".",$number);
$rupee = $split_number[0];
$paise = #$split_number[1];
if(#strlen($rupee)>3)
{
$hundreds = substr($rupee,strlen($rupee)-3);
$thousands_in_reverse = strrev(substr($rupee,0,strlen($rupee)-3));
$thousands = '';
for($i=0; $i<(strlen($thousands_in_reverse)); $i=$i+2)
{
$thousands .= $thousands_in_reverse[$i].$thousands_in_reverse[$i+1].",";
}
$thousands = strrev(trim($thousands,","));
$formatted_rupee = $thousands.",".$hundreds;
}
else
{
$formatted_rupee = $rupee;
}
if((int)$paise>0)
{
$formatted_paise = ".".substr($paise,0,2);
}else{
$formatted_paise = '.00';
}
return $negative.$formatted_rupee.$formatted_paise;
}

Use this function:
function addCommaToRs($amt, &$ret, $dec='', $sign=''){
if(preg_match("/-/",$amt)){
$amts=explode('-',$amt);
$amt=$amts['1'];
static $sign='-';
}
if(preg_match("/\./",$amt)){
$amts=explode('.',$amt);
$amt=$amts['0'];
$l=strlen($amt);
static $dec;
$dec=$amts['1'];
} else {
$l=strlen($amt);
}
if($l>3){
if($l%2==0){
$ret.= substr($amt,0,1);
$ret.= ",";
addCommaToRs(substr($amt,1,$l),$ret,$dec);
} else{
$ret.=substr($amt,0,2);
$ret.= ",";
addCommaToRs(substr($amt,2,$l),$ret,$dec);
}
} else {
$ret.= $amt;
if($dec) $ret.=".".$dec;
}
return $sign.$ret;
}
Call it like this:
$amt = '';
echo addCommaToRs(123456789.123,&$amt,0);
This will return 12,34,567.123.

<?php
function moneyFormatIndia($num)
{
//$num=123456789.00;
$result='';
$sum=explode('.',$num);
$after_dec=$sum[1];
$before_dec=$sum[0];
$result='.'.$after_dec;
$num=$before_dec;
$len=strlen($num);
if($len<=3)
{
$result=$num.$result;
}
else
{
if($len<=5)
{
$result='Rs '.substr($num, 0,$len-3).','.substr($num,$len-3).$result;
return $result;
}
else
{
$ls=strlen($num);
$result=substr($num, $ls-5,2).','.substr($num, $ls-3).$result;
$num=substr($num, 0,$ls-5);
while(strlen($num)!=0)
{
$result=','.$result;
$ls=strlen($num);
if($ls<=2)
{
$result='Rs. '.$num.$result;
return $result;
}
else
{
$result=substr($num, $ls-2).$result;
$num=substr($num, 0,$ls-2);
}
}
}
}
}
?>

heres is simple thing u can do ,
float amount = 100000;
NumberFormat formatter = NumberFormat.getCurrencyInstance(new Locale("en", "IN"));
String moneyString = formatter.format(amount);
System.out.println(moneyString);
The output will be , Rs.100,000.00 .

declare #Price decimal(26,7)
Set #Price=1234456677
select FORMAT(#Price, 'c', 'en-In')
Result:
1,23,44,56,677.00

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