I am trying to do a 2 digit precision in PHP Laravel project but it doesnt work. I have the value 1234666.6666667 that I want to make 1234666.66 but all the results I've seen in here or/and in other search pages.
This is my code:
$value = 1234666.6666667;
return round($value,2);
any other solution?
EDIT:
As I see, you actually want to floor number to 2 decimal points, not to round it, so this answer could help you:
$value = 1234666.6666667;
floor($value * 100) / 100; // returns 1234666.66
If you want 3 decimal points you need to multiple and divide with 1000, for 4 - with 10000 and etc.
You can use number_format, it convert value to string though, so you lose real float value:
$value = 1234666.6666667;
echo number_format($value, 2, '.', ''); // prints 1234666.67
Use this function.
function truncate($i) {
return floor($i*100) / 100.0;
}
Then you can do
$value = truncate(123.5666666); // 123.56
A pragmatic way is to use round($value - 0.05, 2), but even that gets you into hot water with some edge cases. Floating point numbers just don't round well. It's life I'm afraid. The closest double to 1234666.66 is
1234666.65999999991618096828460693359375
That's what $value will be after applying my formula! Really, if you want exact decimal precision, then you need to use a decimal type. Else use integer types and work in multiples of 100.
For the former choice, see http://de2.php.net/manual/en/ref.bc.php
$value = bcadd($value, 0, 2); // 1234666.6666667 -> 1234666.66
Another more exotic way to solve this issue is to use bcadd() with a dummy value for the $right_operand of 0,
This will give you 2 number after decimal.
Related
$quantity = 20;
$product_rate = 66.79;
$total = $quantity * $product_rate;
echo $total;
Output is showing 1335.8000000000002
is there possible to show 1335.8 using php..?
You can use the number_format() function like this:
$firstNum = 1335.8000000000002;
$number = number_format($firstNum, 1, '.', '');
echo $number;
outputs:
1335.8
more on number_format() here: http://php.net/number-format.
You can also multiply the number by 10, then use intval() to convert it to an integer (that way stripping out the decimals) and then divide by 10 like this:
$firstNum = 1335.8000000000002;
$number = 10 * intval($firstNum)/10;
echo $number;
outputs:
1335.8
Note: when using the methods above there will be no rounding, for rounding you would use something like this:
$number = round($firstNum, 1);
echo $number;
which in this case also outputs:
1335.8
Do you really use these variable values? I'm using PHP7 and the output for your given values is 1335.8. If you do a manual calculation it is the same result. It should be 1335.8. Anyway if you need to roundup the value you can use below.
round($total,1);
Please refer the below link and you will be able to grab more details.
http://php.net/manual/en/function.round.php
Because how floating point numbers work, they cannot represent every numbers exactly, so approximations are made.
The closest representation of 20 is 20, it can represent 20 exactly, but 66.79 for instance is approximated to 66.7900000000000062527760746889, that times 20 is 1335.800000000000125055521493778 that again cannot be represented and is approximated to 1335.80000000000018189894035459.
Depending on how you choose to print this number, it may round different ways, in your case for some reason you decided to print 13 decimal places so it rounded to 1335.8000000000002, but if you print only 1 or 2 decimal places it will print as 1335.8 or 1335.80. Just be mindful about that when printing floating point numbers, you may want to specify how many decimal places are relevant to you. For that, use number_format().
Example:
echo number_format($number, 2); // prints 2 decimal places
You can do this simply using echo echo round($total, 1) instead of doing round($total)
I want to round a number and I need a proper integer because I want to use it as an array key. The first "solution" that comes to mind is:
$key = (int)round($number)
However, I am unsure if this will always work. As far as I know (int) just truncates any decimals and since round($number) returns a float with theoretically limited precision, is it possible that round($number) returns something like 7.999999... and then $key is 7 instead of 8?
If this problem actually exists (I don't know how to test for it), how can it be solved? Maybe:
$key = (int)(round($number) + 0.0000000000000000001) // number of zeros chosen arbitrarily
Is there a better solution than this?
To round floats properly, you can use:
ceil($number): round up
round($number, 0): round to the nearest integer
floor($number): round down
Those functions return float, but from Niet the Dark Absol comment: "Integers stored within floats are always accurate, up to around 2^51, which is much more than can be stored in an int anyway."
round(), without a precision set always rounds to the nearest whole number. By default, round rounds to zero decimal places.
So:
$int = 8.998988776636;
round($int) //Will always be 9
$int = 8.344473773737377474;
round($int) //will always be 8
So, if your goal is to use this as a key for an array, this should be fine.
You can, of course, use modes and precision to specify exactly how you want round() to behave. See this.
UPDATE
You might actually be more interested in intval:
echo intval(round(4.7)); //returns int 5
echo intval(round(4.3)); // returns int 4
What about simply adding 1/2 before casting to an int?
eg:
$int = (int) ($float + 0.5);
This should give a predictable result.
Integers stored within floats are always accurate, up to around 253, which is much more than can be stored in an int anyway. I am worrying over nothing.
For My Case, I have to make whole number by float or decimal type
number. By these way i solved my problem. Hope It works For You.
$value1 = "46.2";
$value2 = "46.8";
// If we print by round()
echo round( $value1 ); //return float 46.0
echo round( $value2 ); //return float 47.0
// To Get the integer value
echo intval(round( $value1 )); // return int 46
echo intval(round( $value2 )); // return int 47
My solution:
function money_round(float $val, int $precision = 0): float|int
{
$pow = pow(10, $precision);
$result = (float)(intval((string)($val * $pow)) / $pow);
if (str_contains((string)$result, '.')) {
return (float)(intval((string)($val * $pow)) / $pow);
}
else {
return (int)(intval((string)($val * $pow)) / $pow);
}
}
Round to the nearest integer
$key = round($number, 0);
I want my variable's first decimal to always be rounded up. For example:
9.66 goes to 9.7
9.55 goes to 9.6
9.51 goes to 9.6
9.00000001 goes to 9.1
How do I do this?
Use round() with an optional precision and round type arguments, e.g.:
round($value, 1, PHP_ROUND_HALF_UP)
The optional second argument to round() is the precision argument and it specifies the number of decimal digits to round to. The third optional argument specifies the rounding mode. See the PHP manual for round for details.
Using round() does not always round up, even when using PHP_ROUND_HALF_UP (e.g. 9.00001 is not rounded to 9.1). You could instead try to use multiplication, ceil() and division:
ceil($value * 10.0) / 10.0
Since these are floating-point values, you might not get exact results.
I made couple tests and suggest the following answer with test cases
<?php
echo '9.66 (expected 9.7) => '.myRound(9.66).PHP_EOL;
echo '9.55 (expected 9.6) => '.myRound(9.55).PHP_EOL;
echo '9.51 (expected 9.6) => '.myRound(9.51).PHP_EOL;
echo '9.00000001 (expected 9.1) => '.myRound(9.00000001).PHP_EOL;
echo '9.9 (expected ??) => '.myRound(9.9).PHP_EOL;
echo '9.91 (expected ??) => '.myRound(9.91).PHP_EOL;
function myRound($value)
{
return ceil($value*10)/10;
}
I'm not a php programmer so will have to answer in "steps". The problem you have is the edge case where you have a number with exactly one decimal. (e.g. 9.5)
Here's how you could do it:
Multiply your number by 10.
If that's an integer, then return the original number (that's the edge case), else continue as follows:
Add 0.5
Round that in the normal way to an integer (i.e. "a.5" rounds up).
Divide the result by 10.
For step (2), sniffing around the php documentation reveals a function bool is_int ( mixed $var ) to test for an integer.
You will need a custom ceil() function, your requirements cannot be satisfied by the default function or by the round.
Use this: online test
You can use this technique. Just explode the given number / string, get the number which is next value / digit of the .. after getting this you need to increment that value and check if the value is greater than 9 or nor, if then divide that and add the carry to the first portion of the main number.
$var = '9.96';
$ar = explode(".", $var);
$nxt = substr($ar[1], 0, 1) + 1;
if($nxt > 9){
$tmp = (string) $nxt;
$num = floatval(($ar[0] + $tmp[0]).".".$tmp[1]);
}
else
$num = floatval($ar[0].".".$nxt);
var_dump($num); // float(10)
See, I want to write a function that takes a float number parameter and rounds the float to the nearest currency value (a float with two decimal places) but if the float parameter has a zero fraction (that is, all zeroes behind the decimal place) then it returns the float as an integer (or i.e. truncates the decimal part since they're all zeroes anyways.).
However, I'm finding that I can't figure out how to determine if if a fraction has a zero fraction. I don't know if there's a PHP function that already does this. I've looked. The best I can think of is to convert the float number into an integer by casting it first and then subtract the integer part from the float and then check if the difference equals to zero or not.
if($value == round($value))
{
//no decimal, go ahead and truncate.
}
This example compares the value to itself, rounded to 0 decimal places. If the value rounded is the same as the value, you've got no decimal fraction. Plain and simple.
A little trick with PHPs type juggling abilities
if ($a == (int) $a) {
// $a has a zero fraction value
}
I think the best way:
if ((string)$value == (int)$value){
...
}
Example:
$value = 2.22 * 100;
var_dump($value == (int)$value); // false - WRONG!
var_dump($value == round($value)); // false - WRONG!
var_dump((string)$value == (int)$value); // true - OK!
function whatyouneed($number) {
$decimals = 2;
printf("%.".($number == (int)($number) ? '0' : $decimals)."F", $number);
}
So basically it's either printf("%.2F") if you want 2 decimals and printf("%.2F") if you want none.
Well, the problem is that floats aren't exact. Read here if you're interested in finding out why. What I would do is decide on a level of accuracy, for example, 3 decimal places, and base exactness on that. To do that, you multiply it by 1000, cast it to an int, and then check if $your_number % 1000==0.
$mynumber = round($mynumber *1000);
if ($mynumber % 1000==0)
{ isInt() }
Just so you know, you don't have to write a function to do that, there's already one that exists:
$roundedFloat = (float)number_format("1234.1264", 2, ".", ""); // 1234.13
If you want to keep the trailing .00, just omit the float cast (although it will return a string):
$roundedFloatStr = number_format("1234.000", 2, ".", ""); // 1234.00
When calculating a golf handicap differential you are supposed to truncate the answer to 1 decimal place without rounding. No idea why but...
I know how to do this using TRUNCATE() in mySQL
SELECT TRUNCATE( 2.365, 1 );
// outputs 2.3
but I was wondering if sprintf() could do this? The only way I know to work with decimal places in a float is ...
echo sprintf("%.1f", 2.365);
// outputs 2.4
What language is this in? Assuming it's C or one of its derivatives, and assuming you always want exactly one decimal place, and assuming your values are always non-negative, you can do this:
float val = 12.3456;
val = floor(val*10.0)/10.0;
sprintf("%.1f", val);
Is there a better way? Probably. This is just what comes to mind.
I find it easier to use string operations: let sprintf round to an extra significant digit, trim, then convert back to a float
function truncate_float(float $float, int $significantDigits = 4): float
{
$format = sprintf('%%.%df', $significantDigits + 1);
return (float) substr(sprintf($format, $float), 0, -1);
}