precision issue in the case of left side zero of a number

Question

my function is as follows:

static function applyPrecision($vals) {
        $originalNumber = $vals['reading'];
        if (is_numeric($originalNumber)) {
            $precision = ($vals['precision'] != '' || $vals['precision'] != 0) ? $vals['precision'] : 4;
            $factor = pow(10, $precision);
            $multipliedNumber = $originalNumber * $factor;
            //$integerMultipliedNumber = floor($multipliedNumber);
            $var = explode(".", $multipliedNumber);
            $integerMultipliedNumber = $var[0];
            return $result = (float) ($integerMultipliedNumber / $factor);
        } else {
            return $originalNumber;
        }
    }

using this , we can apply precision for a number without rounding the value for eg:- 45.12345678 precision given is 3,then output is 45.123

but if a number like this, :45.1000000 and precision is 3 then output coming like 45.1 only, that zeros are getting skipped, is there any way to avoid this?

this is a float not a string, if you want it to keep the zeros then you need to convert it to a string and check for the number of digits post period (.) you want -subtracted from number you have and add that many zeros — Shardj, Apr 24 '19 at 11:18
@DimitrisFilippou that function will round the given value, so cannot use that for my function — sarin, Apr 24 '19 at 11:18
@sarin so you understand that 14.100 is the same thing as 14.1 then? — Shardj, Apr 24 '19 at 11:20
@Shardj yes, but is there any way to display that zeros too? — sarin, Apr 24 '19 at 11:24
If that's what you want then this is a string manipulation question really, ill post an answer — Shardj, Apr 24 '19 at 11:25
Do you want to keep only the 3 first digits every time? But without rounding up? Then you can use `number_format(floor($number * 1000) / 1000,3)` — Dimitris Filippou, Apr 24 '19 at 11:28
`echo number_format(floor((float)$number), $precision, '.', '');` works pretty well and removes the rounding problem. — Jay Blanchard, Apr 24 '19 at 11:43
@JayBlanchard This will make the number an int first cause `floor(3.456) = 3`, also the duplicate you posted will round up the value and to my undestanding OP doesn't want to round up the value — Dimitris Filippou, Apr 24 '19 at 11:52

Shardj · Answer 1 · 2019-04-24T11:29:08.523

0

If this is a question about floats, 14.1 is the same as 14.100. If this is a question about strings then you should instead just be using substr(). Answer for that below where $number is the number to cut and $precision is how many decimal places you want.

$number = (string) $number;
$temp = $precision <= 0 ? 0 : $precision+1;
return substr($number, 0, strrpos($number, '.') + $temp);

edited Apr 24 '19 at 11:29

answered Apr 24 '19 at 11:26

Shardj

1,800
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precision can be different based on settings, may be 3,2 ,1,5 etc – sarin Apr 24 '19 at 11:33
we can make a logical solutions, but if there any mathematical solution can be make? – sarin Apr 24 '19 at 11:33
It cannot be a mathematical solution @sarin, because you desire to print a string, not a number once you decide to print the additional zero's. – Jay Blanchard Apr 24 '19 at 11:34
Sorry to say, but this code doesn't work either, it only renders 14.1 – Jay Blanchard Apr 24 '19 at 11:37
You're mistaken Jay, it works perfectly. Just pop it into https://www.runphponline.com/ and define number and precision before this – Shardj Apr 24 '19 at 13:22
this will also return 100.10000000 as 100.1 – sarin Apr 25 '19 at 03:30
That's because you entered a float sarin, please refer back to the first thing I said about floats and strings – Shardj Apr 26 '19 at 10:40

Dimitris Filippou · Answer 2 · 2019-04-24T11:40:06.207

0

In order to get the first 3 subdigits of the float without rounding up you have to use float($number*1000/1000) then you use that with number_format(float($number*1000/1000),3)

example

$var =0.35489;
$var2 = number_format(floor($var * 1000) / 1000,3);
echo $var2;

output

0.354

example2

$var =0.3000;
$var2 = number_format(floor($var * 1000) / 1000,3);
echo $var2;

output2

0.300

If your precision is variable then you will have to use pow(10,$precision)

Example3

$var =0.3000;
$precision =3;
$var2 = number_format(floor($var * pow(10,$precision)) / pow(10,$precision),3);

output3

0.300

edited Apr 24 '19 at 11:40

answered Apr 24 '19 at 11:33

Dimitris Filippou

809
6
12

if value is like 45.568888 ,number_format function will round it to 45.569 – sarin Apr 24 '19 at 11:43
not if you use floor first, read my example carefully – Dimitris Filippou Apr 24 '19 at 11:47
@sarin Check my example [here](https://3v4l.org/n2hpn). Output is `45.568` – Dimitris Filippou Apr 24 '19 at 11:59

precision issue in the case of left side zero of a number

2 Answers2