I have a form, when a user fills in the total number of bottles they have, it inserts into a database and then should sum up how many cases there are.
For example in wine - there are 12 bottle cases, if a user puts in 100 bottles, it should divide this by 12 and give the sum of 8.33333333.
$bottles = "100";
What is the best way to round this down to just the number 8 and then work out how many bottles are left that never made it into a full case?
Hope this makes sense.
You can use floor
$bottles = "100";
$case = floor( $bottles / 12 );
echo $case;
Will result to 8
Documentation: http://php.net/manual/en/function.floor.php
If you want to check the bottles left, you can use modulo
$bottles = "100";
$left = $bottles % 12;
Will result to 4
You can use floor to round down, and the modulo (%) operator to determine how many bottles are left.
$bottles = 100;
$bottles_per_case = 12;
print "There are " . floor($bottles / $bottles_per_case) . " cases...";
print "With " . ($bottles % $bottles_per_case) . " bottles left over";
$bottles = "100";
$case = (int) $bottles / 12 ;
echo $case;
$left = $bottles % 12;
echo '<br>left: ' . $left;
Related
I would like to solve rounding mechanism by using php4,5.2 and below (not 5.3)
Currently I am doing 0.05 rounding, something like this page:
http://www.bnm.gov.my/index.php?ch=209&pg=657&ac=568
before rounding | after rounding
89.90 | 89.90
89.91 | 89.90
89.92 | 89.90
89.93 | 89.95
89.94 | 89.95
89.95 | 89.95
89.96 | 89.95
89.97 | 89.95
89.98 | 90.00
89.99 | 90.00
I try to use string to split it out and manually do adding, but not really a good solution, hoping here can find someone to solve it.
use this function
function rndfunc($x){
return round($x * 2, 1) / 2;
}
Conceptually, the procedure can be done as:
Divide by 0.05
or multiply by (1 / 0.05)
Round to nearest integer
Multiply by 0.05
You basically want to map values to a grid. The grid is defined as a multiple of .05. In general, you need to find the multiplicands your value lies between.
What isn't in the table are the negative numbers. You need to decide on whether to round away from zero (symmetrical) or always in the same direction (i.e. positive).
code:
$step = .05;
$multiplicand = floor( $value / $step );
$rest = $value % $step ;
if( $rest > $step/2 ) $multiplicand++; // round up if needed
$roundedvalue = $step*$multiplicand;
Multiply by two, then round, then divide by two.
Hint:-
$input1 = 24.05;
$things = abs($input * 20 ); // 481 ".05"s
$tenpcnt = abs($things / 10); // 48 ".05"s
$ouput = $tenpcnt / 20;
echo $ouput; // 2.40
function round5Sen ($value) {
return number_format(round($value*20,0)/20,2,'.','');
}
echo round5Sen(155.13);
echo "\n";
echo round5Sen(155.12);
echo "\n";
echo round5Sen(155.0);
echo "\n";
echo round5Sen(155.18);
echo "\n";
I'm sure there are more elegant solutions, but this appears to suit the task:
<?php
// setup test
$start_num = 89.90;
$iterations = 10;
// loop through test numbers
for ($i = 0; $i < $iterations; $i++) {
nickleRound($start_num + (0.01 * $i));
echo "\n\n";
}
//
function nickleRound($num) {
$p = 0.05;
echo "\n" . 'p= ' . $p;
$num = round($num, 2);
echo "\n" . 'num= ' . $num;
$r = ($num / $p);
echo "\n" . 'r= ' . $r;
$r2 = ceil($r) - $r;
echo "\n" . 'r2= ' . $r2;
$a = round($num, 1);
if (($r2 > 0) && ($r2 < 0.5)) {
$a = $a + 0.05;
}
echo "\n" . 'a= ' . $a;
}
Expanding a little on #xtofl to allow for more precise steps (not technically required for this question)
$step = 0.0005;
$multiplicand = floor($value / $step);
$rest = fmod($value, $step);
$value = $step * $multiplicand;
if ($rest > $step / 2) {
$value += $step;
}
//Round to nearest 0.05
echo round ($number * 20, 0) / 20;
//Round Up to nearest 0.05
echo ceil ($number * 20) / 20;
//Round Down to nearest 0.05
echo floor ($number * 20) / 20;
Thank you #mauris for the solution to solve my problem on Malaysia GST rounding mechanism. It also works in SQL.
DECLARE #tempTable AS TABLE(Number Decimal(20,4));
INSERT INTO #tempTable VALUES (89.90),(89.91),(89.92),(89.93),(89.94),(89.95),(89.96),(89.97),(89.98),(89.99)
SELECT Number, round(Number * 2, 1) / 2 AS 'Rounded' FROM #tempTable
PHP has the function round() for the PHP 4, PHP 5, PHP 7, PHP 8
https://www.php.net/manual/en/function.round.php
Hey I have following Problems,
I get ratings and have the average of then but bow I want following:
1,2222222222222 = 1
1,2666666666666 = 1,5
2,3635345435435 = 2.5
2,567435 345345 = 2.5
3.5709 = 3
29,3003453450 = 29
I want all numbers to their most neir .5 but when they have a no decimal then should NOT be displayed 3.0 or 4.0, Just the number without decimals.
at the moment i have this code:
function roundRating($rating) {
return floor($rating * 2) / 2;
}
can someone help me?
greets
Here is a suggestion. See comments for step by step explanation.
<?php
$float = 1.753242342;
// Modulo operator (%) does not return float remainder. Use fmod().
$remainder = fmod($float, 1);
/* Apply rounding logic here.
* It isn't entirely clear to me what your intended behavior is.
* Should 0.75 round up to 1 or down to 0.5?
* I will leave this to you to figure out.
*/
$roundedRemainder = ($remainder >= 0.25)?0.5:0;
$int = $float - $remainder;
$result = $int + $roundedRemainder;
// If remainder is zero, cast to int to remove remainder.
if (fmod($result, 1) == 0)
$result = (int)$result;
echo "Int: " . $int . "<br>";
echo "Remainder: " . $remainder . "<br>";
echo "Rounded remainder: " . $roundedRemainder . "<br>";
echo "Result: " . $result . "<br>";
Outputs:
Int: 1
Remainder: 0.753242342
Rounded remainder: 0.5
Result: 1.5
I want to round up my variable if it's decimal larger than .3 and if it's lower or equal it will round down, for example if i have 1.34 it will round up to 2, if i have 1.29 it will round down to 1, and if i have 1.3 it will round down to 1. I don't know how to do this precisely, right now i'm using the round basic function like this:
$weight = $weight/1000;
if($weight < 1) $weight = 1;
else $weight = round($weight, 0, PHP_ROUND_HALF_DOWN);
If you manipulate the numbers a bit, you can figure out if the decimals are .3 or higher. You achieve this by flooring the value, and subtract that from the original value. Check if the result of that, multiplied by 10, is greater than 3. If it is, you've got something above x.3.
$number = 1.31;
$int = floor($number);
$float = $number-$int;
if ($float*10 > 3.1)
$result = ceil($number);
else
$result = $int;
echo $result; // 2
Live demo
I made you a little hack, here's the code
$weight = 5088;
$weight = $weight/1000;
if($weight < 1) {
$weight = 1;
} else {
// I get the last number (I treat the $weight as a string here)
$last_number = substr($weight, -1, 1);
// Then I get the precision (floating numbers)
$precision = strlen(substr(strrchr($weight, "."), 1));
// Then I convert it to a string so I can use some helpful string functions
$weight_str = (string) $weight;
// If the last number is less then 3
if ($last_number > 3)
// I change it to 9 I could just change it to 5 and it would work
// because round will round up if then number is 5 or greater
$weight_str[strlen($weight_str) -1] = 9;
}
}
// Then the round will round up if it's 9 or round down if it's 3 or less
$weight = round($weight_str, $precision);
echo $weight;
Maybe something like this function?
function roundImproved($value, $decimalBreakPart = 0.3) {
$whole = floor($value);
$decimal = $value - $whole;
$decimalPartLen = strlen($decimal) - 2;
return (number_format($decimal, $decimalPartLen) <= number_format($decimalBreakPart, $decimalPartLen) ? $whole : ceil($value));
}
Proof:
http://sandbox.onlinephpfunctions.com/code/d75858f175dd819de069a8a05611ac9e7053f07a
You can specify "break part" if you want.
I need to echo number(variable) in two ways and i need help with code for this equation.
Example:
Variable is 5003
First echo has to be: 5000 (rounded)
Second echo has to be just the rounded digits: 3
So i want to know if and how can i achieve this equation, im thinking among lines of: variable(5003) minus rounded variable(5000) equals 3
So that way if variable is lets say 15009
Fist will be 15000
Second will be 9
I hope this make sense, thank you for help
You should look into the roundPHP function:
You can have negative decimal points like this:
round(5003, -3); // returns 5000
round(15009, -3); // returns 15000
To figure out the difference you can do like this:
$input = 5003
$x = $input;
$y = round($input, -3);
$z = $x - $y; // z is now 3
PHP is not a mathematical language, so it cannot solve equations for you.
You can make a more general solution like this:
$inputs = [
5003,
15009,
55108,
102010
];
foreach ($inputs as $input) {
$decimals = floor(log10($input)) - 1;
$rounded = round($input, -1 * $decimals);
echo "$input - $rounded = " . ($input - $rounded) . PHP_EOL;
}
Outputs:
5003 - 5000 = 3
15009 - 15000 = 9
55108 - 55000 = 108
102010 - 100000 = 2010
Assuming that you want to round the last three digits:
$input = 5003;
$rounded = (int)(5003 / 1000) * 1000;
$rest = $input - $rounded;
echo($rounded . "\n" . $rest);
This results in:
5000
3
I would like to solve rounding mechanism by using php4,5.2 and below (not 5.3)
Currently I am doing 0.05 rounding, something like this page:
http://www.bnm.gov.my/index.php?ch=209&pg=657&ac=568
before rounding | after rounding
89.90 | 89.90
89.91 | 89.90
89.92 | 89.90
89.93 | 89.95
89.94 | 89.95
89.95 | 89.95
89.96 | 89.95
89.97 | 89.95
89.98 | 90.00
89.99 | 90.00
I try to use string to split it out and manually do adding, but not really a good solution, hoping here can find someone to solve it.
use this function
function rndfunc($x){
return round($x * 2, 1) / 2;
}
Conceptually, the procedure can be done as:
Divide by 0.05
or multiply by (1 / 0.05)
Round to nearest integer
Multiply by 0.05
You basically want to map values to a grid. The grid is defined as a multiple of .05. In general, you need to find the multiplicands your value lies between.
What isn't in the table are the negative numbers. You need to decide on whether to round away from zero (symmetrical) or always in the same direction (i.e. positive).
code:
$step = .05;
$multiplicand = floor( $value / $step );
$rest = $value % $step ;
if( $rest > $step/2 ) $multiplicand++; // round up if needed
$roundedvalue = $step*$multiplicand;
Multiply by two, then round, then divide by two.
Hint:-
$input1 = 24.05;
$things = abs($input * 20 ); // 481 ".05"s
$tenpcnt = abs($things / 10); // 48 ".05"s
$ouput = $tenpcnt / 20;
echo $ouput; // 2.40
function round5Sen ($value) {
return number_format(round($value*20,0)/20,2,'.','');
}
echo round5Sen(155.13);
echo "\n";
echo round5Sen(155.12);
echo "\n";
echo round5Sen(155.0);
echo "\n";
echo round5Sen(155.18);
echo "\n";
I'm sure there are more elegant solutions, but this appears to suit the task:
<?php
// setup test
$start_num = 89.90;
$iterations = 10;
// loop through test numbers
for ($i = 0; $i < $iterations; $i++) {
nickleRound($start_num + (0.01 * $i));
echo "\n\n";
}
//
function nickleRound($num) {
$p = 0.05;
echo "\n" . 'p= ' . $p;
$num = round($num, 2);
echo "\n" . 'num= ' . $num;
$r = ($num / $p);
echo "\n" . 'r= ' . $r;
$r2 = ceil($r) - $r;
echo "\n" . 'r2= ' . $r2;
$a = round($num, 1);
if (($r2 > 0) && ($r2 < 0.5)) {
$a = $a + 0.05;
}
echo "\n" . 'a= ' . $a;
}
Expanding a little on #xtofl to allow for more precise steps (not technically required for this question)
$step = 0.0005;
$multiplicand = floor($value / $step);
$rest = fmod($value, $step);
$value = $step * $multiplicand;
if ($rest > $step / 2) {
$value += $step;
}
//Round to nearest 0.05
echo round ($number * 20, 0) / 20;
//Round Up to nearest 0.05
echo ceil ($number * 20) / 20;
//Round Down to nearest 0.05
echo floor ($number * 20) / 20;
Thank you #mauris for the solution to solve my problem on Malaysia GST rounding mechanism. It also works in SQL.
DECLARE #tempTable AS TABLE(Number Decimal(20,4));
INSERT INTO #tempTable VALUES (89.90),(89.91),(89.92),(89.93),(89.94),(89.95),(89.96),(89.97),(89.98),(89.99)
SELECT Number, round(Number * 2, 1) / 2 AS 'Rounded' FROM #tempTable
PHP has the function round() for the PHP 4, PHP 5, PHP 7, PHP 8
https://www.php.net/manual/en/function.round.php