PHP string to numeric conversion with decimal places - php

I have a string ($maxDeposit) which is a numeric monetary value. So, for example:
123.00
This string is being passed in to jQuery, it needs to be passed in as a numeric data type. I'm achieving this using the following:
$maxDeposit = floatval($maxDeposit);
This loses the last last decimal place however, so my number looks like:
123.0
I have this method of converting the number to two decimal places:
$maxDeposit = sprintf('%0.2f', round($maxDeposit, 2));
However this also converts the number back to a string. Is there a way I can convert the string to a float but keep the last decimal place? Thanks

No, float is a numeric value, and 123.00 is its representation with 2 decimal places. It is responsibility of view layer to format numbers. In your case it is jQuery, e.g. console.log(maxDeposit.toFixed(2)).

I think, You can use floatval/float and number_format.
$maxDeposit = number_format(floatval($maxDeposit), 2);
or
number_format((float)$maxDeposit, 2, '.', '');
http://php.net/manual/pt_BR/function.number-format.php

Related

Floating number upto two decimal PHP

I know the question is very basic but it seems nothing working for me.
I have a number (either or float or integer) which I want to be formatted upto two decimal point. For this purpose I'm using PHP function number_format but it converts my number to string.
To convert it back to float I am using (float) or floatval(). But these functions just truncates the number after converting it to float from string.
Here is my code
$revenue_sum = array_sum(array_column($val2, 'weighted_revenue')); //23722
$test = number_format($revenue_sum, 2); //"23,722.00"
$test = (float)number_format($revenue_sum, 2); //23.0
$test = floatval(number_format($revenue_sum, 2)); //23.0
I want the $test to be 23722.00 for the $revenue_sum = 23722
If $revenue_sum = 2372.2 the $test should be 2372.20
number_format() function can be used as follows:
echo number_format($revenue_sum, 2,'.',''); // will return 23722.00 for the $revenue_sum = 23722
You are trying to type cast with ',' value, it is truncating the string.
you can try this
<?php echo sprintf("%2.2f", 8900.258); ?>
which will output as
8900.26
If you assign a floating point value to a variable, then it is converted to an internal binary format (usually using IEEE 754). Not all possible values has an internal representation. So while scanning a text, the float is rounded to the nearest possible value. So for example 1.23 is rounded to 1.22999999999999998.
Because of the internal representation, there is no difference between 100 or 1e2 or 100.0 or 100.0000.
And when printing a floating point value without any formatting instruction, PHP guess a good format and rounding some digits. So 1.22999999999999 is displayed as 1.23(may varies on different systems).
In general: As long you are calculating, formatting doesn't matter. It is mostly the best, to ignore the decimal fragments on debugging. But when printing (=converting to text), use functions like format_number() or any of the printf() functions.
To be more pragmatic:

Variable and Value Types

I have a simple question about variable type in php. I have two values in my array:
$row['DS'] // type :float (with one decimal like 12.2)
$row['TC'] // type :float (with one decimal like 24.2)
What I'm actually try to do in the make the calculation below:
$row['TC'] / $row['DS'] // $row['DS'] need to be as integer (without point,like 12)
and the result should be with two decimal like (2.32). I tried to do it in that way
$DSF = number_format($row['DS'],0);
$ConF = $row['TC'] / $DSF ;
echo number_format($conF,2);
but it returns the wrong result. for example :
$row['DS'] = 59,009.3 ---> after change the format is change to 59,009
$row['TC'] = 190.0
$ConF = 190.0 / 59,009
it should be 000.223 (something around this number ) and i expect to get 0 (after i change the format using number_format($conF,2) but instead of this the program return me the number 3.22
What am I doing wrong?
The function number_format() is used to format numbers to a comma style representation, not to actually round numbers to what you want.
The function you are looking for is round which returns a float to a specified number of decimal places.
For example:
$yourVar=round($row['TC']/$row['DS'],2);
This means that $yourVar will be the value of the division rounded to two decimal places.
You should use the number_format() function only to display human-friendly numbers at the end.
You can use type casting to convert $row['DS'] to integer right in your calculations, for e.g.:
$row['TC'] / (int)$row['DS']
or
$row['TC'] / intval($row['DS'])

PHP - Convert fixed number into decimal, using last 2 digits as the decimal places

I have a situation where all records in a CSV I'm parsing are currency, but the values are not separated by a decimal point. So for instance, value '1234' is actually '12.34', and '12345' is '123.45'.
I'm struggling to find a way to manually convert these values into decimals. I can't user number_format, because it will give me an output like so:
$original_num = 1234;
$formatted_num = number_format($original_num, '2', '.', '');
$formatted_num = 1234.00; //Output
The other issue is that sometimes I may have a value like '436257.5' after I combine two numbers, which is actually '436.2575' so I can't just manually push in a '.' two places from the end of the string. Should I consider formatting it differently while I'm parsing the file?
Assuming you're using integers to always represent decimals with 2 places of precision after the decimal point, you just divide with 100 to insert the dot in the right place.
What do you mean, "combine"? You mean multiply? You should renormalise after each multiplication, and never get into a situation where you're representing decimals of differing precisions (unless you keep track of the precision, which you can do but it's pain in the ass and normally unnecessary).
function multiply($a, $b) {
return round($a * $b / 100);
}
function format($a) {
return sprintf("%.2f", $a / 100);
}
Since number_format() function always adds 2 00 as decimal, you can divide the value by 100.
number_format($original_num/100,2);

getting an odd division error with PHP number_format()

my script calculates a number X by dividing two other numbers, A by B.
X=A/B
when i use number_format(A,2) on A before calculating X, i get a very odd number. Actual figures:
1,045.00 / 5 = 0.2
but if i don't use number_format on A before the division, i get the correct answer. Is number_format somehow making A into a non-number?
1045 / 5 = 209
number_format should be used only while pretty printing the number. Its return value should not used in calculation as you did.
Example:
If $A = 1045;
then number_format($A,2) will be 1,045.00 now if you treat 1,045.00 as a number it will be 1 as comma and remaining char will be ignored and 1/2 is 0.5 which you are getting.
You want round(A, 2), not number_format() which is for string representations (hence named "format").
The docs show that number_format returns a string. Have you tried casting the result of number_format() to a numeric type before your mathematical manipulation?
I had similar issues. It could be better if we use number format dec_point and thousand_separator parameters. you could use number_format($number, 2, '.', ''); It will help to remove your thousand separator
number_format makes it into a string with commas between thousands, and the comma will be confusing the divisor into thinking that's the decimels based on your locale.

PHP money string conversion to integer error

I have a small financial application with PHP as the front end and MySQL as the back end. I have ancient prejudices, and I store money values in MySQL as an integer of cents. My HTML forms allow input of dollar values, like "156.64" and I use PHP to convert that to cents and then I store the cents in the database.
I have a function that both cleans the dollar value from the form, and converts it to cents. I strip leading text, I strip trailing text, I multiply by 100 and convert to an integer. That final step is
$cents = (integer) ($dollars * 100);
This works fine for almost everything, except for a very few values like '156.64' which consistently converts to 15663 cents. Why does it do this?
If I do this:
$cents = (integer) ($dollars * 100 + 0.5);
then it consistently works. Why do I need to add that rounding value?
Also, my prejudices about storing money amounts as integers and not floating point values, is that no longer needed? Will modern float calculations produce nicely rounded and accurate money values adequate for keeping 100% accurate accounting?
If you want precision, you should store your money values using the DECIMAL data type in MySQL.
Your "prejudices" about floats will never be overcome - it's fundamental to the way they work. Without going into too much detail, they store a number based on powers of two and since not all decimal number can be presented this way, it doesn't always work. Your only reliable solution is to store the number as a sequence of digits and the location of the decimal point (as per DECIMAL type mentioned above).
I'm not 100% on the PHP, but is it possible the multiplication is converting the ints to floats and hence introducing exactly the problem you're trying to avoid?
Currency/money values should never be stored in a database (or used in a program) as floats.
Your integer method is fine, as is using a DECIMAL, NUMERIC or MONEY type where available.
Your problem is caused by $dollars being treated as a float and PHP doesn't have a better type to deal with money. Depending on when $dollars is being assigned, it could be being treated as a string or a float, but is certainly converted to a float if it's still a string for the * 100 operation if it looks like a float.
You might be better off parsing the string to an integer "money" value yourself (using a regex) instead of relying on the implicit conversions which PHP is doing.
The code you posted does the multiplication first, forcing a floating point calculation that introduces error, before converting the value to an integer. Instead, you should avoid floating point arithmetic entirely by reversing the order. Convert to integer values first, then perform the arithmetic.
Assuming previous code already validated and formatted the input, try this:
list($bills, $pennies) = explode('.', $dollars);
$cents = 100 * $bills + $pennies;
Your prejudice against floating point values to represent money is well founded because of truncation and because of values being converted from base-10 to base-2 and back again.
Casting does not round() as in round-to-nearest, it truncates at the decimal: (int)3.99 yields 3. (int)-3.99 yields -3.
Since float arithmetic often induces error (and possibly not in the direction you want), use round() if you want reliable rounding.
You should never ever store currency in floating point, because it always get results you don't expect.
Check out php BC Maths, it allow you to store your currency as string, then perform very high precision arithmetic on them.
Instead of using
$cents = (integer) ($dollars * 100);
you may want to try to use:
$cents = bcmul($dollars, 100, 2);
When converting from float to integer, the number will be rounded towards zero (src).
Read the Floating point precision warning.
There's no point in storing money as integer if you enter it through a floating point operation (no pun intended). If you want to convert from string to int and be consistent with your "prejudice" you can simply use string functions.
You can use an arbitrary precision library to divide by 10 (they handle numbers internally as strings), e.g. bcdiv() or gmp_div_q(), but of course, you could have also used it from the beginning for all the math.
Or you can use plain string functions:
<?php
// Quick ugly code not fully tested
$input = '156.64';
$output = NULL;
if( preg_match('/\d+(\.\d+)?/', $input) ){
$tmp = explode('.', $input);
switch( count($tmp) ){
case 1:
$output = $tmp[0];
break;
case 2:
$output = $tmp[0] . substr($tmp[1], 0, 2);
break;
default:
echo "Invalid decimal\n";
}
}else{
echo "Invalid number\n";
}
var_dump($output);
?>

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