I need execute json_encode() and convert my original number from:
50610101800060384093800100001010000000056199999999
to
"50610101800060384093800100001010000000056199999999"
But it return
5.061010180006E+49
I tried this:
ini_set('precision', 30); //With 1, 30, 50, 100, 1000
ini_set('serialize_precision', -1);
'content' => json_encode($params, JSON_NUMERIC_CHECK)
but doesn't work. Can you help me?
50610101800060384093800100001010000000056199999999 exceeds the value of the maximum integer in PHP and so it is promoted to a float and expressed in scientific notation. The float result may be problematic for various reasons as the Manual explains in warning about floating point precision.
If you wish to express the value as if it were an integer you must encapsulate it in a string. That string you may add zero to it but when you do so the result in scientific notation will refer to a float, as follows:
<?php
$s = "50610101800060384093800100001010000000056199999999";
echo $s,"\n";
$x = $s + 0;
echo $x, "\n",is_float($x);
See here.
For more info in re PHP and floats, see here.
On the other hand, if there were an array of numbers whose digits corresponded to the numerical display in the OP's post, you could write code as follows:
<?php
$a = [5,0,6,1,0,1,0,1,8,0,0,0,6,0,3,8,4,0,9,3,8,0,0,1,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,5,6,1,9,9,9,9,9,9,9,9];
foreach($a as $e) {
$e = (string) $e;
}
$foo = join($a);
var_dump($foo);
$foo = bcadd($foo, 1);
var_dump($foo);
See live code.
The reason this example works is because each array value is converted to a numerical string and then the individual elements are joined to form one very long numerical string. BC Math is an extension in PHP which supports arbitrary precision. In this case, the bcadd() function adds one to the numerical string which results in the display of an incremented numerical string value.
Try This [https://3v4l.org/biNJG][1]
If you want this output "50610101800060384093800100001010000000056199999999"
you may want to pass this Value as string after encoding the value to JSON using json_encode
An integer data type is a non-decimal number between -2,147,483,648 and 2,147,483,647.
Rules for integers:
An integer must have at least one digit
An integer must not have a decimal point
An integer can be either positive or negative
Integers can be specified in three formats: decimal (10-based), hexadecimal (16-based - prefixed with 0x) or octal (8-based - prefixed with 0)
Related
Number format adds the commas I want but removes the decimals.
echo number_format("1000000.25");
This returns 1,000,000
I want it to return 1,000,000.25
I need both the commas and the decimals, without any number rounding. All my values are decimals. They vary in length.
Do I have to use something other than number format?
In case what you meant by they vary in length is related to the decimal part, take a look at this code:
function getLen($var){
$tmp = explode('.', $var);
if(count($tmp)>1){
return strlen($tmp[1]);
}
}
$var = '1000000.255555'; // '1000000.25'; // '1000000';
echo number_format($var, getLen($var));
Some tests
Output for 1000000.255555:
1,000,000.255555
Output for 1000000.25:
1,000,000.25
Output for 1000000:
1,000,000
It counts how many chars there are after the . and uses that as argument in the number_format function;
Otherwise just use a constant number instead of the function call.
And some reference...
From the manual -> number_format():
string number_format ( float $number [, int $decimals = 0 ] )
And you can see in the description that
number
The number being formatted.
decimals
Sets the number of decimal points.
And a bit more:
[...]If two parameters are given, number will be formatted with decimals
decimals with a dot (".") in front, and a comma (",") between every
group of thousands.
$number = '1000000.25';
echo number_format($number, strlen(substr(strrchr($number, "."), 1)));
Explanation:
Number Format takes a second parameter which specifies the number of decimal places required as pointed out in the docs. This Stack overflow answer tells you how to get the number of decimal places of your provided string
The docs for number_format() indicate the second parameter is used to specify decimal:
echo number_format('1000000.25', 2);
Ref: http://php.net/manual/en/function.number-format.php
I have a problem with this function always my mt_rand() give me the same number:
$hex = 'f12a218a7dd76fb5924f5deb1ef75a889eba4724e55e6568cf30be634706bd4c'; // i edit this string for each request
$hex = hexdec($hex);
mt_srand($hex);
$hex = sprintf("%04d", mt_rand('0','9999'));
$hex is always changed, but the result is always the same 4488.
Edit
$hex = str_split($hex);
$hex = implode("", array_slice($hex, 0, 7));
mt_srand($hex);
$number = sprintf("%04d", mt_rand('0','9999'));
http://php.net/manual/en/function.mt-srand.php
Your problem is, that you always end up with a float value in your variable $hex. And the function mt_srand() as you can also see in the manual:
void mt_srand ([ int $seed ] )
Expects an integer. So what it does is, it simply tries to convert your float value to an integer. But since this fails it will always return 0.
So at the end you always end up with the seed 0 and then also with the same "random" number.
You can see this if you do:
var_dump($hex);
output:
float(1.0908183557664E+77)
And if you then want to see in which integer it will end up if it gets converted you can use this:
var_dump((int)$hex);
And you will see it will always be 0.
Also if you are interested in, why your number ends up as float, it's simply because of the integer overflow, since your number is way too big and accoding to the manual:
If PHP encounters a number beyond the bounds of the integer type, it will be interpreted as a float instead. Also, an operation which results in a number beyond the bounds of the integer type will return a float instead.
And if you do:
echo PHP_INT_MAX;
You will get the max value of int, which will be:
28192147483647 //32 bit
9223372036854775807 //64 bit
EDIT:
So now how to fix this problem and still make sure to get a random number?
Well the first thought could be just to check if the value is bigger than PHP_INT_MAX and if yes set it to some other number. But I assume and it seems like you will always have such a large hex number.
So I would recommend you something like this:
$arr = str_split($hex);
shuffle($arr);
$hex = implode("", array_slice($arr, 0, 7));
Here I just split your number into an array with str_split(), to then shuffle() the array and after this I implode() the first 7 array elements which I get with array_slice().
After you have done this you can use it with hexdec() and then use it in mt_srand().
Also why do I only get the first 7 elements? Simply because then I can make sure I don't hit the PHP_INT_MAX value.
When I run the code below:
<?
echo rand(0, 0xfffffffffffffbff);
echo rand(0, 0xfffffffffffffc00);
echo rand(0, $something_bigger_than_0xfffffffffffffbff);
I got something like:
-828
0
0
(the 2nd and 3rd number will always be zero)
mt_rand() has the same behavior.
so, why zero?
rand() function takes two integers as its arguments. Integers usually have the maximum range of 2^32. The supplied argument, when converted to integers, will be larger than this, hence causing an integer overflow.
This is actually documented behavior. From the PHP manual documentation:
When converting from float to integer, the number will be rounded
towards zero.
If the float is beyond the boundaries of integer (usually +/- 2.15e+9
= 2^31 on 32-bit platforms and +/- 9.22e+18 = 2^63 on 64-bit platforms), the result is undefined, since the float doesn't have
enough precision to give an exact integer result. No warning, not even
a notice will be issued when this happens!
You need to pass integer value to the rand() as the above $a = 0xfffffffffffffbff; is a double but not an integer value, the double value 0xfffffffffffffbff will be cast to an integer and eventually you will be getting a zero.
An illustration for your understanding.
<?php
$a = 0xfffffffffffffbff;
echo gettype($a); //"prints" double
if(is_int($a))
{
echo "Yes";
}
else
{
echo "Nope.. casting that to int results in..".intval($a)."<br>";
}
echo rand(0, $a);
OUTPUT :
double
Nope.. casting that to int results in..0
0
When I do this typecasting:
(float) '0.00';
I get 0. How do I get 0.00 and still have the data type as a float?
A float doesn't have 0 or 0.00 : those are different string representations of the internal (IEEE754) binary format but the float is the same.
If you want to express your float as "0.00", you need to format it in a string, using number_format :
$numberAsString = number_format($numberAsFloat, 2);
As far as i know there is no solution for PHP to fix this. All other (above and below) answers given in this thread are nonsense.
The number_format function returns a string as result as written in PHP.net's own specification.
Functions like floatval/doubleval do return integers if you give as value 3.00 .
If you do typejuggling then you will get an integer as result.
If you use round() then you will get an integer as result.
The only possible solution that i can think of is using your database for type conversion to float. MySQL for example:
SELECT CAST('3.00' AS DECIMAL) AS realFloatValue;
Execute this using an abstraction layer which returns floats instead of strings and there you go.
JSON output modification
If you are looking for a solution to fix your JSON output to hold 2 decimals then you can probably use post-formatting like in the code below:
// PHP AJAX Controller
// some code here
// transform to json and then convert string to float with 2 decimals
$output = array('x' => 'y', 'price' => '0.00');
$json = json_encode($output);
$json = str_replace('"price":"'.$output['price'].'"', '"price":'.$output['price'].'', $json);
// output to browser / client
print $json;
exit();
Returns to client/browser:
{"x":"y","price":0.00}
0.00 is actually 0. If you need to have the 0.00 when you echo, simply use number_format this way:
number_format($number, 2);
You can show float numbers
with a certain number of decimals
with a certain format (localised)
i.e.
$myNonFormatedFloat = 5678.9
$myGermanNumber = number_format($myNonFormatedFloat, 2, ',', '.'); // -> 5.678,90
$myAngloSaxonianNumber = number_format($myNonFormatedFloat, 2, '.', ','); // -> 5,678.90
Note that, the
1st argument is the float number you would like to format
2nd argument is the number of decimals
3rd argument is the character used to visually separate the decimals
4th argument is the character used to visually separate thousands
Use the number_format() function to change how a number is displayed. It will return a string, the type of the original variable is unaffected.
try this
$nom="5695.5";
number_format((float)($nom), 2, '.', ','); // -> 5,695.50
$nom="5695.5215";
number_format((float)($nom), 2, '.', ','); // -> 5,695.52
$nom="5695.12";
number_format((float)($nom), 0, '.', ','); // -> 5,695
//use round()
$nom="5695.12";
number_format((float)round($nom), 2, '.', ','); // -> 5,695.00
$nom="5695.52";
number_format((float)round($nom), 2, '.', ','); // -> 5,696.00
you can try this,it will work for you
number_format(0.00, 2)
A number of comments on this page have missed the fundamental point that the question is ill-formed. Floating point is a binary representation, designed for efficient calculations; it fundamentally has no notion of decimal digits of any sort.
So asking this:
How do I get "0.00" instead of "0" and still have the data type as a float?
Is like asking this:
How do I get "deux" instead of "zwei" and still have the language as English?
Whether you specify the input as "2", or "2.0", or "2.000000000", if you ask for a floating point value, what will be stored in memory is this (in IEEE 754 double-precision):
0100000000000000000000000000000000000000000000000000000000000000
If you convert to an integer, the value stored in memory is this (assuming a 64-bit system):
0000000000000000000000000000000000000000000000000000000000000010
(Note that "integer" in this context is not just a synonym for "whole number", it is a specific data type, with its own rules for how values should be represented in memory.)
By contrast, the string "2" would look like this:
00110010
And the string "2.00" would look like this:
00110010001011100011000000110000
(In a PHP program, there would actually be additional information in memory, such as an indicator of the type, but that's not really relevant here.)
So, the question can only be answered by rephrasing it as a conversion: given the input of a floating point number, how do I choose a string representation which has a fixed number of decimals.
As others have pointed out, the answer to that is to use number_format.
The question doesn't mention JSON, but several comments do, so I will also point out that PHP's json_encode function has an option JSON_PRESERVE_ZERO_FRACTION, which will format a floating point number that happens to be a whole number with a trailing ".0", for instance:
$example = ['int' => 2, 'float' => 2.0];
echo json_encode($example);
# => {"int":2,"float":2}
echo json_encode($example, JSON_PRESERVE_ZERO_FRACTION);
# => {"int":2,"float":2.0}
Again, note that this is a string representation of the value.
You can use round function
round("10.221",2);
Will return 10.22
You can use floatval()
floatval()
try this
$result = number_format($FloatNumber, 2);
You can use this simple function.
number_format ()
$num = 2214.56;
// default english notation
$english_format = number_format($num);
// 2,215
// French notation
$format_francais = number_format($num, 2, ',', ' ');
// 2 214,56
$num1 = 2234.5688;
// English notation with thousands separator
$english_format_number = number_format($num1,2);
// 2,234.57
// english notation without thousands separator
$english_format_number2 = number_format($num1, 2, '.', '');
// 2234.57
When we format any float value, that means we are changing its data type to string. So when we apply the formatting on any amount/float value then it will set with all possible notations like dot, comma, etc. For example
(float)0.00 => (string)'0.00',
(float)10000.56 => (string) '10,000.56'
(float)5000000.20=> (string) '5,000,000.20'
So, logically it's not possible to keep the float datatype after formatting.
I am not familiar with PHP at all and had a quick question.
I have 2 variables pricePerUnit and InvoicedUnits. Here's the code that is setting these to values:
$InvoicedUnits = ((string) $InvoiceLineItem->InvoicedUnits);
$pricePerUnit = ((string) $InvoiceLineItem->PricePerUnit);
If I output this, I get the correct values. Lets say 5000 invoiced units and 1.00 for price.
Now, I need to show the total amount spent. When I multiply these two together it doesn't work (as expected, these are strings).
But I have no clue how to parse/cast/convert variables in PHP.
What should I do?
$rootbeer = (float) $InvoicedUnits;
Should do it for you. Check out Type-Juggling. You should also read String conversion to Numbers.
You want the non-locale-aware floatval function:
float floatval ( mixed $var ) - Gets the float value of a string.
Example:
$string = '122.34343The';
$float = floatval($string);
echo $float; // 122.34343
Well, if user write 1,00,000 then floatvar will show error. So -
floatval(preg_replace("/[^-0-9\.]/","",$input));
This is much more reliable.
Usage :
$input = '1,03,24,23,434,500.6798633 this';
echo floatval(preg_replace("/[^-0-9\.]/","",$input));
Dealing with markup in floats is a non trivial task. In the English/American notation you format one thousand plus 46*10-2:
1,000.46
But in Germany you would change comma and point:
1.000,46
This makes it really hard guessing the right number in multi-language applications.
I strongly suggest using Zend_Measure of the Zend Framework for this task. This component will parse the string to a float by the users language.
you can follow this link to know more about How to convert a string/number into number/float/decimal in PHP.
HERE IS WHAT THIS LINK SAYS...
Method 1: Using number_format() Function. The number_format() function is used to convert a string into a number. It returns the formatted number on success otherwise it gives E_WARNING on failure.
$num = "1000.314";
//Convert string in number using
//number_format(), function
echo number_format($num), "\n";
//Convert string in number using
//number_format(), function
echo number_format($num, 2);
Method 2: Using type casting: Typecasting can directly convert a string into a float, double, or integer primitive type. This is the best way to convert a string into a number without any function.
// Number in string format
$num = "1000.314";
// Type cast using int
echo (int)$num, "\n";
// Type cast using float
echo (float)$num, "\n";
// Type cast using double
echo (double)$num;
Method 3: Using intval() and floatval() Function. The intval() and floatval() functions can also be used to convert the string into its corresponding integer and float values respectively.
// Number in string format
$num = "1000.314";
// intval() function to convert
// string into integer
echo intval($num), "\n";
// floatval() function to convert
// string to float
echo floatval($num);
Method 4: By adding 0 or by performing mathematical operations. The string number can also be converted into an integer or float by adding 0 with the string. In PHP, performing mathematical operations, the string is converted to an integer or float implicitly.
// Number into string format
$num = "1000.314";
// Performing mathematical operation
// to implicitly type conversion
echo $num + 0, "\n";
// Performing mathematical operation
// to implicitly type conversion
echo $num + 0.0, "\n";
// Performing mathematical operation
// to implicitly type conversion
echo $num + 0.1;
Use this function to cast a float value from any kind of text style:
function parseFloat($value) {
return floatval(preg_replace('#^([-]*[0-9\.,\' ]+?)((\.|,){1}([0-9-]{1,3}))*$#e', "str_replace(array('.', ',', \"'\", ' '), '', '\\1') . '.\\4'", $value));
}
This solution is not dependant on any locale settings. Thus for user input users can type float values in any way they like. This is really helpful e.g. when you have a project wich is in english only but people all over the world are using it and might not have in mind that the project wants a dot instead of a comma for float values.
You could throw javascript in the mix and fetch the browsers default settings but still many people set these values to english but still typing 1,25 instead of 1.25 (especially but not limited to the translation industry, research and IT)
I was running in to a problem with the standard way to do this:
$string = "one";
$float = (float)$string;
echo $float; : ( Prints 0 )
If there isn't a valid number, the parser shouldn't return a number, it should throw an error. (This is a condition I'm trying to catch in my code, YMMV)
To fix this I have done the following:
$string = "one";
$float = is_numeric($string) ? (float)$string : null;
echo $float; : ( Prints nothing )
Then before further processing the conversion, I can check and return an error if there wasn't a valid parse of the string.
For the sake of completeness, although this question is ancient, it's worth mentioning the filter_var() set of functions, which should not only handle the formatting bits itself, but also validate or sanitise the output, thus being safer to use in the context of a form being filled in by users (or, eventually, a database that might have some corrupted/inconsistent fields):
$InvoicedUnits = (float) filter_var($InvoiceLineItem->InvoicedUnits,
FILTER_SANITIZE_NUMBER_FLOAT, FILTER_FLAG_ALLOW_FRACTION));
$pricePerUnit = (float) filter_var($InvoiceLineItem->PricePerUnit,
FILTER_SANITIZE_NUMBER_FLOAT, FILTER_FLAG_ALLOW_FRACTION));
printf("The total is: %.2f\n", $InvoicedUnits * $pricePerUnit); // both are now floats and the result is a float, formatted to two digits after the decimal sign.
This sanitises the output (which will still remain a string) and will accept the current locale's setting of the decimal separator (e.g. dot vs. comma); also, there are more options on the PHP manual for validation (which will automatically convert the result to a float if valid). The results will be slightly different for different scenarios — e.g. if you know in advance that the $InvoiceLineItem will only have valid digits and symbols for floating-point numbers, or if you need to 'clean up' the field first, getting rid of whitespace, stray characters (such as currency symbols!), and so forth.
Finally, if you wish to have nicely-formatted output — since the total is expressed in a currency — you should also take a look at the built-in NumberFormatter class, and do something like:
$InvoicedUnits = (float) filter_var($InvoiceLineItem->InvoicedUnits,
FILTER_SANITIZE_NUMBER_FLOAT, FILTER_FLAG_ALLOW_FRACTION));
$pricePerUnit = (float) filter_var($InvoiceLineItem->PricePerUnit,
FILTER_SANITIZE_NUMBER_FLOAT, FILTER_FLAG_ALLOW_FRACTION));
$fmt = new NumberFormatter('de_DE', NumberFormatter::CURRENCY);
echo 'Total is: ' . $fmt->formatCurrency($InvoicedUnits * $pricePerUnit, 'EUR') . PHP_EOL;
This will also handle thousand separators (spaces, dots, commas...) according to the configured locale, and other similar fancy things.
Also, if you wish, you can use '' (the empty string) for the default locale string (set either by the server or optionally by the browser) and $fmt->getSymbol(NumberFormatter::INTL_CURRENCY_SYMBOL) to get the default 3-letter currency code (which might not be what you want, since usually prices are given in a specific currency — these functions do not take currency exchange rates into account!).
If you need to handle values that cannot be converted separately, you can use this method:
try {
// use + 0 if you are accounting in cents
$doubleValue = trim($stringThatMightBeNumeric) + 0.0;
} catch (\Throwable $th) {
// bail here if you need to
}