intval

(PHP 4, PHP 5, PHP 7)

intval변수의 정수값을 얻습니다

설명

int intval ( mixed $var [, int $base ] )

지정한 base 값에 따른 varinteger값을 반환합니다. (기본 진법은 10입니다)

인수

var

정수로 변환할 스칼라 값

base

변환에 사용할 진법 (기본값은 10진법)

반환값

성공시엔 var의 정수 값, 실패시엔 0. 빈 배열과 객체는 0을 반환하고, 비어있지 않은 배열과 객체는 1을 반환합니다.

최대값은 시스템에 의존합니다. 32비트 시스템은 부호 있는 정수의 최대값 범위로 -2147483648에서 2147483647을 가집니다. 그러한 시스템에서 intval('1000000000000') 등은 2147483647을 반환합니다. 64비트 시스템에서 부호 있는 정수의 최대값은 9223372036854775807입니다.

문자열은 대부분 0을 반환하지만, 이는 문자열의 가장 왼쪽 문자들에 의해 결정됩니다. 정수형 변환의 기본 규칙이 적용됩니다.

예제

Example #1 intval() 예제

다음 예제는 32비트 시스템입니다.

echo intval(42);                      // 42
echo intval(4.2);                     // 4
echo intval('42');                    // 42
echo intval('+42');                   // 42
echo intval('-42');                   // -42
echo intval(042);                     // 34
echo intval('042');                   // 42
echo intval(1e10);                    // 1410065408
echo intval('1e10');                  // 1
echo intval(0x1A);                    // 26
echo intval(42000000);                // 42000000
echo intval(420000000000000000000);   // 0
echo intval('420000000000000000000'); // 2147483647
echo intval(42, 8);                   // 42
echo intval('42', 8);                 // 34
?>

주의

Note:

var 인수가 문자열이 아니면 base 인수는 아무런 영향을 주지 않습니다.

참고

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User Contributed Notes 17 notes

up
287
Ken
10 years ago
Not mentioned elsewhere: intval(NULL) also returns 0.
up
63
leon at leonidasjp dot nl
4 years ago
It seems intval is interpreting valid numeric strings differently between PHP 5.6 and 7.0 on one hand, and PHP 7.1 on the other hand.

<?php
echo intval('1e5');
?>

will return 1 on PHP 5.6 and PHP 7.0,
but it will return 100000 on PHP 7.1.
up
115
winbill at hotmail dot com
10 years ago
Be careful :

<?php
$n
="19.99";
print
intval($n*100); // prints 1998
print intval(strval($n*100)); // prints 1999
?>
up
7
Anonymous
1 year ago
PHP 7.2

$test = intval(150.20*100); //15019
$test2 = intval(15020); //15020
$test3 = intval(15020.0); //15020
$test4 = 150.20*100; //15020.0
up
44
zak at php dot net
21 years ago
intval converts doubles to integers by truncating the fractional component of the number.

When dealing with some values, this can give odd results.  Consider the following:

print intval ((0.1 + 0.7) * 10);

This will most likely print out 7, instead of the expected value of 8.

For more information, see the section on floating point numbers in the PHP manual (http://www.php.net/manual/language.types.double.php)

Also note that if you try to convert a string to an integer, the result is often 0.

However, if the leftmost character of a string looks like a valid numeric value, then PHP will keep reading the string until a character that is not valid in a number is encountered.

For example:

"101 Dalmations" will convert to 101

"$1,000,000" will convert to 0 (the 1st character is not a valid start for a number

"80,000 leagues ..." will convert to 80

"1.4e98 microLenats were generated when..." will convert to 1.4e98

Also note that only decimal base numbers are recognized in strings.

"099" will convert to 99, while "0x99" will convert to 0.

One additional note on the behavior of intval.  If you specify the base argument, the var argument should be a string - otherwise the base will not be applied.

For Example:

print intval (77, 8);   // Prints 77
print intval ('77', 8); // Prints 63
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9
Anthony
3 years ago
The binary notation is NOT supported until php7.2
<?php
                        
// PHP <7.2 | PHP >=7.2
echo intval(0b11);       //    3     |     3
echo intval(-0b11);      //   -3     |    -3
echo intval('0b11');     //    0     |     0
echo intval('-0b11');    //    0     |     0
echo intval('0b11', 0);  //    0     |     3
echo intval('-0b11', 0); //    0     |    -3
?>
up
29
Ben Laurienti
15 years ago
You guys are going to love this.  I found something that I found quite disturbing.

$test1 = intVal(1999);

$amount = 19.99 * 100;
$test2 = intVal($amount);
$test3 = intVal("$amount");

echo $test1 . "<br />\n";
echo $test2 . "<br />\n";
echo $test3 . "<br />\n";

expected output:
1999
1999
1999

actual output
1999
1998
1999

Appears to be a floating point issue, but the number 1999 is the only number that I was able to get to do this.  19.99 is the price of many things, and for our purpose we must pass it as 1999 instead of 19.99.
up
35
spoon_reloaded at gmail dot com
12 years ago
Here is a really useful undocumented feature:

You can have it automatically deduce the base of the number from the prefix of the string using the same syntax as integer literals in PHP ("0x" for hexadecimal, "0" for octal, non-"0" for decimal) by passing a base of 0 to intval():

<?php
echo intval("0x1a", 0), "\n"; // hex; prints "26"
echo intval("057", 0), "\n"; // octal; prints "47"
echo intval("42", 0), "\n"; // decimal; prints "42"
?>
up
6
chinmay235 at gmail dot com
3 years ago
<?php
echo intval("10 days");          //10
echo intval("days 10");          //0
?>
up
2
Anony Moose
1 year ago
As a warning, do not use this function alone for input validation.

Vulnerable example:
<?php
if(isset($_GET['id']) && intval($_GET['id']) > 0){
    echo
$id;
}
?>

The following requests would pass this filter:

/page.php?id=10
/page.php?id=10oops
/page.php?id=10<script>alert(1)</script>
/page.php?id=1' OR '1'='1
/page.php?id[]=<script>alert(1)</script>

Instead use the is_numeric() function for integer validation:

<?php
echo intval("10oops"); // 10
echo is_numeric("10oops");  // false
?>

Secure example:
<?php
if(isset($_GET['id']) && is_numeric($_GET['id']) && intval($_GET['id']) > 0){
    echo
$id;
}
?>
up
15
espertalhao04 at hotmail dot com
8 years ago
if you want to take a number from a string, no matter what it may contain, here is a good solution:

<?php
function int($s){return(int)preg_replace('/[^\-\d]*(\-?\d*).*/','$1',$s);}

echo
int('j18ugj9hu0gj5hg');
//output: 18
?>
this example returns an int, so it will follow the int rules, and has support for negative values.

<?php
function int($s){return($a=preg_replace('/[^\-\d]*(\-?\d*).*/','$1',$s))?$a:'0';}

echo
int('j-1809809808908099878758765ugj9hu0gj5hg');
//output: -1809809808908099878758765
?>

this one returns a string with just the numeric value.
it also supports negative values.

the latter is better when you have a 32 bit system and you want a huge int that is higher than PHP_MAX_INT.
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6
tuxedobob at mac dot com
17 years ago
Sometimes intval just won't cut it. For example if you want to use an unsigned 32-bit int and need all 32 bits. Recently, I wrote a little script that took and integer and converted it to an IP address. After realizing I couldn't just mod the whole thing, since the sign bit throws it off (and compensating for that), we ran into a problem where if it was entered into a form, the value somehow wasn't converted to an integer properly, at least not implicitly. The solution for this, and the way I recommend converting a string to an integer, is:

$num = $num + 0;

and PHP will leave your number alone; it'll just know it's a number. Such is the fun of a loosely-typed language. :)
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4
mkamerma at science dot uva dot nl
15 years ago
As addendum, the "if ($int > 0)" check in the encode function is redundant. It doesn't do anything bad to keep it in since it will always be true when reaching that point, but it's a meaningless conditional this way. It's a remnant from when I tried to write the function in terms of bitshifts, which could lead to negative ints when shifting if the 32nd bit was set (instead of always padding with 0's when using >> it pads with 1's leading to negative ints).
up
4
pfreet at gmail dot com
8 years ago
Do not use intval() when you really want round(). This is due to how PHP handles precision.

echo number_format(8.20*100, 20), "<br />";
echo intval(8.20*100), "<br />";
echo floor(8.20*100), "<br />";
echo round(8.20*100), "<br />";

819.99999999999988631316
819
819
820
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3
yves
10 years ago
The behaviour of intval() is interesting when supplying a base, and you better check your intval base-based expressions, as it is counter-intuitive.
As the example shows
<?php
  intval
('42', 8); // => 34
 
intval(42, 8);   // => 42 !
?>
PHP considers the 42 as being already an integer, and doesn't apply any conversion. And supplying
<?php
  intval
(49, 8);  // => 49 !
?>
produces no error and no warning.
up
-1
taylorsarrafian at gmail dot com
6 years ago
beware:

<?php

  
// observe the following
  
echo intval( strval( -0.0001 ) ); // 0
  
echo intval( strval( -0.00001 ) ); // -1

   // this is because
  
echo strval( -0.0001 ); // -.0001
  
echo strval( -0.00001 ); // -1.0E-5

   // thus beware when using
  
function trunc2_bad( $n ) {
      return
intval( strval( $n * 100 ) / 100 );
   }

  
// use this instead
  
function trunc2_good( $n ) {
      return
intval( floatval( strval( $n * 100 )  ) / 100 );
   }

?>
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-3
simon at npkk dot cz
15 years ago
Still have on mind, that if you convert big numbers by adding zero, PHP makes automatic "to a float" conversion, so it is same as floatVal(). So if the number is realy big (over 13 digits), you can lose preciosity. Do not use it for such long numbers, if all bits do matter (IPv6 addresses and similar).
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