PHP: Convert a string into an integer.
In this tutorial, we will show you how to convert a string into an integer using PHP.
There are two ways of doing this.
You can either cast the string as an int, or you can use the function intval.
Casting a string to an int.
To cast a string to an integer value, you can use what is known as “type casting”:
//Our string, which contains the //number 2. $num = "2"; //Cast it to an integer value. $num = (int) $num; //var_dump it out to the page var_dump($num);
If you run the code snippet above, you will see that the output is: int 2
Using the intval function.
You can also use the built-in PHP function intval, which returns a variable’s integer value:
//Our string variable. $num = "2"; //Use the intval function to convert it $num = intval($num); //var_dump it out to the page var_dump($num);
The snippet above will print out the exact same result as the code in the first “type casting” example.
Where are my decimal places gone?
An integer is a whole number. It is not a fraction. Therefore, any decimal places will be lost during the conversion.
That means that 2.22 will become 2 and 7.67 will become 7.
Note that intval will not round the decimals to the closest integer. Instead, it will always round it down. For example, 9.99 will become 9, not 10.
If you need to convert your string into a number that retains its decimal places, then you can check out our guide on parsing a string as a float value.
Do I really need to convert my PHP strings into integers?
In most cases, you will not need to convert your strings into integer values.
This is because, unlike other programming languages, PHP is weakly typed and will automatically “figure it out” for you.
For example, if you are doing simple math, then PHP will automatically convert the variable for you.
Take the following example, where we get the sum of two string variables that contain numbers:
//Number A $numA = "9"; //Number B $numB = "2"; //Get the sum of our two numbers. $sum = $numA + $numB; //Print out the result var_dump($sum);
If you run the code above, you will see that the output is: int 11.
This is because PHP looked at our two variables and presumed that we were trying to get the sum of two integer values.
Note that you can technically use this “feature” to convert a string into an int:
$numA = "2"; $numA = $numA + 0; var_dump($numA); //becomes int 2
Although this code will work, we recommend that you use one of the other two methods instead.
The problem with this approach is that the intent of your code will be far less clear to others.
In other words, it may confuse other developers.
Php convert string to long
Cast a string to binary using PHP < 5.2.1
I found it tricky to check if a posted value was an integer.
is_int ( $_POST [ ‘a’ ] ); //false
is_int ( intval ( «anything» ) ); //always true
?>
A method I use for checking if a string represents an integer value.
$foo [ ‘ten’ ] = 10 ; // $foo[‘ten’] is an array holding an integer at key «ten»
$str = » $foo [ ‘ten’]» ; // throws T_ENCAPSED_AND_WHITESPACE error
$str = » $foo [ ten ] » ; // works because constants are skipped in quotes
$fst = (string) $foo [ ‘ten’ ]; // works with clear intention
?>
It seems (unset) is pretty useless. But for people who like to make their code really compact (and probably unreadable). You can use it to use an variable and unset it on the same line:
$hello = ‘Hello world’ ;
print $hello ;
unset( $hello );
$hello = ‘Hello world’ ;
$hello = (unset) print $hello ;
?>
Hoorah, we lost another line!
It would be useful to know the precedence (for lack of a better word) for type juggling. This entry currently explains that «if either operand is a float, then both operands are evaluated as floats, and the result will be a float» but could (and I think should) provide a hierarchy that indicates, for instance, «between an int and a boolean, int wins; between a float and an int, float wins; between a string and a float, string wins» and so on (and don’t count on my example accurately capturing the true hierarchy, as I haven’t actually done the tests to figure it out). Thanks!
May be expected, but not stated ..
Casting to the existing (same) type has no effect.
$t = ‘abc’; // string ‘abc’
$u=(array) $t; // array 0 => string ‘abc’ $v=(array) $u; // array 0 => string ‘abc’
Correct me if I’m wrong, but that is not a cast, it might be useful sometimes, but the IDE will not reflect what’s really happening:
class MyObject /**
* @param MyObject $object
* @return MyObject
*/
static public function cast ( MyObject $object ) return $object ;
>
/** Does nothing */
function f () <>
>
class X extends MyObject /** Throws exception */
function f () < throw new exception (); >
>
$x = MyObject :: cast (new X );
$x -> f (); // Your IDE tells ‘f() Does nothing’
?>
However, when you run the script, you will get an exception.
In my much of my coding I have found it necessary to type-cast between objects of different class types.
More specifically, I often want to take information from a database, convert it into the class it was before it was inserted, then have the ability to call its class functions as well.
The following code is much shorter than some of the previous examples and seems to suit my purposes. It also makes use of some regular expression matching rather than string position, replacing, etc. It takes an object ($obj) of any type and casts it to an new type ($class_type). Note that the new class type must exist:
Looks like type-casting user-defined objects is a real pain, and ya gotta be nuttin’ less than a brain jus ta cypher-it. But since PHP supports OOP, you can add the capabilities right now. Start with any simple class.
class Point protected $x , $y ;
public function __construct ( $xVal = 0 , $yVal = 0 ) $this -> x = $xVal ;
$this -> y = $yVal ;
>
public function getX () < return $this ->x ; >
public function getY () < return $this ->y ; >
>
$p = new Point ( 25 , 35 );
echo $p -> getX (); // 25
echo $p -> getY (); // 35
?>
Ok, now we need extra powers. PHP gives us several options:
A. We can tag on extra properties on-the-fly using everyday PHP syntax.
$p->z = 45; // here, $p is still an object of type [Point] but gains no capability, and it’s on a per-instance basis, blah.
B. We can try type-casting it to a different type to access more functions.
$p = (SuperDuperPoint) $p; // if this is even allowed, I doubt it. But even if PHP lets this slide, the small amount of data Point holds would probably not be enough for the extra functions to work anyway. And we still need the class def + all extra data. We should have just instantiated a [SuperDuperPoint] object to begin with. and just like above, this only works on a per-instance basis.
C. Do it the right way using OOP — and just extend the Point class already.
class Point3D extends Point protected $z ; // add extra properties.
public function __construct ( $xVal = 0 , $yVal = 0 , $zVal = 0 ) parent :: __construct ( $xVal , $yVal );
$this -> z = $zVal ;
>
public function getZ () < return $this ->z ; > // add extra functions.
>
$p3d = new Point3D ( 25 , 35 , 45 ); // more data, more functions, more everything.
echo $p3d -> getX (); // 25
echo $p3d -> getY (); // 35
echo $p3d -> getZ (); // 45
?>
Once the new class definition is written, you can make as many Point3D objects as you want. Each of them will have more data and functions already built-in. This is much better than trying to beef-up any «single lesser object» on-the-fly, and it’s way easier to do.
Re: the typecasting between classes post below. fantastic, but slightly flawed. Any class name longer than 9 characters becomes a problem. SO here’s a simple fix:
function typecast($old_object, $new_classname) if(class_exists($new_classname)) // Example serialized object segment
// O:5:»field»:9: $old_serialized_prefix = «O:».strlen(get_class($old_object));
$old_serialized_prefix .= «:\»».get_class($old_object).»\»:»;
$old_serialized_object = serialize($old_object);
$new_serialized_object = ‘O:’.strlen($new_classname).’:»‘.$new_classname . ‘»:’;
$new_serialized_object .= substr($old_serialized_object,strlen($old_serialized_prefix));
return unserialize($new_serialized_object);
>
else
return false;
>
Thanks for the previous code. Set me in the right direction to solving my typecasting problem. 😉
If you have a boolean, performing increments on it won’t do anything despite it being 1. This is a case where you have to use a cast.
I have 1 bar.
I now have 1 bar.
I finally have 2 bar.
Checking for strings to be integers?
How about if a string is a float?
/* checks if a string is an integer with possible whitespace before and/or after, and also isolates the integer */
$isInt = preg_match ( ‘/^\s*(3+)\s*$/’ , $myString , $myInt );
echo ‘Is Integer? ‘ , ( $isInt ) ? ‘Yes: ‘ . $myInt [ 1 ] : ‘No’ , «\n» ;
/* checks if a string is an integer with no whitespace before or after */
$isInt = preg_match ( ‘/^5+$/’ , $myString );
echo ‘Is Integer? ‘ , ( $isInt ) ? ‘Yes’ : ‘No’ , «\n» ;
/* When checking for floats, we assume the possibility of no decimals needed. If you MUST require decimals (forcing the user to type 7.0 for example) replace the sequence:
4+(\.2+)?
with
8+\.3+
*/
/* checks if a string is a float with possible whitespace before and/or after, and also isolates the number */
$isFloat = preg_match ( ‘/^\s*(3+(\.3+)?)\s*$/’ , $myString , $myNum );
echo ‘Is Number? ‘ , ( $isFloat ) ? ‘Yes: ‘ . $myNum [ 1 ] : ‘No’ , «\n» ;
/* checks if a string is a float with no whitespace before or after */
$isInt = preg_match ( ‘/^8+(\.8+)?$/’ , $myString );
echo ‘Is Number? ‘ , ( $isFloat ) ? ‘Yes’ : ‘No’ , «\n» ;