is_scalar

(PHP 4 >= 4.0.5, PHP 5)

is_scalar Prüft ob eine Variable skalar ist

Beschreibung

bool is_scalar ( mixed $var )

Prüft ob die gegebene Variable skalar ist.

Unter skalaren Variablen versteht man solche die einen integer, float, string oder boolean Wert enthalten. Variablen vom Typ array, object und resource sind nicht skalar.

Hinweis:

is_scalar() betrachtet Variablen vom Typ resource nicht als skalar da Resourcen abstrakte Datentypen darstellen auch wenn sie zur Zeit als Integer repräsentiert sind. Sie sollten sich nicht auf dieses Implementationsdetail verlassen da es sich ändern kann.

Parameter-Liste

var

Die zu prüfende Variable.

Rückgabewerte

Liefert TRUE wenn var skalar ist, sonst FALSE.

Beispiele

Beispiel #1 is_scalar() Example

<?php
function show_var($var
{
    if (
is_scalar($var)) {
        echo 
$var;
    } else {
        
var_dump($var);
    }
}
$pi 3.1416;
$proteins = array("Hämoglobin""Cytochrom C Oxidase""Ferredoxin");

show_var($pi);
show_var($proteins)

?>

Das oben gezeigte Beispiel erzeugt folgende Ausgabe:

3.1416
array(3) {
  [0]=>
  string(10) "Hämoglobin"
  [1]=>
  string(20) "Cytochrome C Oxidase"
  [2]=>
  string(10) "Ferredoxin"
}

Siehe auch

  • is_float() - Prüft, ob eine Variable vom Typ float ist
  • is_int() - Prüft, ob eine Variable vom Typ int ist
  • is_numeric() - Prüft, ob eine Variable eine Zahl oder ein numerischer String ist
  • is_real() - Alias von is_float
  • is_string() - Prüft, ob Variable vom Typ string ist
  • is_bool() - Prüft, ob eine Variable vom Typ boolean ist
  • is_object() - Prüft, ob eine Variable vom Typ object ist
  • is_array() - Prüft, ob die Variable ein Array ist

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User Contributed Notes 6 notes

up
8
Anonymous
7 years ago
Another warning in response to the previous note:
> just a warning as it appears that an empty value is not a scalar.

That statement is wrong--or, at least, has been fixed with a later revision than the one tested.  The following code generated the following output on PHP 4.3.9.

CODE:
<?php
   
echo('is_scalar() test:'.EOL);
    echo(
"NULL: "      . print_R(is_scalar(NULL),     true) . EOL);
    echo(
"false: "    . print_R(is_scalar(false),   true) . EOL);
    echo(
"(empty): "  . print_R(is_scalar(''),      true) . EOL);
    echo(
"0: "         . print_R(is_scalar(0),       true) . EOL);
    echo(
"'0': "      . print_R(is_scalar('0'),     true) . EOL);
?>

OUTPUT:
is_scalar() test:
NULL:
false: 1
(empty): 1
0: 1
'0': 1

THUS:
   * NULL is NOT a scalar
   * false, (empty string), 0, and "0" ARE scalars
up
6
Dr K
8 years ago
Having hunted around the manual, I've not found a clear statement of what makes a type "scalar" (e.g. if some future version of the language introduces a new kind of type, what criterion will decide if it's "scalar"? - that goes beyond just listing what's scalar in the current version.)

In other lanuages, it means "has ordering operators" - i.e. "less than" and friends.

It (-:currently:-) appears to have the same meaning in PHP.
up
0
efelch at gmail dot com
8 years ago
A scalar is a single item or value, compared to things like arrays and objects which have multiple values. This tends to be the standard definition of the word in terms of programming. An integer, character, etc are scalars. Strings are probably considered scalars since they only hold "one" value (the value represented by the characters represented) and nothing else.
up
-1
webmaster at oehoeboeroe dot nl
5 years ago
Here's a little function that will test whether a variable can be used as offset to an array.

<?php
function is_offset(&$var) {
    return (
is_scalar($var) || is_null($var)) && !is_resource($var);
}
?>

The resource check is currently redundant, but according to the manual that may change in the future.
up
-1
popanowel HAT hotmailZ DOT cum
10 years ago
Hi ... for newbees here, I just want to mention that reference and scalar variable aren't the same. A reference is a pointer to a scalar, just like in C or C++.

<? php  // simple reference to scalar

 
$a = 2;
 
$ref = & $a;

  echo
"$a <br> $ref";

?>
this should print out: "2 <br> 2".

Scalar class also exists. Look below:
<? php

 
class Object_t {

     var
$a;

     function
Object_t ()  // constructor
    
{
       
$this->a = 1;
     }

  }

 
$a = new Object_t; // we define a scalar object

 
$ref_a = &a;

  echo
"$a->a <br> $ref->a";

?>
again, this should echo: "1 <br> 1";

Here is another method isued in OOP to acheive on working only over reference to scalar object. Using this, you won't ever have to  ask yourself if you work on a copy of the scalar or its reference. You will only possess reference to the scalar object. If you want to duplicate the scalar object, you will have to create a function for that purpose that would read by the reference the values and assign them to another scalar of the same type... or an other type, it is as you wish at that moment.
<?php

 
class objet_t {
     var
$a;

     function
object_t
    
{
       
$this->a = "patate_poil";
     }
  }

   function &
get_ref($object_type)
   {
     
// here we create a scalar object in memory
      // and we return it by reference to the calling
      // control scope.
     
return &new $object_type;
   }

  
$ref_object_t = get_ref(object_t);

   echo
"$ref_object_t->a <br>";
 
?>
this should echo: "patate_poit <br>".

The only thing that I try to demonstrate is that scalar variable ARE object in memory while a reference is usualy a variable (scalar object) that contain the address of another scalar object, which contain the informations you want by using the reference.

Good Luck!

otek is popanowel HAT hotmailZ DOT cum
up
-3
bps7j at yahoNOSPAMo.com
10 years ago
is_scalar(null) is false.  Apparently a variable needs to have a value to be considered a scalar.
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