settype

(PHP 4, PHP 5, PHP 7)

settype设置变量的类型

说明

settype ( mixed &$var , string $type ) : bool

将变量 var 的类型设置成 type

参数

var

要转换的变量。

type

type 的可能值为:

  • "boolean" (或为"bool",从 PHP 4.2.0 起)
  • "integer" (或为"int",从 PHP 4.2.0 起)
  • "float" (只在 PHP 4.2.0 之后可以使用,对于旧版本中使用的"double"现已停用)
  • "string"
  • "array"
  • "object"
  • "null" (从 PHP 4.2.0 起)

返回值

成功时返回 TRUE, 或者在失败时返回 FALSE

范例

Example #1 settype() 示例

<?php
$foo 
"5bar"// string
$bar true;   // boolean

settype($foo"integer"); // $foo 现在是 5   (integer)
settype($bar"string");  // $bar 现在是 "1" (string)
?>

注释

Note:

Maximum value for "int" is PHP_INT_MAX.

参见

User Contributed Notes

geoffsmiths at hotmail dot com 24-Dec-2019 10:23
Please note:

When using settype to convert indexed arrays to objects, the properties of the typed object will be integers:

A brief example:

$a = ['1', '2'];

settype($a, 'object');

var_dump($a);

// output
object(stdClass)#1 (2) {
  ["0"]=>
  string(1) "1"
  ["1"]=>
  string(1) "2"
}
Anonymous 04-Jan-2019 01:41
If you attempt to convert the special $this variable from an instance method (only in classes) :
* PHP will silently return TRUE and leave $this unchanged if the type was 'bool', 'array', 'object' or 'NULL'
* PHP will generate an E_NOTICE if the type was 'int', 'float' or 'double', and $this will not be casted
* PHP will throw a catchable fatal error when the type is 'string' and the class does not define the __toString() method
Unless the new variable type passed as the second argument is invalid, settype() will return TRUE. In all cases the object will remain unchanged.
<?php
   
// This was tested with PHP 7.2
   
class Foo {
        function
test() {
           
printf("%-20s %-20s %s\n", 'Type', 'Succeed?', 'Converted');
           
           
// settype() should throw a fatal error, as $this cannot be re-assigned
           
printf("%-20s %-20s %s\n", 'bool', settype($this, 'bool'), print_r($this, TRUE));
           
printf("%-20s %-20s %s\n", 'int', settype($this, 'int'), print_r($this, TRUE));
           
printf("%-20s %-20s %s\n", 'float', settype($this, 'float'), print_r($this));
           
printf("%-20s %-20s %s\n", 'array', settype($this, 'array'), print_r($this, TRUE));
           
printf("%-20s %-20s %s\n", 'object', settype($this, 'object'), print_r($this, TRUE));
           
printf("%-20s %-20s %s\n", 'unknowntype', settype($this, 'unknowntype'), print_r($this, TRUE));
           
printf("%-20s %-20s %s\n", 'NULL', settype($this, 'NULL'), print_r($this, TRUE));
           
printf("%-20s %-20s %s\n", 'string', settype($this, 'string'), print_r($this, TRUE));
        }
    }
   
$a = new Foo();
   
$a->test();
?>
Here is the result :
Type                 Succeed?             Converted
bool                 1                    Foo Object
(
)

Notice: Object of class Foo could not be converted to int in C:\php\examples\oop-settype-this.php on line 9

int                  1                    Foo Object
(
)

Notice: Object of class Foo could not be converted to float in C:\php\examples\oop-settype-this.php on line 10

float                1                    Foo Object
(
)

array                1                    Foo Object
(
)

object               1                    Foo Object
(
)

Warning: settype(): Invalid type in C:\php\examples\oop-settype-this.php on line 14

unknowntype                               Foo Object
(
)

NULL                 1                    Foo Object
(
)

Catchable fatal error: Object of class Foo could not be converted to string in C:\php\examples\oop-settype-this.php on line 15

If the class Foo implements __toString() :
<?php
   
class Foo {
       
// ...
       
function __toString() {
            return
'Foo object is awesome!';
        }
       
// ...
   
}
?>
So the first code snippet will not generate an E_RECOVERABLE_ERROR, but instead print the same string as for the other types, and not look at the one returned by the __toString() method.

Hope this helps !  :)
Skayo 26-Jul-2017 08:59
$foo = "1";
settype($foo, "bool");
var_dump($foo); // Outputs: bool(true)

$bar = "0";
settype($bar, "bool");
var_dump($bar); // Outputs: bool(false)
marjune 10-Nov-2016 05:31
any digit except 0 or -0 are considered true in boolean, and any string except '0' or '' are also considered true.

<?php

$foo
= '0';
settype($foo, 'boolean');
var_dump($foo); // false

$foo = 0;
settype($foo, 'boolean');
var_dump($foo); // false
Special Notes 26-Feb-2012 12:00
Note that you can't use this to convert a string 'true' or 'false' to a boolean variable true or false as a string 'false' is a boolean true. The empty string would be false instead...

<?php
$var
= "true";
settype($var, 'bool');
var_dump($var); // true

$var = "false";
settype($var, 'bool');
var_dump($var); // true as well!

$var = "";
settype($var, 'bool');
var_dump($var); // false
?>
Chris Sullins 18-Apr-2010 11:59
settype() has some really strange, potentially buggy behavior.

As noted by Michael Benedict, using settype() on a variable will initialize that variable.  What is stranger is that using settype() on an uninitialized variable that you are treating as an array or object will also initialize the variable.  So:

<?php
settype
($foo->bar,"integer"); // stdClass Object ( [test] => 0 )
?>

This works for a chain of any length: $foo->bar['baz']->etc

Next we look at what happens if $foo is already set.

<?php
$foo
= false;
settype($foo->bar,"integer"); // stdClass Object ( [test] => 0 )
?>

In and of itself, this wouldn't be problematic.  It might even make sense.  But in all other cases where $foo is defined, even if (boolean) $foo === false, it will throw an error unless $foo->bar is valid (i.e. $foo is an object already).

<?php
$foo
= true;
settype($foo->bar,"integer"); // Notice: Trying to get property of non-object
?>
NWdev 17-Nov-2008 09:38
In trying to convert an array of strings to an array of ints,
I attempted to use settype with array_walk.

<?php
//$numArray is generated by another process
$numArray = array('13','14','33');

var_dump($numArray);

//my conversion function
function str_to_int($val){
 
//remember: settype($x, 'int') returns boolean (1=success, 0=failure)
  //--> so return $x to return new value
   
settype($val,'int');
    echo
"<br />gettype = ".gettype($val)."<br />";
    return
$val;
}

array_walk($numArray,'str_to_int');

var_dump($numArray);
?>

The var_dumps both return the following:
<?php
array(3) { [0]=> string(2) "13" [1]=> string(2) "14" [2]=> string(2) "33" }
?>

The gettype echo will show the value as an integer.

So it seems that settype($val,'int') makes the conversion,
but the function return value remains a string.
Since settype returns a boolean, using
<?php $val = settype($val, 'int'); ?>
is not a option.

I resolved my array value conversion using this instead:
<?php
$numArray
=
     
array_map(create_function('$value', 'return (int)$value;'),$numArray);
?>
Thanks to the posting here:
http://usrportage.de/archives/
808-Convert-an-array-of-strings-into-an-array-of-integers.html

Perhaps this will save someone else spinning wheels a bit.

Also thanks to robin at barafranca dot com for
pointing out the boolean return value of settype.
robin at barafranca dot com 05-Mar-2008 04:20
Just a quick note, as this caught me out very briefly:

settype() returns bool, not the typecasted variable - so:

$blah = settype($blah, "int"); // is wrong, changes $blah to 0 or 1
settype($blah, "int"); // is correct

Hope this helps someone else who makes a mistake.. ;)
ludvig dot ericson gmail.dot com 09-Apr-2006 08:36
To matt:
This function accepts a paremeter, which does not imply you using hardcoded stuff, instead you can let the user choose! \o/

As a part of a framework or something.

Plus, you can probably call this with call_user_func
matt at mattsoft dot net 06-Dec-2005 09:49
using (int) insted of the settype function works out much better for me. I have always used it. I personally don't see where settype would ever come in handy.
Michael Benedict 29-Oct-2005 10:55
note that settype() will initialize an undefined variable.  Therefore, if you want to preserve type and value, you should wrap the settype() call in a call to isset().

<?php
settype
($foo, "integer");
echo(
"|$foo|");
?>

prints "|0|", NOT "||".

To get the latter, use:
<?php
if(isset($foo)) settype($foo, "integer");
echo(
"|$foo|");
?>
nospamplease at veganismus dot ch 22-Jul-2005 07:38
you must note that this function will not set the type permanently! the next time you set the value of that variable php will change its type as well.
reinier_deblois at hotmail dot com 13-Mar-2005 09:18
Instead of settype you could use:
<?php

$int
=593// $int is a integer

$int.="";   // $int is now a string
memandeemail at gmail dot com 09-Dec-2004 05:17
/**
    * @return bool
    * @param array[byreference] $values
    * @desc Convert an array or any value to Escalar Object [not tested in large scale]
    */
    function setobject(&$values) {
        $values = (object) $values;
        foreach ($values as $tkey => $val) {
            if (is_array($val)) {
                setobject($val);
                $values->$tkey = $val;
            }
        }
        return (bool) $values;
    }
sdibb at myway dot com 06-Sep-2003 10:03
Using settype is not the best way to convert a string into an integer, since it will strip the string wherever the first non-numeric character begins.  The function intval($string) does the same thing.

If you're looking for a security check, or to strip non-numeric characters (such as cleaning up phone numbers or ZIP codes),  try this instead:

<?
     $number=ereg_replace("[^0-9]","",$number);
?>
ns at canada dot com 05-May-2000 04:38
This settype() behaviour seems consistent to me. Quoting two sections from the manual:

"When casting from a scalar or a string variable to an array, the variable will become the first element of the array: "
<pre>
2 $var = 'ciao';
3 $arr = (array) $var;
4 echo $arr[0];  // outputs 'ciao'
</pre>

And if (like your code above) you do a settype on an empty variable, you'll end up with a one element array with an empty (not unset!) first element. So appeanding to it will start appending at index 1. As for why reset() doesn't do anything:

"When you assign a value to an array variable using empty brackets, the value will be added onto the end of the array."

It doesn't matter where the array counter is; values are added at the end, not at the counter.