CODEWITHSUNDEEP
phpoopwordpress
andrew mclagan
andrew mclagan

Reputation: 337

dynamic class names in php

I have a base class called field and classes that extend this class such as text, select, radio, checkbox, date, time, number, etc.

Classes that extend field class are dynamically called in a directory recursively using include_once(). I do this so that I ( and others) can easily add a new field type only by adding a single file

What I want to know: Is there a way to substantiate a new object from one of these dynamically included extending classes from a variable name?

e.g. a class with the name checkbox :

$field_type = 'checkbox';

$field = new {$field_type}();

Maybe this would work? but it does not?

$field_type = 'checkbox';

$field = new $$field_type();

Upvotes: 32

Views: 68692

Answers (7)

zstate
zstate

Reputation: 2055

Spent some time figuring this out. From PHP documentation Namespaces and dynamic language features:

Note that because there is no difference between a qualified and a fully qualified Name inside a dynamic class name, function name, or constant name, the leading backslash is not necessary.

namespace namespacename;
class classname
{
    function __construct()
    {
        echo __METHOD__,"\n";
    }
}
function funcname()
{
    echo __FUNCTION__,"\n";
}
const constname = "namespaced";

/* note that if using double quotes, "\\namespacename\\classname" must be used */
$a = '\namespacename\classname';
$obj = new $a; // prints namespacename\classname::__construct
$a = 'namespacename\classname';
$obj = new $a; // also prints namespacename\classname::__construct

$b = 'namespacename\funcname';
$b(); // prints namespacename\funcname
$b = '\namespacename\funcname';
$b(); // also prints namespacename\funcname

echo constant('\namespacename\constname'), "\n"; // prints namespaced
echo constant('namespacename\constname'), "\n"; // also prints namespaced

Upvotes: 1

sg3s
sg3s

Reputation: 9567

This should work to instantiate a class with a string variable value:

$type = 'Checkbox'; 
$field = new $type();
echo get_class($field); // Output: Checkbox

So your code should work I'd imagine. What is your question again?

If you want to make a class that includes all extended classes then that is not possible. That's not how classes work in PHP.

Upvotes: 55

Sam_Benne
Sam_Benne

Reputation: 615

If you are using a namespace you will need to add it even if you are within the namespace.

namespace Foo;

$my_var = '\Foo\Bar';
new $my_var;

Otherwise it will not be able to get the class.

Upvotes: 26

shesek
shesek

Reputation: 4682

You can also use reflection, $class = new ReflectionClass($class_name); $instance = $class->newInstance(arg1, arg2, ...);

Upvotes: 1

RiaD
RiaD

Reputation: 47619

$field_type = 'checkbox';
$field = new $field_type;

If you need arguments:

$field_type = 'checkbox';
$field = new $field_type(5,7,$user);

Upvotes: 0

Mchl
Mchl

Reputation: 62369

This should be enough:

$field_type = 'checkbox';
$field = new $field_type();

Code I tested it with in PHP 5.3

$c = 'stdClass';

$a = new $c();

var_dump($a);

>> object(stdClass)#1 (0) {
}

Upvotes: 0

genesis
genesis

Reputation: 50976

just 

$type = 'checkbox';
$filed = new $type();

is required. you do not need to add brackets

Upvotes: 7

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