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();

Here is Solutions:

We have many solutions to this problem, But we recommend you to use the first solution because it is tested & true solution that will 100% work for you.

Solution 1

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.

Solution 2

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.

Solution 3


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

is required. you do not need to add brackets

Solution 4

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

Solution 5

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

Solution 6

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();


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

Solution 7

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

If you need arguments:

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

Note: Use and implement solution 1 because this method fully tested our system.
Thank you 🙂

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

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