How to Use PHP’s is_object() to Distinguish Objects from Other Types

Overview

In PHP, the is_object() function is used to check whether a variable is an object.

Syntax

bool is_object (mixed $var)

Parameters

$var

: the variable to be checked.

Return Value

Returns true if $var is an object; otherwise returns false.

Example Code

// Define a class
class Person {
    public $name;

    public function __construct($name) {
        $this->name = $name;
    }
}

// Create an object
$person = new Person('John');

// Check object variable
if (is_object($person)) {
    echo 'Variable $person is an object';
} else {
    echo 'Variable $person is not an object';
}

// Define an array
$fruit = array('apple', 'banana', 'orange');

// Check array variable
if (is_object($fruit)) {
    echo 'Variable $fruit is an object';
} else {
    echo 'Variable $fruit is not an object';
}

Output

Variable $person is an object
Variable $fruit is not an object

Explanation

The code first defines a Person class with a public property $name and a constructor __construct(). An instance $person is created with the name “John”. Using is_object() on $person returns true, so the script outputs that $person is an object.

Next, an array $fruit is defined. Since $fruit is an array, is_object() returns false, and the script outputs that $fruit is not an object.

Conclusion

The is_object() function can be used to verify whether a variable is an object, allowing developers to ensure correct variable types at runtime and avoid unexpected type errors.