is_resource

(PHP 4, PHP 5, PHP 7, PHP 8)

is_resource — Определяет, относится ли переменная к типу ресурса

Описание

is_resource(mixed $value): bool

Функция проверяет, принадлежит ли переменная типу resource, или хранит ли переменная ссылку на открытый ресурс.

Список параметров

value: Переменная, которую оценит функция.

Возвращаемые значения

Функция возвращает true, если значение value принадлежит типу resource, иначе возвращает false.

Примеры

Пример #1 Пример определения принадлежности переменной типу ресурса функцией is_resource()

<?php

$handle = fopen("php://stdout", "w");

if (is_resource($handle)) {
    echo '$handle — ресурс';
}

?>

Результат выполнения приведённого примера:

$handle — ресурс

Примечания

Замечание:
Функция is_resource() — не строгий метод проверки типа; она вернёт false, если в параметр value передали переменную, которая хранит ссылку на закрытый ресурс.

Смотрите также

Тип resource
get_resource_type() - Возвращает тип ресурса

Улучшение страницы

Инструкция • Исправление • Сообщение об ошибке

＋Добавить

Примечания пользователей 3 notes

down

btleffler [AT] gmail [DOT] com ¶

13 years ago

I was recently trying to loop through some objects and convert them to arrays so that I could encode them to json strings.

I was running into issues when an element of one of my objects was a SoapClient. As it turns out, json_encode() doesn't like any resources to be passed to it. My simple fix was to use is_resource() to determine whether or not the variable I was looking at was a resource.

I quickly realized that is_resource() returns false for two out of the 3 resources that are typically in a SoapClient object. If the resource type is 'Unknown' according to var_dump() and get_resource_type(), is_resource() doesn't think that the variable is a resource!

My work around for this was to use get_resource_type() instead of is_resource(), but that function throws an error if the variable you're checking isn't a resource.

So how are you supposed to know when a variable is a resource if is_resource() is unreliable, and get_resource_type() gives errors if you don't pass it a resource?

I ended up doing something like this:

<?php

function isResource ($possibleResource) { return !is_null(@get_resource_type($possibleResource)); }

?>

The @ operator suppresses the errors thrown by get_resource_type() so it returns null if $possibleResource isn't a resource.

I spent way too long trying to figure this stuff out, so I hope this comment helps someone out if they run into the same problem I did.

down

Anonymous ¶

7 years ago

Note that is_resource() is unreliable. It considers closed resources as false:

<?php

$a = fopen('http://www.google.com', 'r');
var_dump(is_resource($a)); var_dump(is_scalar($a));
//bool(true)
//bool(false)

fclose($a);
var_dump(is_resource($a)); var_dump(is_scalar($a));
//bool(false)
//bool(false)

?>

That's the reason why some other people here have been confused and devised some complex (bad) "solutions" to detect resources...

There's a much better solution... In fact, I just showed it above, but here it is again with a more complete example:

<?php

$a = fopen('http://www.google.com', 'r');
var_dump(is_resource($a)); var_dump(is_scalar($a)); var_dump(is_object($a)); var_dump(is_array($a)); var_dump(is_null($a));
//bool(true)
//bool(false)
//bool(false)
//bool(false)
//bool(false)

?>

So how do you check if something is a resource?

Like this!

<?php

$a = fopen('http://www.google.com', 'r');
$isResource = is_resource($a) || ($a !== null && !is_scalar($a) && !is_array($a) && !is_object($a));
var_dump($isResource);
//bool(true)

fclose($a);

var_dump(is_resource($a));
//bool(false)

$isResource = is_resource($a) || ($a !== null && !is_scalar($a) && !is_array($a) && !is_object($a));
var_dump($isResource);
//bool(true)

?>

How it works:

- An active resource is a resource, so check that first for efficiency.
- Then branch to check what the variable is NOT:
- A resource is never NULL. (We do that check via `!== null` for efficiency).
- A resource is never Scalar (int, float, string, bool).
- A resource is never an array.
- A resource is never an object.
- Only one variable type remains if all of the above checks succeeded: IF it's NOT any of the above, then it's a closed resource!

Just surfed by and saw the bad and hacky methods other people had left, and wanted to help out with this proper technique. Good luck, everyone!

PS: The core problem is that is_resource() does a "loose" check for "living resource". I wish that it had a $strict parameter for "any resource" instead of these user-workarounds being necessary.

down

CertaiN ¶

10 years ago

Try this to know behavior:



<?php

function resource_test($resource, $name) {

    echo 

        '[' . $name. ']',

        PHP_EOL,

        '(bool)$resource => ',

        $resource ? 'TRUE' : 'FALSE',

        PHP_EOL,

        'get_resource_type($resource) => ',

        get_resource_type($resource) ?: 'FALSE',

        PHP_EOL,

        'is_resource($resource) => ',

        is_resource($resource) ? 'TRUE' : 'FALSE',

        PHP_EOL,

        PHP_EOL

    ;

}

 

$resource = tmpfile();

resource_test($resource, 'Check Valid Resource');

 

fclose($resource);

resource_test($resource, 'Check Released Resource');

 

$resource = null;

resource_test($resource, 'Check NULL');

?>



It will be shown as...



[Check Valid Resource]

(bool)$resource => TRUE

get_resource_type($resource) => stream

is_resource($resource) => TRUE



[Check Released Resource]

(bool)$resource => TRUE

get_resource_type($resource) => Unknown

is_resource($resource) => FALSE



[Check NULL]

(bool)$resource => FALSE

get_resource_type($resource) => FALSE

Warning:  get_resource_type() expects parameter 1 to be resource, null given in ... on line 10

is_resource($resource) => FALSE

＋Добавить