is_resource
(PHP 4, PHP 5, PHP 7)
is_resource — Проверяет, является ли переменная ресурсом
Список параметров
-
var -
Проверяемая переменная.
Возвращаемые значения
Возвращает TRUE, если var является ресурсом,
или FALSE в противном случае.
Примеры
Пример #1 Пример использования is_resource()
<?php
$db_link = @mysql_connect('localhost', 'mysql_user', 'mysql_pass');
if (!is_resource($db_link)) {
die('Ошибка соединения : ' . mysql_error());
}
?>
- PHP Руководство
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- Расширения, относящиеся к переменным и типам
- Функции для работы с переменными
- boolval
- debug_zval_dump
- doubleval
- empty
- floatval
- get_defined_vars
- get_resource_type
- gettype
- import_request_variables
- intval
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- is_callable
- is_double
- is_float
- is_int
- is_integer
- is_long
- is_null
- is_numeric
- is_object
- is_real
- is_resource
- is_scalar
- is_string
- isset
- print_r
- serialize
- settype
- strval
- unserialize
- unset
- var_dump
- var_export
Коментарии
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 $possibleResource isn't a resource.
I spent way too long trying to figure this stuff out, so I hope this comment helps someone out if they run into the same problem I did.
Try this to know behavior:
<?php
function resource_test($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
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.