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    is_resource

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

    is_resource — Finds whether a variable is a resource

    Description

    is_resource(mixed $value): bool

    Finds whether the given variable is a resource.

    Parameters

    value

    The variable being evaluated.

    Return Values

    Returns true if value is a resource, false otherwise.

    Examples

    Example #1 is_resource() example

    <?php

    $handle     = fopen("php://stdout", "w");
    if (    is_resource($handle)) {
    echo     '$handle is a resource';
    }

    ?>

    The above example will output:

    $handle is a resource

    Notes

    Note:

    is_resource() is not a strict type-checking method: it will return false if value is a resource variable that has been closed.

    See Also

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    User Contributed Notes 3 notes

    up
    22
    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.
    up
    12
    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.
    up
    5
    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
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