JavaScript is_resource

Returns true if variable is a resource

function is_resource (handle) {
    // Returns true if variable is a resource  
    // 
    // version: 1008.1718
    // discuss at: http://phpjs.org/functions/is_resource    // +   original by: Brett Zamir (http://brett-zamir.me)
    // +   improved by: Luis Salazar (http://www.freaky-media.com/)
    // *     example 1: is_resource('a');
    // *     returns 1: false
    var getFuncName = function (fn) {        var name = (/\W*function\s+([\w\$]+)\s*\(/).exec(fn);
        if (!name) {
            return '(Anonymous)';
        }
        return name[1];    };
    return !(!handle || typeof handle !== 'object' || !handle.constructor || getFuncName(handle.constructor) !== 'PHPJS_Resource');
}

external links: original PHP docs | raw js source

Examples

Running

1	is_resource('a');

Should return

false

Dependencies

No dependencies, you can use this function standalone.

Open syntax issues

php.js uses JsLint to help us keep our code consistent and prevent some common bugs.

Eventually we want all code to pass or at least take into consideration most fixes suggested by JsLint, following this JsLint configuration we’ve decided on.

Authors

Thanks to the following developers, you get to have is_resource goodness in JavaScript.

Brett Zamir ( original )
Luis Salazar ( improved )

Comments

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Use:

[CODE]
your_stuff('here');
[/CODE]

for proper code formatting
By submitting code here you are allowing us to use it in php.js hence dual licensing it under the MIT and GPL licenses

Brett Zamir
29 May '09

Good one! Changed in SVN.

Luis Salazar
28 May '09

Hi, great!!

Just one thing. Is there any special reason for using the brackets in this case, or can we just do :

return !(!handle || typeof handle !== 'object' || !handle.constructor || handle.constructor.name !== 'PHPJS_Resource');

Thanks a lot!!

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JavaScript is_resource

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