PHP convert array to resource

Question

For visual representation, for simplicity and of course to feed my curiosity, I'm wondering how to convert a PHP array into a valid PHP resource.

See the below example: enter image description here (example image created with dBug component available at http://dbug.ospinto.com/)

I've made 3 examples:

resource: this is the typical representation of a MySQL resource, visualized as a grid
object: a handmade create object from an array
array: a handmade multidimensional array

As you can see, the resource is a visual beauty, while the object and array are constructed by using multidimensional arrays, using poor numeric array indexes to bind them together :(

What I'm looking for, would probably be something like this:

$resource_var = (resource) $array_var;

This seems to me it is not a PHP-related question, but rather a dbug-related question. Is your question "How to make dbug display the content of an object or an array the same way it displays the content of a resource?" — Jocelyn
– Jocelyn, Commented Aug 24, 2012 at 14:31
@Jocelyn I read that too quickly and thought you said "...rather a drug-related question." I don't think you would have been too far off. — Matt
– Matt, Commented Aug 24, 2012 at 14:35
Only you have a resource does not mean you will get beautiful output. Try with a resource handle of a CSV file and your dbug extension. — hakre
– hakre, Commented Aug 24, 2012 at 14:40
Thanks for your replies guys! This is giving me new clues to work on. However, I really like the data to be sorted internally into the query grid (with rows and columns) but maybe I'm pushing to hard :) — hardcoder
– hardcoder, Commented Aug 24, 2012 at 14:48

Community · Accepted Answer · 2020-06-20 09:12:55Z

8

What I'm looking for, would probably something like this:
$resource_var = (resource) $array(var)

You will never find that. A resource is an internal data-type in PHP. If (and only if) you write yourself a PHP extension and load it, you could do the following:

$resource = array_resource_create($array);

Your PHP extension then would create that resource (as the mysql extension for example creates its specific resource type) within that array_resource_create function. However, it would be useless, because there is no other function so far that could deal with that resource.

edited Jun 20, 2020 at 9:12

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answered Aug 24, 2012 at 14:28

hakre

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2 Comments

hardcoder Over a year ago

Thanks for your replies guys! This is giving me new clues to work on. However, I really like the data to be sorted internally into the query grid (with rows and columns) but maybe I'm pushing to hard :)

hakre Over a year ago

Write your own a function for that and/or extend the dbug library you use. You would only need to check if it is an array and if so it contains only of arrays that have the same keys -or- objects that have the same properties. That is also a nice tutorial you will get in extending existing code.

Gérald PLUSQUELLEC · Accepted Answer · 2015-07-02 16:06:41Z

2

You can't create resource. but you can use native one.

Try with curl for example.

function makeResourceFromArray($array) {

    $resource = curl_init();

    curl_setopt($resource, CURLOPT_PRIVATE, serialize($array));

    return $resource;
}

function makeArrayFromResource($resource) {
    return unserialize(curl_getinfo($resource, CURLINFO_PRIVATE));
}

$resource = makeResourceFromArray(['name' => 'test']);

$array = makeArrayFromResource($resource);

answered Jul 2, 2015 at 16:06

Gérald PLUSQUELLEC

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1 Comment

kaiser Over a year ago

That will lead to a resource of type curl. Not really what the OP wants.

IMSoP · Accepted Answer · 2012-08-24 14:32:06Z

1

The output you show there is nothing to do with it being a resource as such, but the pretty-print function you're using noticing that the variable you've given it points at a database result set, and fetching and displaying the results.

What PHP means by a resource is that the variable doesn't actually hold data within PHP, but is a pointer or reference usable by some lower-level module of code - in this case, a DB library which can use that reference to retrieve the results of the executed query.

If you just want the pretty-print to look similar for an array with a DB-resultset-like structure, then you should simply modify the pretty-print function to do so - you don't need to do anything to the array itself.

answered Aug 24, 2012 at 14:32

IMSoP

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Comments

tfont · Accepted Answer · 2017-02-23 09:15:09Z

0

A resource is a special type. And a resource is specific to a source that's external. Therefore going backwards wouldn't be possible.

Theoretically, an interface with an instance of the resource would help manage the type - but this is just nonsense theoretical talk that is impossible in PHP.

answered Feb 23, 2017 at 9:15

tfont

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