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Home/ Questions/Q 6362823
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Editorial Team
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Editorial Team
Asked: May 24, 20262026-05-24T23:57:31+00:00 2026-05-24T23:57:31+00:00

I need to create a variable of type resource, which is not a stream.

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I need to create a variable of type resource, which is not a stream. This means that resources created by fopen() etc. are out. Ideally it should be possible to create the resource without the help of extensions.

I would use an SQLite in-memory database handle, but sqlite is an optional PHP extension (even though it is enabled by default) which is not guaranteed to be installed on any given system (in current Ubuntu installations, for example, the sqlite extension is NOT installed by default).

Here’s an example of how the test case might be structured:

function is_stream($resource)
{
  // some code that determines whether input is stream
}

class StreamResourceTest extends PHPUnit_Framework_TestCase
{
  public function testStreamResource()
  {
    $stream = fopen(__FILE__, 'r');
    $nonStream = ???; // how can this be created?

    $this->assertTrue(is_resource($stream));
    $this->assertTrue(is_stream($stream));

    $this->assertTrue(is_resource($nonStream));
    $this->assertFalse(is_stream($nonStream));

    fclose($stream);
  }
}
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  1. Editorial Team
    Editorial Team
    2026-05-24T23:57:31+00:00Added an answer on May 24, 2026 at 11:57 pm

    stream_context_create()Docs creates a resource that is not strictly a stream resource.

    They could be considered to not require extensions since

    Streams are an integral part of PHP as of version 4.3.0. No steps are required to enable them.

    …although the same could possibly be said of sqlite, so whether you want to accept that as better alternative would be up to you.

    Also, this page might help.

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