GuzzlePHP mock response content

Question

I want to mock a response to the Guzzle request:

 $response = new Response(200, ['X-Foo' => 'Bar']);

 //how do I set content of $response to--> "some mocked content"

 $client = Mockery::mock('GuzzleHttp\Client');
 $client->shouldReceive('get')->once()->andReturn($response);

I noticed I need to add as third parameter the interface:

 GuzzleHttp\Stream\StreamInterface

but there are so many implementations of it, and I want to return a simple string. Any ideas?

Edit: now I use this:

 $response = new Response(200, [], GuzzleHttp\Stream\Stream::factory('bad xml here'));

but when I check this:

$response->getBody()->getContents()

I get an empty string. Why is this?

Edit 2: this happened to me only when I used xdebug, when it runs normally it works great!

Why not just mock the `Response` as well?! I suppose that the behaviour of the class under test depends on the response. So you'd want to mock it too, so you'd be sure that each time the input data for your CUT is the same. — Havelock, Dec 16 '14 at 14:30

score 32 · Answer 1 · answered Jul 06 '15 at 11:52

32

We'll just keep doing this. The previous answer is for Guzzle 5, this is for Guzzle 6:

use GuzzleHttp\Psr7;

$stream = Psr7\stream_for('{"data" : "test"}');
$response = new Response(200, ['Content-Type' => 'application/json'], $stream);

answered Jul 06 '15 at 11:52

tomvo

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Looking at the [Guzzle source](https://github.com/guzzle/psr7/blob/1.2.1/src/Response.php#L96), Response does this for you automatically. Just pass a string. – Tim Nov 11 '15 at 12:44

score 12 · Accepted Answer · answered Dec 16 '14 at 22:21

12

The previous answer is for Guzzle 3. Guzzle 5 uses the following:

<?php
$body = GuzzleHttp\Stream\Stream::factory('some mocked content');
$response = new Response(200, ['X-Foo' => 'Bar'], $body);

answered Dec 16 '14 at 22:21

Michael Dowling

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I added some of the code to my post, can you please look at it? – Tzook Bar Noy Dec 17 '14 at 14:21
\GuzzleHttp\Stream\Stream::factory (backslash or use **use** statement) – Bogdan Jul 18 '15 at 14:00

score 7 · Answer 3 · answered Jan 18 '17 at 16:53

Using @tomvo answer and the comment from @Tim - this is what I did for testing Guzzle 6 inside my Laravel app:

use GuzzleHttp\Psr7\Response;

$string = json_encode(['data' => 'test']);
$response = new Response(200, ['Content-Type' => 'application/json'], $string);

$guzzle = Mockery::mock(GuzzleHttp\Client::class);
$guzzle->shouldReceive('get')->once()->andReturn($response);

score 3 · Answer 4 · answered Dec 16 '14 at 15:37

Guzzle\Http\Message\Response allows you to specify the third parameter as a string.

$body = '<html><body>Hello world!</body></html>';
$response = new Response(200, ['X-Foo' => 'Bar'], $body);

If you'd prefer a solution that implements Guzzle\Stream\StreamInterface, then I recommend using Guzzle\Http\EntityBody for the most straightforward implementation:

$body = Guzzle\Http\EntityBody::fromString('<html><body>Hello world!</body></html>');
$response = new Response(200, ['X-Foo' => 'Bar'], $body);

score 1 · Answer 5 · answered Feb 28 '22 at 10:46

1

For Guzzle 7, you can use the GuzzleHttp\Psr7\Utils::streamFor() method as follows:

$data = json_encode(['X-Foo' => 'Bar']);
$stream = Utils::streamFor($data);

And then you can pass the $stream object to the andReturn method of the mocked client.

answered Feb 28 '22 at 10:46

M074554N

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GuzzlePHP mock response content

5 Answers5