Transform a CURL POST request to GuzzleHttp Post

Question

I've got this CURL that I'm executing on the command line and it successfully creates content in my CMS (Drupal 9).

curl \
--user username:9aqW72MUbFQR4EYh \
--header 'Accept: application/vnd.api+json' \
--header 'Content-type: application/vnd.api+json' \
--request POST http://www.domain.drupal/jsonapi/node/article \
--data-binary @payload.json

and the JSON file as:

{
  "data": {
    "type": "node--article",
    "attributes": {
      "title": "My custom title",
      "body": {
        "value": "Custom value",
        "format": "plain_text"
      }
    }
  }
}

Working like a charm and data is being created. I've been trying to do this in GuzzleHttp but couldn't get it working.

Get is working: require 'vendor/autoload.php';

use GuzzleHttp\Client;
use GuzzleHttp\Psr7\Request;

$url = 'http://www.domain.drupal';

$content_client = new GuzzleHttp\Client([
    'base_uri' => $url,
    'timeout'  => 20.0,
]);

$res = $content_client->request('GET', '/jsonapi/node/article/71adf560-044c-49e0-9461-af593bad0746');

For POST, I've got probably around 10 versions trial and error but nothing worked. How can I POST my JSON/Content to Drupal or how can I correctly implement the CURL in Guzzle?

did the answer below worked out for you? – bhucho Nov 03 '20 at 15:36 — bhucho, Nov 03 '20 at 15:36

score 0 · Answer 1 · answered Nov 02 '20 at 08:13

If you want a simple post request to send json body with your headers you can do it simply without using Psr7 Request. Guzzle utilizes PSR-7 as the HTTP message interface.

use GuzzleHttp\Client;

$url = 'http://www.domain.drupal';

$content_client = new Client([
    'base_uri' => $url,
    'timeout'  => 20.0,
]);
$headers = [
    'Content-type' => 'application/vnd.api+json',
    'Accept' => 'application/vnd.api+json'
];
$payload['data'] = [
        'type' => 'node--article',
        'attributes' => [
                "title" => "My custom title",
                "body" => [
                        "value" => "Custom value",
                        "format" => "plain_text"
                    ]
            ]
    ]; 
$guzzleResponse = $content_client->post('/jsonapi/node/article/71adf560-044c-49e0-9461-af593bad0746', [
                'json' => json_encode($payload),
                'headers' => $headers
            ]);

if ($guzzleResponse->getStatusCode() == 200) {
                $response = json_decode($guzzleResponse->getBody());
}

You can write it in try catch block with using RequestException (see this Catching exceptions from Guzzle to know more about it.)

If you want to use Guzzle Pool then you can use psr7 requests directly — bhucho, Nov 02 '20 at 08:15

Transform a CURL POST request to GuzzleHttp Post

1 Answers1