G
G
ganjo8882018-12-03 14:39:58
PHP
ganjo888, 2018-12-03 14:39:58

How to generate swagger via console?

I am making documentation on swagger, my code is using the Swagger-PHP library

/**
 * @OA\Info(
 *     version="1.0",
 *     title="Parser images links"
 * )
 */
/**
 * @OA\Get(
 *     path="/images",
 *     summary="Get links images",
 *         @OA\MediaType(
 *             mediaType="application/json",
 *             @OA\Schema(
 *                 @OA\Property(
 *                     property="url",
 *                     type="string"
 *                 ),
 *                 example={"url": "http://okozorko.ru"}
 *             )
 *     ),
 *     @OA\Response(
 *         response=200,
 *         description="Successful operation"
 *     ),
 *     @OA\Response(
 *         response=404,
 *         description="Not found"
 *     ),
 *     @OA\Response(
 *         response= 400,
 *         description="Bad Request"
 *     ),
 * )
 */

how to make swagger.yml file generated via console? That is, I enter the path command and I get swagger.yml

Answer the question

In order to leave comments, you need to log in

Didn't find what you were looking for?

Ask your question

Ask a Question

731 491 924 answers to any question