Question

我无法使用我的模型;两者都以“评论”的方式和代码方式。

任何人都可以帮帮我吗？操作注释中的“类型”字段不会产生任何影响。

    $swagger = $this->app["swagger"];

    $swagger->models["user"] = new \Swagger\Annotations\Model([
        "id" => "user",
        "properties" => array(
            new \Swagger\Annotations\Property([
                "name" => "test", "type"=>"string"
            ])
        )
    ]);

    $path="/users";
    $name="user";
    $serivce = "Users";

    $swagger->registry[$path] = new \Swagger\Annotations\Resource([
        'resourcePath' => $path,
        'basePath'=>"/".$this->app['api.version'],
        'apis' => array(
            new \Swagger\Annotations\API([
                'path' => $path,
                'operations' => array(
                    new \Swagger\Annotations\Operation([
                        'method' => "GET",
                        'summary' => "Returns a list of ".ucfirst($service),
                        'nickname' => "GetAll".ucfirst($service),
                        'type' => "user",
                    ]),
                )
            ])
        )
    ]);

Answer 1

是的，我明白了..

$swagger->registry[$path] = new \Swagger\Annotations\Resource([
            'resourcePath' => $path,
            'basePath'=>"/api/".$this->app['api.version'],
            'models' => ($swagger->models[$name]->id ? array($name=>$swagger->models[$name]) : []),
            ....

Swagger PHP，模型不起作用

1 个答案: