如何在Swashbuckle中为包含OWIN的WebApi 2生成Swagger文档中包含类和属性描述？

Question

在你想到它之前,这是不一样的.

我认为这应该是非常自我解释的.我想在Swagger文档中包含类描述.我的Swagger配置如下所示:

config.EnableSwagger(c =>
{
    c.SingleApiVersion("v1", "My Api Name");
    c.OperationFilter<AddAuthorizationHeaderParameterOperationFilter>();
    c.IncludeXmlComments(GetXmlCommentsPath());

}).EnableSwaggerUi(c => { });

而且MyAwesomeController看起来是这样的:

/// <summary>
/// Controller description (is included by Swashbuckle)
/// </summary>
public class MyAwesomeController : ApiController
{
    /// <summary>
    /// Method description (is included by Swashbuckle)
    /// </summary>
    public IHttpActionResult Get()
    {
        return Ok("hello... from the other side");
    }

    public IHttpActionResult Post([FromBody]MyAwesomeModel model)
    {
        return Ok("hello... from the other side");
    }
}

我MyAwesomeModel看起来像这样:

/// <summary>
/// **I would like this to be included in the Swagger description of the parameter**
/// </summary>
public class MyAwesomeModel
{
    /// <summary>
    /// **I would like this to be included in the Swagger description of the parameter**
    /// </summary>
    public string MyProperty { get; set; }
}

这可能不雇用Sr.Skeet吗？

Answer 1

嗯...所以也许如果其他人碰到这个.

基本上我找到了一种方法可以完成,我意识到为什么默认情况下没有这样做.不确定它是否是最好的方法,但在这里它.

在我的解决方案中,POCO位于与实际API分开的项目中,因此MyAwesomeModel未包含注释描述,因为没有为类和属性生成XML节点.因此,在我的POCO所在的单独项目中,我修改了属性以生成XML.

1. 为POCO所在的项目生成XML

2. 确保将XML复制到您Swashbuckle想要查找的任何路径.我Post-build event command line在项目属性中使用过;

copy "$(SolutionDir)MyAwesomeProjectWithPocos\bin\MyAwesomeProjectWithPocos.xml" "$(ProjectDir)\bin\MyAwesomeProjectWithPocos.xml"

3. 修改SwaggerConfig以包含此XML

即

config.EnableSwagger(c =>
{
    c.SingleApiVersion("v1", "My Api Name");
    c.OperationFilter<AddAuthorizationHeaderParameterOperationFilter>();
    c.IncludeXmlComments(GetXmlCommentsPath());
    c.IncludeXmlComments(GetXmlCommentsPathForModels());

}).EnableSwaggerUi(c => { });

现在,在Swagger页面上,如果我切换Model Schema到ModelI,现在可以阅读整个模型和属性描述.

当然,没有要求复制XML文件,可能只是指向步骤#3中的正确位置,GetXmlCommentsPathForModels());但这是我的选择.