SwashBuckle 定义错误时的响应主体

SwashBuckle define Response Body on Error

我的应用程序是 ASP.NET Core 1.0 Web API。

我有以下控制器:

    [HttpGet("{someData:MinLength(5):MaxLength(5)}")]
    [Produces("application/json")]
    public async Task<IActionResult> GetSomeData(string someData)
    {
        return this.Ok(JsonConvert.SerializeObject("Data is: " + someData));
    }

每当我传递例如字符串“111”时,swagger 都会向我显示以下消息:

我怎样才能获得像这样的响应主体:

"Please enter 5 numbers"

谢谢

您可以使用 [ProducesResponseType(typeof(ModelStateDictionary), (int)HttpStatusCode.OK)] 到 return 注释您的操作,典型的字典类型错误消息就像使用 return BadRequest(ModelState).

时一样

但是您不能 return 模式定义中的文本。它用于 json 结构,而不用于错误消息。相反,您应该使用 xmldoc(并大摇大摆地启用它)来为参数添加描述。

更新:添加参数及其含义文档的替代方法

/// <summary>
/// Returns some data based on <paramref name="someData"/> parameter.
/// </summary>
/// <param name="someData">Some data. (Must be exactly 5 characters wide)</param>
/// <response code="200">Returns indexed tags on success</response>
/// <response code="400">Invalid data sent</response>
/// <returns>A paged list of results</returns>
[HttpGet("{someData:MinLength(5):MaxLength(5)}")]
[ProducesResponseType(typeof(MyReturnType), (int)HttpStatusCode.OK)]
[ProducesResponseType(typeof(void), (int)HttpStatusCode.BadRequest)]
public async Task<IActionResult> GetSomeData(string someData)
{
}

您还需要在项目的属性中启用 xmldocs 的构建。

并添加到您的启动:

services.AddSwaggerGen(options =>
{
    ...
    var appEnv = PlatformServices.Default.Application;

    options.IncludeXmlComments(Path.Combine(appEnv.ApplicationBasePath, $"{appEnv.ApplicationName}.xml"));
    ...
});

我正在使用 ApiBadRequestResponse class ,它从 modelState 而不是直接 ModelState 构造字符串列表。 class 复制自一篇不错的文章 https://www.devtrends.co.uk/blog/handling-errors-in-asp.net-core-web-api

使用示例:

 [ProducesResponseType(typeof(ApiBadRequestResponse), (int)HttpStatusCode.BadRequest)]
 public async Task<IActionResult> GetSomeData(string someData)
 {
   return BadRequest(new ApiBadRequestResponse(ModelState));
 }

实施(基础class描述in the article):

/// <summary>
    ///  an ApiBadRequestResponse class to handle validation errors from modelState or exception.
    /// </summary>
    public class ApiBadRequestResponse : ApiResponse
    {
        public IEnumerable<string> Errors { get; }

        public ApiBadRequestResponse(ModelStateDictionary modelState)
            : base(HttpStatusCode.BadRequest) //400)
        {
            if (modelState.IsValid)
            {
                throw new ArgumentException("ModelState must be invalid", nameof(modelState));
            }

            Errors = modelState.SelectMany(x => x.Value.Errors)
                .Select(x => x.ErrorMessage).ToArray();
        }

        public ApiBadRequestResponse(Exception exception, string message=null)
            : base(HttpStatusCode.BadRequest, message) //400)
        {
            Errors = new List<string>() {exception.ToString()};
        }
    }

另一个类似的解决方案在文章 http://www.khalidabuhakmeh.com/a-better-validation-result-for-asp-net-webapi