是否有一种方法显示所有枚举作为他们的字符串值在swagger而不是他们的int值?
我希望能够提交POST动作，并根据它们的字符串值放置枚举，而不必每次都查看enum。
我尝试了DescribeAllEnumsAsStrings，但服务器然后接收字符串而不是enum值，这不是我们要寻找的。
有人解决了吗?
编辑:
public class Letter
{
[Required]
public string Content {get; set;}
[Required]
[EnumDataType(typeof(Priority))]
public Priority Priority {get; set;}
}
public class LettersController : ApiController
{
[HttpPost]
public IHttpActionResult SendLetter(Letter letter)
{
// Validation not passing when using DescribeEnumsAsStrings
if (!ModelState.IsValid)
return BadRequest("Not valid")
..
}
// In the documentation for this request I want to see the string values of the enum before submitting: Low, Medium, High. Instead of 0, 1, 2
[HttpGet]
public IHttpActionResult GetByPriority (Priority priority)
{
}
}
public enum Priority
{
Low,
Medium,
High
}
要在swagger中以字符串形式显示枚举，请在ConfigureServices中添加以下行来配置JsonStringEnumConverter:
services.AddControllers().AddJsonOptions(options =>
options.JsonSerializerOptions.Converters.Add(new JsonStringEnumConverter()));
如果你想以字符串和int值的形式显示枚举，你可以尝试创建一个EnumSchemaFilter来改变模式，如下所示:
public class EnumSchemaFilter : ISchemaFilter
{
public void Apply(OpenApiSchema model, SchemaFilterContext context)
{
if (context.Type.IsEnum)
{
model.Enum.Clear();
Enum.GetNames(context.Type)
.ToList()
.ForEach(name => model.Enum.Add(new OpenApiString($"{Convert.ToInt64(Enum.Parse(context.Type, name))} = {name}")));
}
}
}
配置SwaggerGen使用上面的SchemaFilter:
services.AddSwaggerGen(c =>
{
c.SwaggerDoc("v1", new OpenApiInfo
{
Version = "v1",
Title = "ToDo API",
Description = "A simple example ASP.NET Core Web API",
TermsOfService = new Uri("https://example.com/terms"),
Contact = new OpenApiContact
{
Name = "Shayne Boyer",
Email = string.Empty,
Url = new Uri("https://twitter.com/spboyer"),
},
License = new OpenApiLicense
{
Name = "Use under LICX",
Url = new Uri("https://example.com/license"),
}
});
c.SchemaFilter<EnumSchemaFilter>();
});
为了得到Swagger和. net版本6的答案，我遇到了一些问题。x工作。我想继续为枚举使用整数值，但也要显示可能值的列表(以可读的格式)。所以这是我的修改版本(包括一些答案的部分)，也许它可以为你们节省一些时间;)
附注:还有一些改进的空间，你也应该检查方法“GetEnumTypeByName”的逻辑是否适合你。在我的情况下，我想主要更新描述仅为项目内部和唯一的枚举。
using System;
using System.Collections.Generic;
using System.Linq;
using Microsoft.OpenApi.Any;
using Microsoft.OpenApi.Models;
using Models.Api;
using Swashbuckle.AspNetCore.SwaggerGen;
/// <summary>
/// Add enum value descriptions to Swagger
/// </summary>
public class SwaggerEnumDocumentFilter : IDocumentFilter
{
public void Apply(OpenApiDocument swaggerDoc, DocumentFilterContext context)
{
// add enum descriptions to result models
foreach (var property in swaggerDoc.Components.Schemas)
{
var propertyEnums = property.Value.Enum;
if (propertyEnums is { Count: > 0 })
{
property.Value.Description += DescribeEnum(propertyEnums, property.Key);
}
}
if (swaggerDoc.Paths.Count <= 0)
{
return;
}
// add enum descriptions to input parameters
foreach (var pathItem in swaggerDoc.Paths.Values)
{
DescribeEnumParameters(pathItem.Parameters);
var affectedOperations = new List<OperationType> { OperationType.Get, OperationType.Post, OperationType.Put };
foreach (var operation in pathItem.Operations)
{
if (affectedOperations.Contains(operation.Key))
{
DescribeEnumParameters(operation.Value.Parameters);
}
}
}
}
private static void DescribeEnumParameters(IList<OpenApiParameter> parameters)
{
if (parameters == null) return;
foreach (var param in parameters)
{
if (param.Schema.Reference != null)
{
param.Description += DescribeEnum(param.Schema.Reference.Id);
}
}
}
private static Type GetEnumTypeByName(string enumTypeName)
{
if (string.IsNullOrEmpty(enumTypeName))
{
return null;
}
try
{
var projectNamespaceRoot = "MyProject.";
return AppDomain.CurrentDomain
.GetAssemblies()
.SelectMany(x => x.GetTypes())
.Single(x => x.FullName != null
&& x.FullName.StartsWith(projectNamespaceRoot)
&& x.Name == enumTypeName);
}
catch (InvalidOperationException _)
{
throw new ApiException($"SwaggerDoc: Can not find a unique Enum for specified typeName '{enumTypeName}'. Please provide a more unique enum name.");
}
}
private static string DescribeEnum(IEnumerable<IOpenApiAny> enums, string propertyTypeName)
{
var enumType = GetEnumTypeByName(propertyTypeName);
if (enumType == null)
{
return null;
}
var parsedEnums = new List<OpenApiInteger>();
foreach (var @enum in enums)
{
if (@enum is OpenApiInteger enumInt)
{
parsedEnums.Add(enumInt);
}
}
return string.Join(", ", parsedEnums.Select(x => $"{x} = {Enum.GetName(enumType, x.Value)}"));
}
}
正如其他人已经提到的，你必须在你的Swagger设置中注册这个过滤器:
services.AddSwaggerGen(options =>
{
options.SwaggerDoc("v1", new OpenApiInfo
{
// some configuration
});
options.DocumentFilter<SwaggerEnumDocumentFilter>();
});
